Expert: Paul Klarreich - 11/23/2009

Question
Have you ever solved related rates problems?

Answer
Yes.  Here is a collection of them:

Questioner:   Sarah
Category:  Calculus
Private:  No
 
Subject:  related rates!
Question:  Hi, I'm having a hard time with a related rates problem and i can't seem to find out how to do it anywhere. It's...

A railroad bridge is 20 feet above, and at right angles to, a river. A man in a train traveling 60 mi/hr passes over the center of the bridge at the same instant that a man in a boat traveling 20 mi/hr passes under the center of the bridge. How fast are the two men moving away from each other 10 sec later?

So I converted 60 mi/hr to 88 ft/sec and 20 mi/hr to 29.3 ft/sec, but after that I'm completely lost. Thank you so much!
....................................
Hi, Sarah,
I'll try to get you started and you can finish up.


    train
+------------------> 88 ft/sec
|       x
|y
|
V
29.3 f/sec

Let  x = distance traveled by train
    y = """""""""""""""""''' boat
    r = distance between two men

dx/dt = speed of train, GIVEN as 88
dy/dt = speed of boat, GIVEN AS 29.3
dr/dt = rate at which they separate, TO BE FOUND.

Relation:  (Since one is higher by 20 ft.)

x^2 + y^2 + 20^2 = r^2  [Pythag Thm in 3 dimensions.]

x dx/dt + y dy/dt = r dr/dt

Values at t = 10

x = 880
y = 293

We need r, so back to the P.T.:

(880)^2 + (293)^2 + 20^2 = r^2

and we start our calculator:

r^2 = 860649

r = 927

Ready to go:
x dx/dt + y dy/dt = r dr/dt

(880) (88) + (293)(29.3) = (927) dr/dt

Ok, you can do the rest, I think.

===============================
Questioner:   Ruedi
 
Subject:  Simple Related Rates problem
Question:  Hi. the I got stuck with the following question:

A ship K, is sailing due north at 16 km/h, and a second ship R, which is 44 km north of K, is sailing due east at 10 km/h. At what rate is the distance between K and R changing 90 minutes later? Are they approaching each other or separating at this time? Explain.
----------------------------------------
Hi, Ruedi,

Is this your first attempt at Related-Rate problems?  If so, the scheme is something like this:

1. Identify the variables in the problem -- the things that change.  Give them names.
2. Write their rates of change as derivatives WITH RESPECT TO time.  Note which are known and which is to be found.
3. Determine a relationship (yes, it is called 'related rates' for a reason) between the variables.  Use a diagram, use your life experience, your general knowledge and brilliance, whatever you have to.  This may be the hard part.
4. Differentiate implicitly, THEN substitute the known quantities and rates, and solve for the unknown rate.


Your problem looks like this:
      x  
A+-------R-->
 |
 |
 |
 |
44|44 - y
 |
 |
 |
 |
 |
 ^K, going up
 |
 |
y|
 |
 +



In this case, your variables:

x = horizontal position of R.= AR
y = vertical position of K. = 44 - y
r = distance between R,K.

dx/dt = speed of R, given as 10
dy/dt = speed of K, given as 16
dr/dt = rate of change of distance between them, TO FIND.

At t = 0,  R is at (0,44), and  K is at (0,0).

As they move,  (44 - y)^2 + x^2 = r^2

Now diff:

2(44 - y)(-dy/dt) + 2x dx/dt = 2r dr/dt.
(44 - y)(-dy/dt) + x dx/dt = r dr/dt.

Now where is everyone after 90 min? or 1.5 hrs?

K has moved up 24 miles, so y = 24
R .........right 15 miles, so x = 15

We need an r, so:

(44 - y)^2 + x^2 = r^2, and:
(44 - 24)^2 + (15)^2 = r^2

(20)^2 + (15)^2 = r^2

-->  r = 25

Now subst:
(44 - 24)(-16) + 15(10) = (25) dr/dt.

(20)(-16) + 15(10) = (25) dr/dt.

(4)(-16) + 3(10) = (5) dr/dt.

-64 + 30  = (5) dr/dt.

-34  = (5) dr/dt.

dr/dt = -34/5;  decreasing.
=======================================
Questioner:   Korey
Category:  Calculus
Private:  No
 
Subject:  Calculus
Question:  The wind is carrying a balloon west at a constant speed of 100 ft/min. Meanwhile the balloon is losing altitude at a constant rate
of 50 ft/min.

If the balloon started 300 ft straight above your head, how fast is
the angle of elevation of the balloon changing, one minute later?

Is there a generalized method that will help with problems of this type?
.................................
Hi, Korey,

Glad you asked me that.  [You may soon be sorry you asked me that.]
..........................
In your example, the variables are:

x = the horizontal distance -- distance East. [I don't like West.]
y = the vertical distance -- the altitude.
theta [write @] = the angle of elevation.

The rates are:

dx/dt = rate of movement East, GIVEN as 100
dy/dt = rate of loss of altitude, GIVEN as -50.
d@/dt = rate of change of angle, TO BE FOUND.

Relations:

tan @ = y/x  or   

@ = arctan(y/x)

Differentiate with respect to t:  (t = time)

          1           (x)(dy/dt) - (dx/dt)(y)
d@/dt = ------------   -----------------------
       1 + (y/x)^2              (x)^2

        (x)(dy/dt) - (dx/dt)(y)
d@/dt = -------------------------
            x^2 + y^2     

ow we need some values to substitute.

At t = 0, we HAD x = 0 and y = 300
One minute later, at  t = 1, we HAVE  x = 100  and  y = 250.
Also, we can find that:

cos @ = x/r =

We have  dx/dt = 100,  dy/dt = -50
 
        (100)(-50) - (100)(250)
d@/dt = -------------------------
            (100)^2 + (250)^2

Basically, that's your answer, before a little simplification, like taking out and cancelling two 50's from every term:

        (2)(-1) - (2)(5)
d@/dt = -------------------
         (2)^2 + (5)^2

         -2 - 10
d@/dt = -----------
         4 + 25

        -12
d@/dt = -----
         29
========================================
Questioner:   Bryce
Category:  Calculus
Private:  No
 
Subject:  Graphs
Question:  Please help.  A particle is moving along the graph of y=^3square root of x.  

>>I think you mean  y = x^1/3, or cube root of x.

When x=8, the y component of its position is increasing at the rate of 1 centimeter per second.  
a.) How fast is the x component changing at this moment?
b.) How fast is the distance from the origin changing at this moment?
c.) How fast is the angle of inclination theta changing at this moment?
..................................................

Hi, Bryce,

the variables are:

x = the x-coordinate of the particle at point P.
y = the y-coordinate of the particle at point P.
D = the distance from the origin to P.
@ = the angle of inclination of the line  OP

The rates of change are:

dx/dt = rate of change of x, TO BE FOUND.
dy/dt = rate of change of y, given as  1 cm/sec.
dD/dt = rate of change of D, TO BE FOUND.
d@/dt = rate of change of @, TO BE FOUND.

So you need relations.  Start with:

a) y = x^1/3

dy/dt =  1/3 x^-2/3  dx/dt

*********************
Substitute known values:
dy/dt = 1
x = 8
********************

 1   =  1/3 (8)^-2/3  dx/dt

 1   =  1/3 (1/4)  dx/dt

dx/dt = 12 cm/sec

b)  x^2 + y^2 = D^2

2x dx/dt + 2y dy/dt = 2D dD/dt

x dx/dt +  y dy/dt = D dD/dt

We have  x = 8, dx/dt = 12,  dy/dt = 1.  We need  y, D.

But y = x^1/3 = (8)^1/3 = 2  and

x^2 + y^2 = D^2

8^2 + 2^2 = D^2

64 + 4 = D^2

D = sqrt(68) = 2 sqrt(17)

Ready to go.  Use that relation:  

x dx/dt +  y dy/dt = D dD/dt

*********************
Substitute known values:
dy/dt = 1
x = 8
dx/dt = 12
y = 2
********************

(8) (12) +  2 (1)  = 2 sqrt(17) dD/dt

   96 + 2  = 2 sqrt(17) dD/dt

   98  = 2 sqrt(17) dD/dt
           98
dD/dt = ------------
       2 sqrt(17)

           49
dD/dt = ----------
        sqrt(17)


c) Now you need a relation for theta.  How about:

tan @ = y/x
                (x)(dy/dt) - (dx/dt)(y)
sec^2 @  d@/dt = -----------------------
                    (x)^2

Now  cos @ = x/D = 8/2 sqrt(17) = 4/ sqrt(17),
so  sec @ = sqrt(17)/4

*********************
Substitute known values:
dy/dt = 1
x = 8
dx/dt = 12
y = 2
D = 2 sqrt(17)
sec @ = sqrt(17)/4
********************

                     (8)(1) - (12)(2)
(sqrt(17)/4)^2  d@/dt = -----------------
                           (8)^2


             8 - 24
17/16 d@/dt = -------
               64

             - 16
17/16 d@/dt = -----
               64

             - 1
17/16 d@/dt = -----
               4

d@/dt = -16/68 = -4/17

=====================================
Questioner:   Bryce
Category:  Calculus
Private:  No
 
Subject:  Decreasing rate of a cube
Question:  Could you help me with this question. If the edge of a metal cube is decreasing at a rate of 0.2cm/hr, how fast is the volume changing when the edge is 60 cm? thanks
..............................................
Hi, Bryce,

variables:

e = the length of an edge of the cube.
V = volume of the cube.

Rates:

de/dt = the rate at which the edge is decreasing, GIVEN as  -0.2.
dV/dt = the rate of change of the volume, TO BE FOUND.

Relationship:

V = e^3

Differentiate:

dV/dt = 3e^2 de/dt

Substitute:

e = 60,  de/dt = -0.2

dV/dt = 3(60)^2 (-0.2)

dV/dt = 3(3600)(-0.2)

dV/dt = -0.6(3600)

dV/dt = - 2160 cm^3/hr
===============================
Questioner:   Alex
Category:  Calculus
Private:  No
 
Subject:  Calculus-derivatives
Question:  I have two questions:
1) The rate of change of electric charge with respect to time is called current. Suppose that 1/3t^3 + t coulombs of charge flow through a wire in t seconds. Find the current in amperes after 3 seconds. When will a 20-ampere fuse in the line blow?

2)The radius of a circular oil spill is growing at a constant rate of 2 km/day. At what rate is the area of the spill growing 3 days after it began?

Could you please tell me how to figure these out? Thanks!
.............................................................

Hi, Alex,

These are all about representing rates of change as derivatives.  Once you figure out the variables and their derivatives WITH RESPECT TO TIME, the rest flows.

1) The rate of change of electric charge with respect to time is called current. Suppose that 1/3t^3 + t coulombs of charge flow through a wire in t seconds. Find the current in amperes after 3 seconds. When will a 20-ampere fuse in the line blow?

The sentence "The rate of change of electric charge with respect to time is called current" says:

If  E = electric charge, and  c = current, then  dE/dt = c.

The sentence: "1/3t^3 + t coulombs of charge flow through a wire in t seconds" says that:

E = 1/3t^3 + t

Now put those together.  If E = 1/3t^3 + t  and  dE/dt = c,  then  c = t^2 + 1, and c(3) = 10.

Finally, the question: "When will a 20-ampere fuse in the line blow?" says "at what t is c >= 20?"

So solve  c = 20, or  t^2 + 1 = 20.  Looks like  t = sqrt(19), or about 4.4 seconds.

...................................
2)The radius of a circular oil spill is growing at a constant rate of 2 km/day. At what rate is the area of the spill growing 3 days after it began?

Ah -- your classic Related Rates problem.  I suggest you look over previous questions in the AllExperts database for subject lines that say Related Rates.  There probably aren't more than a couple of hundred.

A. Identify the variables.  Give them names.

r = radius of the spill.
A = area of the spill.

B. Identify their rates of change.  Say which are known and which is to be found.

dr/dt = rate of growth of the radius.  Given as  2 km/day.
dA/dt = rate of growth of the area.  TO BE FOUND.

C. Find a relation between the variables.

A = pi r^2

D. Differentiate implicitly.

dA/dt = 2 pi r dr/dt

E. Substitute known values.

dr/dt = 2,  t = 3, so  r = 6.

dA/dt = 2 pi (6) (2) = 24 pi  km^2/day
============================================
Todd Glickman Asks in Category Calculus:
Subject:  Related Rates Revenue
Question:  Recall that revenue is defined R=pq. Suppose you work for a
company that has revenue this month of $27,000 with a sales rate
of  q= 9000 units of their main product. Suppose that in addition
to an ongoing price increase of $0.12 a month, your company’s
sales are increasing at a rate of 30 units a month. Your boss asks
you to write a report describing how fast is the company’s revenue
is increasing. Set this up as a related rates problem (i.e., no credit
for just winging the solution), expressing all the relevant quantities
in terms of the given variables and their derivatives (with respect to
time   in months).  

1. The variables are:
p = price per unit?
q = monthly sales.
R = monthly revenue

2. The rates of change are:

dp/dt = price change per month = 0.12
dq/dt = increase of sales per month = 30
dR/dt = increase in revenue, TO BE FOUND.

3. The relation is

R = pq.

4. Differentiate: (product rule, of course)

dR/dt = p dq/dt + q dp/dt

Now put in the appropriate numbers (your text doesn't make this totally clear.) which could be, that in month zero,  q = 9000, p = 3.
======================================
Hi, John,

Subject:  Related Rates
Question:  Suppose that water is being poured into a cubical tank at the rate of 100 cubic feet per minute. If the side of the cube measures 20 feet, Find the rate at which the depth of the water in the tank is changing with respect to time.
----------------

V = volume of water in the tank
h = height or depth ....................

dV/dt = rate water is poured in, which is 100 ft^3/min
dh/dt = rate of change of depth, TO BE FOUND.

Relation:

The tank may be a cube, but the water in the tank at any time will be a rectangular parallelepiped (I'll need a new ink cartridge after that.) whose volume is given by  V = L W H.  Now the L and W are each 20 feet and the H is the depth of the water.  So we have:

V = 20*20 h = 400 h

Differentiate:

dV/dt = 400 dh/dt

Specific values:  none -- in this case the relation between the rates does not actually depend on the depth of the water as it frequently does in these problems.

Substitute:

100 = 400 dh/dt

dh/dt = 1/4 foot/minute
==============================
Hi, John,

Subject:  Related Rates
Question:  A plane flying north at 640 miles/hour passes over a certain town at noon, and a second plane going east at 600 miles/hour is directly over the same town 15 minutes later. If the planes are flying at the same altitude, how fast will they be separating at 1:15 p.m. ?  

In this case, you have:

Variables:

y = north distance to first plane.
x = east distance to second plane.
z = distance between planes.

Rates of change:
dy/dt = first plane speed = 640 mph
dx/dt = second............= 600 mph
dz/dt = rate of increase of distance between planes, TO BE FOUND.

Relation:

Right triangle:  x^2 + y^2 = z^2

Differentiate:

2x dx/dt + 2y dy/dt = 2z dz/dt

x dx/dt + y dy/dt = z dz/dt

Determine instantaneous values.  At 1:15 hours, we have:

First plane traveled  1.25 hours at 640 = 800 miles = y.
Second............... 1    hour at  600 = 600 miles = x.
Distance is now  1000 miles, by the P.T. again = z.

Substitute:

600(600) + 800(640) = 1000 dz/dt

360000 + 512000 = 1000 dz/dt

360 + 512 = dz/dt
dz/dt = 872 mph.
==================================================
Keegan Asks in Category Calculus:
Subject:  Calculus
Question:  My name is Keegan and I am studying Calculus in the field of Mechanical Enginneering

The Problem is A metal Cube dissolves in Acid such than an edge of the cube decreases by .5 mm/min. How fast is the volume of the cube changing when the edge is 9.2 min.

I have tried deriving from the formula x^3 -.5 but I am sure this is not it
--------------------------------------
Your instincts are correct; that is not it.

In this example:
Part A:
s = length of a side (or edge) of the cube
V = volume of the cube.

Part B:
ds/dt = rate at which the edge is decreasing (given as -0.5 mm/min)
dV/dt = rate at which volume is decreasing (to be found)

Part C:

V = s^3

Part D:

dV/dt = 3s^2 ds/dt

Part E:

Use ds/dt = -0.5  and  s = 9.2

dV/dt = 3(9.2)^2 (-0.5)

You can take it from there.
====================================================
Mjay08 Asks in Category Calculus:
Subject:  Related Rates
Question:  The ends of a water trouph are equilateral triangles whose sides are 2m long. The length of the trouph is 15m. Water is being poured into the trouph at the rate of 3m(cubed)/hour. Find the rate at which the water level is rising when the depth of the water is 1 m.
-------------------------------------
Hi, 08,

I think it is called a 'trough'.  

In this case, the variables are:

h = the height of the water in the trough.
dh/dt is the rate at which the water level is rising, and IS TO BE FOUND.

V = the volume of the water in the trough.
dV/dt = the rate at which it is being filled, and is 3 cubic meters per hour.

Now you want a relation between h and V.  In other words, find V in terms of h.

For the volume. we have something called a prism, which has a base and height.  The base is the area of the sides.  (Remember from your geometry -- base has nothing to do with bottom.)  The height (of the prism) is the length of the trough, 15 m.

For an equilateral triangle, you get the area from some high school geometry.  The height is h, which is half the hypotenuse over sqrt(3).  So the hypotenuse, which is also the base (of the triangle) is  2h/sqrt(3).  Then the area of the triangle is  (1/2) h 2h/sqrt(3) = h^2/sqrt(3)

So your volume is given by :

V = 15 h^2/sqrt(3)

Now you are ready.  Differentiate implicitly with respect to t, and you have:

dV/dt = 30 h/sqrt(3) dh/dt

To find dh/dt, put in dV/dt = 3, and also  h = 1.  (You thought we forgot about the value of h, perhaps.  No, we didn't; it comes in at the end.)

15 = 30(1)/sqrt(3)  dh/dt

dh/dt = sqrt(3)/2

In the future, please:
1. Give me a real name to respond to, not a 'handle'.  Anonymity is for chat rooms; this isn't one of them.
2. Let me know what you already tried to do; maybe I can find where you made a mistake.

===================================================
Nicole Asks in Category Calculus:
Subject:  related rates
Question:  Ok, here's another one.

A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 8 m from the dock?
--------------------------
I searched my database of previous questions and to my amazement I could not find the solution to this problem.  I cannot believe you are the first to send it; this problem is SO old.

In this case your diagram looks like this:

A horizontal line from the bow (B) of the boat to the base (D) of the dock.
A vertical line from D to the pulley (P).
A 'hypotenuse' line from P to B.

Oops-- that word 'hypotenuse' gives it away, doesn't it?  So what do we have?

DP = 1 meter, and never changes.
PB is the length of the rope out, which we will call r, and this changes.
BD is the distance from the boat to the dock, which we will call x, and this also changes.  So the variables, etc, are:

x = BD
dx/dt is to be found
r = PB
dr/dt = -1 m/sec.

(And later we insert the value  x = 8)

The relation between the variables is:

x^2 + 1^2 = r^2

Differentiate:

2x dx/dt = 2r dr/dt

x dx/dt = r dr/dt

Now we should be ready to substitute  dr/dt = -1, x = 8.
Oops -- we don't have a value of r to put in.
Not to worry.  The right triangle gives it to us.

8^2 + 1^2 = r^2
65 = r^2
r = sqrt(65)

Now we are in business:

8 dx/dt = sqrt(65)(-1)

dx/dt = - sqrt(65)/8
================================
Hi, Nicole,

You wrote (sort of)
Subject:  related rates
Question:  If a snowball melts so that its surface area decreases at a rate of 1 cm^2/min, find the rate at which the radius decreases when the radius is 5 cm.

[Yes, I know - you said diameter.  But the usual formulas involve the radius; we can always change back.]

In this problem:

The variables are:

A = surface area of the sphere.  (Did you grow up in a warm climate and never saw a snowball?  Well, it's a sphere.)

r = radius of the sphere. [DON'T WORRY; WE'LL TAKE CARE OF THE DIAMETER LATER.]

The rates are:

dA/dt is the rate of decrease of area, and is 1 cm^2/sec.
dr/dt is the rate of decrease of radius and is to be found.

The relation is:

A = 4 pi r^2    [Surface area of a sphere.  Look it up.]

Differentiate implicitly w.r.t. time:

dA/dt = 8 pi r  dr/dt

Substitute, using the fact that  r = 5:

- 1 = 8 pi (5) dr/dt

Solve:

dr/dt = -1/40pi


-----------------------------
Now what about diameter?  You can do either of two things:

I. Since  d = 2r,  dd/dt = 2 dr/dt = 2(-1/40pi) = -1/20pi

II. Modify the problem this way:

The variables are:

A = surface area of the sphere.  (Did you grow up in a warm climate and never saw a snowball?  Well, it's a sphere.)

r = radius of the sphere.

d = diamater of the sphere.

The rates are:

dA/dt is the rate of decrease of area, and is 1 cm^2/sec.
dr/dt is the rate of decrease of radius.
dd/dt is the rate of decrease of radius and is to be found.

The relations ARE:

A = 4 pi r^2,  and  d = 2r,  so r = d/2

Then  A = 4 pi(d/2)^2 = pi d^2

Differentiate implicitly w.r.t. time:

dA/dt = 2 pi d  dd/dt

Substitute, using the fact that  d = 10:

- 1 = 2 pi (10) dd/dt

Solve:

dr/dt = -1/20pi
================================================
Daniel Asks in Category Calculus:

The base of a pyramid-shaped tank is a square with sides 9 feet in length, and the vertex of the pyramid is 12 feet above the base. The tank is filled to a depth of 4 feet, and water is flowing into the tank at the rate of 3 cubic feet per second. Find the rate of change of the depth of water in the tank. (Hint: the volume of a pyramid is given by V = Bh, where B is the base area and h is the height of the pyramid.)

For your pyramid-shaped tank, you need the formula for the volume of a pyramid, which is:

V = 1/3 base * height.

In this case, the entire pyramid has a base that is a square 12 X 12, and height 10.

Assuming the base is at the bottom (not necessarily, you understand) then the volume of the water is a TRUNCATED pyramid -- a pyramid with its top cut off.  Its volume is just the entire pyramid minus the empty part at the top.  


Let V = the volume of the water in the tank.
Then dV/dt is the rate water is flowing in, and this is many cubic feet per minute.  

Let h = the height (depth) of the empty part of the tank.
Then dh/dt is its rate of change and to be found.

Let s = the width of the base of the empty part.

How is V related to h?

Volume of the tank = 1/3 12^2 (10) = 1/3(144)(10) = 480
Volume of the empty pyramid = 1/3 s^2 h

Uh, oh, we have to do something -- express s in terms of h.  However, similar triangles tell us that

h/10 = s/12,  so s = 6h/5

                         36 h^2 h   12 h^3
Now  V = 1/3(6h/5)^2 h =  -------- = ------
                          3(5^2)     25

OK, now.  Using

    12 h^3
V = ------
      25
we differentiate with respect to t, implicitly:

dV   36 h^2 dh
-- = ------ --
dt     25   dt

Now, finally, we can use the remaining facts in the problem:

h = 6 feet.  (10 - 4, that is)

dV/dt = many cubic feet per second.  Oh, well:
      36 (6)^2 dh
some = -------- --
        25     dt

And now you solve for dh/dt
=====================================
Questioner:  ridralev2
Category:  Calculus
 
Subject:  Related Rates
Question:  A conical tank (with vertex down) is 10 feet across the top and 12 feet deep.  If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.
................................................
Hi, Ridralev2, [Is that what your mother calls you?]

In this case,

Variables:
h = height (depth) of water in the tank
V = volume of water in the tank.

Rates:
dh/dt = rate of change of depth, TO BE FOUND
dV/dt = rate of increase of volume, given as 10 cu.ft/min

Relation:

The water in the tank is in the form of a cone, vertex down, and its volume is given by

V = 1/3 pi r^2 h

We have to get rid of the 'r' variable, and for that, we use similar triangles.  The water and tank, in cross section, look like this:

10 feet across
---------------
\     |     /  ^
 \    | r  /   |
  \---|---/   12 feet deep.
    \ |h /     |
     \|/       V

Proportions:

r   10
- = --
h   12

r = 5h/6

So V = 1/3 pi (5h/6)^2 h
   25 pi h^3
V = ---------
     108

Now we can go ahead:

dV   75 pi h^2 dh
-- = --------- --
dt     108     dt

dV   25 pi h^2 dh
-- = --------- --
dt     36      dt

Now we can put our values:

dV/dt = 10,  and  h = 8

    25 pi (8)^2 dh
10 = ----------- --
      36        dt

     5 pi(64)   dh
2 = ----------- --
      36        dt

     5 pi(64)   dh
2 = ----------- --
      36        dt

dh    72
-- = -------
dt   320 pi


dh    9
-- = -----
dt   40 pi
=============================
Questioner:  Stephanie
Category:  Calculus
Private:  no
 
Subject:  Related rates
Question:  A square is inscribed in a circle.  As the square expands, the circle expands to maintain the four parts of intersection.  The perimeter of the square is increasing at the rate of 8 inches per second.

a) Find the rate at which the circumference of the circle is increasing.

b) At the instant when the area of the square is 16 square inches, find the rate at which the area enclosed between the square and the circle is increasing.
................................
Hi, Stephanie,

I have reworded your Question:  Four points are marked on a circle to form a square.  As the circle expands, one side of that square increases at the rate of 2 inches per second.

a) Find the rate at which the circumference of the circle is increasing.

b) At the instant when one side of the square is 4 square inches, find the rate at which the area enclosed between the square and the circle is increasing.

...........................

In this case, (part a, that is) the variables are:

s = length of one side of the square.
r = radius of the circle.
C = circumference of the circle.

The rates are:

ds/dt = the rate of increase of a side, GIVEN as  2 in/sec.
dr/dt = the rate of increase of the radius.
dC/dt = the rate of increase of the circumference, TO BE FOUND.

Now, how about a relationship:

C = 2 pi r.

But we want that in terms of s, don't we?  Alas, I can't draw diagrams for you, so you will have to go with my description.  

Draw the square and mark one side as s.
Mark the diagonal as  2r.  Yes, 2r, because the diagonal of the square is the same as the diameter of the circle, or 2r.

Now s^2 + s^2 = (2r)^2
   2 s^2 = 4 r^2
     s^2 = 2 r^2
       r = s/sqrt(2)

OK, now:  C = 2 pi (s/sqrt(2))

C = sqrt(2) pi s

Ready to go.  Differentiate:

dC/dt = sqrt(2) pi ds/dt

Substitute numbers:

dC/dt = sqrt(2) pi (2) = 2 sqrt(2) pi.
............... DONE WITH PART a ..............

For part b, we have these variables:

s = length of one side of the square.
r = radius of the circle.
A = area between the square and circle.

The rates are:

ds/dt = the rate of increase of a side, GIVEN as  2 in/sec.
dr/dt = the rate of increase of the radius.
dA/dt = the rate of increase of that area, TO BE FOUND.

Relationships:

A = circle area - square area
A = pi r^2 - s^2

Get rid of one of those:

s^2 = 2 r^2, remember?

A = pi r^2 - 2r^2 = (pi - 2)r^2

Differentiate:

dA/dt = (pi - 2)(2r)

dA/dt = 2(pi - 2)r

Substitute values.  We have  s = 4.

Small difficulty.  We don't have a value of r.  
Resolution:  Use one of the relations to find it.

r = s/sqrt(2) = 4/sqrt(2) = 2 sqrt(2)

Ready now:

dA/dt = 2(pi - 2)(2 sqrt(2))
dA/dt = 4 sqrt(2)(pi - 2)

I think that does it.
============================================================
Questioner:  jino
Category:  Calculus
 
Subject:  Related Rates
Question:  can u help me out with this question

The volume of the cylindrical trunk of a tree is proportional to the cube of its diameter and the diameter increases uniformly from year to year, prove that the rate of increase in the volume  when the diameter is equal to 90 cm is 25 times the rate when it is 18 cm
................................
Hi, Jino,

In this case,
Variables:
D = the diameter of the trunk.
V = the volume of the trunk.

Rates:
dD/dt = rate of increase of the diameter.  STATED AS BEING CONSTANT.
dV/dt = rate of increase of the volume.  TO BE FOUND.

Releationship: You stated that  V is proportional to  D^3.  That means:

V = kD^3, where k is some constant, depending on climate, soil conditions, whatever.

dV         dD
-- = 3kD^2 --
dt         dt

Now you said that the diameter increases uniformly, so it is a constant, which we will call G, the growth rate.  That is to say:

dD
-- = G.
dt

So

dV         
-- = 3kD^2 G =  3kG D^2
dt         

Now put in data:

When the diameter is  18 cm:

dV(18)/dt = 3kG (18)^2 = (whatever)

When the diameter is  90 cm:

dV(90)/dt = 3kG (90)^2

dV(90)/dt = 3kG (5*18)^2

dV(90)/dt = 3kG (5)^2(18)^2

dV(90)/dt = 3kG (25)(18)^2

dV(90)/dt = 25 [3kG (18)^2]

dV(90)/dt = 25 dV(18)/dt

That's what you were supposed to prove.
====================================
Questioner:  Joyce
Category:  Calculus
Private:  no
 
Subject:  Calculus: Related Rates

Let @ (in radians) be an acute angle in a right triangle, and let x and y, respectively, be the lengths of the sides adjacent to and opposite @.  At a certain instant, x= 2 units and is increasing at 1 unit/sec, while y = 2 units and is decreasing at 1/4 unit/sec.  How fast is @ decreasing or increasing at that instant?
...............................
Hi, Joyce,

In this example, your variables are:

@ is the angle in the triangle.
x is one leg.
y is the other leg.

The rates are:

d@/dt is the rate of increase of the angle, TO BE FOUND.
dx/dt is the rate of increase of the first leg, given as  1.
dy/dt is the rate of increase of the second leg, given as -1/4

The relation is (as you noted)

tan @ = y/x

Differentiate with respect to time:(yes, you will use the quotient rule)
               x dy/dt - y dx/dt
sec^2 @ d@/dt = -----------------
                      x^2

Now substitute known values:

x = 2,  y = 2.
dx/dt = 1,  dy/dt = -1/4

What about @?  

sec^2 @ = 1 + tan^2 @ = 1 + (y/x)^2 = 1 + 2^2 = 5

We're in business:

               x dy/dt - y dx/dt
sec^2 @ d@/dt = -----------------  (repeated)
                      x^2


           2 (-1/4) - 2(1)
(5) d@/dt = -----------------
                 2^2

         -1/2 - 2
5 d@/dt = --------
             4

         -1 - 4
5 d@/dt = --------
            8

         -5
5 d@/dt = ---
          8


       -1
d@/dt = --- radians/sec
        8
==============================
Questioner:  Jarred
Category:  Calculus
Private:  no
 
Subject:  Related Rates

The demand for a shirt is 2px-65p-4950=0, where x hundreds of shirts are demanded per week when p dollars is the price of a shirt.  If the shirt is selling this week at $30 and the price is increasing at the rate of $0.20 per week, find the rate of change in the demand.

I have tried to do this problem like a rate of expansion problem but i have never done a problem like this before.
...........................
Hi, Jarred,

Variables:
x = the demand, in hundreds of shirts.
p = the price.

Rates:
dx/dt = rate of change of demand, TO BE FOUND.
dp/dt = rate of change of price, given as  0.20

Relation. (Normally you have to do some work to find it, but here it is given.)

2px - 65p - 4950 = 0

Differentiate implicitly with respect to t:

2p dx/dt + 2x dp/dt - 65 dp/dt = 0

Substitute known values:
dp/dt = 0.2
p = 30

Difficulty:  We don't have a current value of x.
Resolution:  Use the relation above to get it:

2(30)x - 65(30) - 4950 = 0
60x - 1950 - 4950 = 0
60 x = 6900
x = 6900/60 = 115

Ready to go:

2(30) dx/dt + 2(115)(0.2) - 65 (0.2) = 0

60 dx/dt + 46 -  13 = 0
60 dx/dt + 33 = 0

dx/dt = 33/60 = 11/20
=================================
Questioner:  Gary
Category:  Calculus
Private:  no
 
The hypotenuse of a right triangle is growing at the rate of "a" cm/sec and one leg is decreasing at the rate of "b" cm/sec.  How fast is the acute angle between the hypotenuse and the other leg changing at the instant when both legs are 1 cm?

>> changed to say "How fast is the angle opposite that leg changing..."
 
------------------------------

In this case, you will have a diagram of triangle XYZ, with the right angle at Z.  We will also have:

Variables:
z = side XY, the hypotenuse.
x = side YZ, which is opposite angle X, and is the one that is decreasing.
y = side XZ, which is opposite angle Y.

Rates: Now you didn't say this, but I will have to assume that the triangle remains a right triangle.  In that case, if one leg decreases, the other will have to increase.

Rates: (I will write  'h' for 'theta', the angle at X.)

dz/dt = rate of increase of the hypotenuse, given as a.
dx/dt = rate of decrease of side YZ, given as b.
dh/dt = rate of decrease of angle X, TO BE FOUND.

Relation:  In a right triangle,

sin h = x/z

Differentiate w.r.t. t:
             (z)(dx/dt) - (x)(dz/dt)
cos h dh/dt = -----------------------
                    z^2

Now substitute known values:

x = 1 and y = 1 are given.
dx/dt = b,  dz/dt = a.

Difficulty: We don't have current values of  z and h.
Resolution: Compute them from the diagram.

x^2 + y^2 = z^2
1 + 1 = z^2,  so  z = sqrt(2)

cos h = y/z = 1/sqrt(2)

Now we should be in business:

  1            sqrt(2)(b) - (1)(a)
------- dh/dt = -----------------------
sqrt(2)                 2

dh   2b - a sqrt(2)
-- = --------------
dt         2


============================
Questioner:  Gallic
Category:  Calculus
Private:  no
 
Subject:  Related Rates.
Question:  I have attempted this question, but I got stuck because I'm not sure what the formula for the Volume of a Right triangular prism is.

Bob's pool is in the shape of a right triangular prism with the triangle in the horizontal plane. The triangle sides are a = 6.0m, b = 8.0m, c = 10.0m. Bob rented a fire hose to fill the pool at 10.0 m^3 per minute. Find the rate at which the height of the water is rising. Include a sketch.
..................................
Hi, Gallic,

Sorry I can't send a sketch through the site interface -- you'll have to do that yourself.  HOWEVER,

Any prism has the volume given by:

V = bh, where

b = the area of the base (in this case the triangle)
h = the height (in this case it will be the depth of the water in meters.)

Since the base is your right triangle, its area is just 1/2(6 * 8) = 24 meters^2.

After that, you shouldn't have any difficulty.  If you do, let me know and I'll send you the standard stuff on R-R problems.  

OK-- here it goes:

Variables are:
V = volume of water in the pool.
h = height of water in the pool.
Rates are:
dV/dt = rate at which water is filling the pool.  Given as 10 m^3/min
dh/dt = rate at which the water is rising.  TO BE FOUND.

Relation:

V = 24h    (as we said before.)

Differentiate:

dV/dt = 24 dh/dt

Substitute:

10 = 24 dh/dt
dh/dt = 10/24 = 5/12 m/sec
============================================================
Questioner:  Leah Parker
 
Two sides of triangle have lengths 12 m and 15 m.  The angle between them is increasing at a rate of 2 degrees/min.  How fast is the length of the third side increasing when the angle between the sides of fixed length is 60 degrees?

..................................
Hi, Leah,

In this problem:

1. The variables are:

c = the length of the third side.
h = the angle between the fixed sides.  (theta, but I will use 'h')

2. The rates of change are:

dc/dt = the rate of increase of the third side.  TO BE FOUND.
dh/dt = the rate of increase of the angle. GIVEN as 2 degrees/min, WHICH WE WILL CONVERT TO RADIAN MEASURE.

3. The relation between the objects is given by the law of cosines:

c^2 = a^2 + b^2 - 2ab cos C   <-- which is cos h in our case.

c^2 = 12^2 + 15^2 - 2(12)(15) cos h

c^2 = 144 + 225 - 360 cos h

c^2 = 369 - 360 cos h

4. Differentiate implicitly:

2c dc/dt = 360 sin h  dh/dt

Now we have to put in some values:

A. The value of dh/dt, given as  pi/90 radians/min
B. The value of h, given as 60 degrees. [We won't have to change this.]
C. The current value of c.

Difficulty:  We don't have a current value of c.
Resolution:  Use the L.of C. to find it when  h = 60 degrees.

c^2 = 144 + 225 - 360 cos 60

c^2 = 144 + 225 - 360(1/2)

c^2 = 144 + 225 - 180

c^2 = 369 - 180 = 189

c = sqrt(189)   << it isn't a whole number, but we can handle it.

Now we are ready.  Back to:

2c dc/dt = 360 sin h  dh/dt

2(sqrt(189)) dc/dt = 360 sin(60) pi/90

2 sqrt(189) dc/dt = 4 (sqrt(3)/2) pi

2 sqrt(189) dc/dt = 2 pi sqrt(3)

sqrt(189) dc/dt = pi sqrt(3)
       pi sqrt(3)
dc/dt = ----------
       sqrt(189)

You can use your calculator on that.
=========================================
Questioner:  Rachel
Category:  Calculus
 
Subject:  Related Rates

1)As a circular steel griddle is being heated, it's diameter changes at a rate of .01 cm/min.  Find the rate at which the area of one side is changing when the diameter is 30 cm.
2)A girl starts at a point A and runs east at a rate of 10 ft/s.  One minutes later, another girl starts at a point A and runs north at a rate of 8 ft/s.  At what rate is the distance between them changing one minute after the second girl starts?
3)As sand leaks out of a hole in a container, it forms a conical pile whose height is always equal to the radius of the pile.  If the height of the pile is increasing at a rate of 6 inches per minute, find the rate at which the sand is leaking out of the container when the height of the pile is 10 inches.
4)A model rocket enthusiast launches a rocket vertically from a point that is 5 miles away from his tracking device.  Assume the launch site and the tracking device are at the same elevation.  For the first 20 seconds of flight, the angle of elevation of theta changes at a constant rate of 2 degrees/s.  Find the velocity of the rocket in miles per hour when the angle of elevation is 30 degrees.

..........................................
Hi, Rachel,

Your examples:

1)As a circular steel griddle is being heated, it's diameter changes at a rate of .01 cm/min.  Find the rate at which the area of one side is changing when the diameter is 30 cm.

Variables:
D = diameter of the griddle.
a = area of one side.

Rates:
dD/dt = rate of increase of diameter, GIVEN as 0.01 cm/min
da/dt = rate of increase of area, TO BE FOUND.

Relationship:

We know the formula for area of a circle:

a = pi r^2,  and  r = D/2,  so this would be  a = pi D^2/4

Differentiate:

da/dt = 2pi D/4  dD/dt

da/dt = pi D/2  dD/dt

Substitute:  D = 30,  dD/dt = 0.01

da/dt = pi(30)/2 (0.01) = pi 15(0.01) = 0.15 pi

..............................
2)A girl starts at a point A and runs east at a rate of 10 ft/s.  One minutes later, another girl starts at a point A and runs north at a rate of 8 ft/s.  At what rate is the distance between them changing one minute after the second girl starts?

Variables:

x = distance from A for first girl.
y = distance from A for second girl.
z = distance between the two.

Rates:

dx/dt = speed of first girl,  GIVEN as  10 ft/sec
dy/dt = speed of second girl,  GIVEN as  8 ft/sec
dz/dt = rate of increase of distance,  TO BE FOUND.

Relation:  There is a right triangle here, and  x^2 + y^2 = z^2.

Differentiate:

2x dx/dt + 2y dy/dt = 2z dz/dt

x dx/dt + y dy/dt = z dz/dt

Substitute.  We have rates but:

Difficulty encountered:  We don't have current values for  x,y,z.
Resolution of this issue:  Use  t = 60 and the relation to find them.

x(60) = 60*10 = 600 feet.
y(60) = 60(8) = 480 feet.

x^2 + y^2 = z^2

600^2 + 480^2 = z^2

z = sqrt(600^2 + 480^2) = sqrt(120^2(5^2 + 4^2)) = 120 sqrt(41)

Ready to go.

x dx/dt + y dy/dt = z dz/dt

600(10) + 480(8) = 120 sqrt(41) dz/dt

5(10) + 4(8) = sqrt(41) dz/dt

50 + 32 = sqrt(41) dz/dt
         82
dz/dt = --------, which you can simplify a little.
       sqrt(41)

......................................
I think you now know enough to do 3,4 yourself.  If you still have trouble, send me another query, but INCLUDE what you already managed to do on them -- don't just repeat the problems.

=============================
Questioner:  Rachel
Category:  Calculus
Private:  no
 
Subject:  Related Rates
Question:  Thank you so much for your help!!! I feel like I'm beginning to really grasp the concept.  I do have a couple questions about number 3:
3)As sand leaks out of a hole in a container, it forms a conical pile whose height is always equal to the radius of the pile.  If the height of the pile is increasing at a rate of 6 inches per minute, find the rate at which the sand is leaking out of the container when the height of the pile is 10 inches.

I think the formala that will relate all the variable together is V = pi/3 * r^2 * h, but i'm not positive if the rate of change of the volume of the pile is equal to the rate at which the sand is leaking out of the container.

.......................
That would seem logical.  If the volume of the pile is increasing, that means it's getting more sand.  Where do you suppose that sand is coming from?

Nevertheless, the conditions of this problem are a bit strange, because normally the sand leaks out at a constant rate and the height of the pile increases at a varying rate -- here it seems the other way around.  No matter, though; let's do our standard stuff.

Variables:

h = the height of the conical pile of sand.
r = the radius of the conical pile, WHICH IS ALWAYS EQUAL TO h, says the problem.
V = the volume of the conical pile.

Rates:

dh/dt = the rate of increase of the height, GIVEN as  6 in/min.
dr/dt, which is irrelevant, since we will eliminate r.
dV/dt = the rate of increase of volume, TO BE FOUND.

Relation:

V = 1/3 pi r^2 h = 1/3 pi r^3,  since  r = h.

Differentiate:

dV/dt = pi r^2 dr/dt

Substitute:

dr/dt = 6,  h = 10

dV/dt = pi (10)^2 (6) = 600 pi  cubic inches/min
............................................................

4)A model rocket enthusiast launches a rocket vertically from a point that is 5 miles away from his tracking device.  Assume the launch site and the tracking device are at the same elevation.  For the first 20 seconds of flight, theta, the angle of elevation, changes at a constant rate of 2 degrees/s.  Find the velocity of the rocket in miles per hour when the angle of elevation is 30 degrees.

For this, a diagram is a good idea.  Put the observer at O, the launch point at L, and the position of the rocket at R, with  OLR being a right triangle, and theta = angle ROL.

Variables:  (I will write T for theta in any expressions.)

T = angle ROL
y = the height LR of the rocket.

Rates:

dT/dt = the rate of increase of theta, GIVEN as 2 degrees/sec.
dy/dt = the velocity of the rocket, TO BE FOUND.

Now I see some obvious issues here.  

dT/dt is in degrees/second, but dy/dt will be in miles per hour.  So we have to use a common unit here.  I'll do everything in things/hour.  

T should never be in degrees -- it should be in radians.  2 degrees = pi/90 radians.

Putting those together, we have:

dT/dt = pi/90 * 3600 degrees/hour. = 40 pi deg/hour

Relation:

tan T = y/5,  from the right triangle.

Differentiate:

sec^2(T) dT/dt = 1/5 dy/dt

Substitute:  T = 30 degrees,  dT/dt = 40 pi degrees/hour.

sec(30) = 1/cos 30 = 1/(sqrt(3)/2)) = 2/sqrt(3),
sec^2(30) = 4/3

Finally:

4/3 (40 pi) = 1/5 dy/dt
160pi/3 = 1/5 dy/dt

dy/dt = 800 pi/3 miles per hour, which is about 850 or so, I guess.
============================================================
Questioner:  Tommy
 

Question:  Hi paul my question is,
A 4-foot-long manger has a cross section which is an equilateral triangle with side length of 18 inches.  If rain is accumulating in the manger at the rate of 250 cubic inches/hour, how fast is the water level rising in the manger when it is 10 inches deep?

..................................
Hi, Tommy,

About your example:  I have never seen a manger, so I will have to refer to it as a 'rough'.

1. Let

V = the volume of the water in the trough.
h = the height of the water in the trough.

2. Let

dV/dt = the rate at which the water is accumulating.  Given as  250 cu.in./hr
dh/dt = the rate at which the water level is rising.  TO BE FOUND.

3. The water in the trough is in the shape of a prism laid on its side, with a cross-section that is an equilateral triangle, with  height = h.


    / \
   / | \
s /  |  \ s
 /  h|   \
/    |    \
/     |     \
--------------
     s

Since  h = s/2 sqrt(3),  s = 2h/sqrt(3)

Area = 1/2 base * height = 1/2 (2h/sqrt(3))(h) = h^2/sqrt(3)

The volume of water = triangle * length =  h^2/sqrt(3) * 48 inches.
   48 h^2
V = -------
   sqrt(3)

OK, we have our relation between h and V.

Part 4:  Diff:

dV     96 h  dh
-- = ------- --
dt   sqrt(3) dt

Now we can put in:  dV/dt = 250,  h = 10,  and find dh/dt:
     96(10)  dh
250 = ------- --
     sqrt(3) dt

       96    dh
25  = ------- --
     sqrt(3) dt

dh   25 sqrt(3)
-- = ----------
dt      96

=====================================
mike Asks in Category Calculus ...
 

"A kite 100 feet above the ground moves horizontally at a constant speed of 8
ft/sec. At what rate is the angle between the string and the horizontal
decreasing at the moment when 200 feet of string is let out? write to the
nearest 0.01 rad/sec."

So far I've figured out that y=200  h=100  x=sqrt(200^2 - 100^2)    and
dy/dx=8ft/sec. I'm trying to figure out what the angle/dt is. any suggestions
would be appreciated.
......................................
Hi, Mike,

Your variables are:

(y is not a variable; it's always 100)
x = The kite's horizontal distance from you.
z = the length of string let out.
H (theta) = the angle between the string and the horizontal.

The rates are:

dx/dt = the rate the kite moves horizontally, GIVEN = 8 ft/sec
dz/dt = the rate the string is let out.
dH/dt = the rate of decrease of the angle.  TO BE FOUND.

The relationships are:

x/100 = cot H

x = 100 cot H

Differentiate that:

dx/dt = 100 (- csc H cot H) dH/dt

dx/dt = -100 csc H cot H dH/dt

Now you need an instantaneous value of H.  Actually, values of the trigonometric functions will do nicely.  At this point you have:

y = 100,  z = 200.

x = 100 sqrt(3), by your usual geometry.

Now in the triangle, we have:

sin H = y/z = 1/2
cos H = x/z = sqrt(3)/2

csc H = 1/sin H = 2
cot H = cos H/sin H = sqrt(3)

dx/dt = 8  (given)

8 = -100 (2)(sqrt(3)) dH/dt
         - 8
dH/dt = -----------
       200 sqrt(3)

The rest I leave to you and your calculator.  The answer WILL come out in radians.  Note that it is negative, as we suspected.
=======================================
kelsey Brannan Asks in Category Calculus ...
 
 
Question:  I have been working on this related rate problem forever,and I don't seem to be getting anywhere. This is how it goes.

Two trucks, one traveling west and the other traveling south, are approaching an intersecton,

-- BETTER GET OUT OF THERE!

If both trucks are traveling at the rate of k km/hr, show that they are approaching each other at
the rate of k(2)^(1/2) ( k* the square root of 2) km/hr when they are each m kilometers from the intersection.

I need help setting up a question, I don't know if I am supposed to come up with easy numbers to plug into the equation or if I can prove it using variables. Thanks, and hope it all works!

-kels

-------------------------------------
OK, Kels,

Variables:  Let:

x = the distance from the first truck (going W)  to the crash point, er..., sorry, intersection.
y = the distance from the second truck (going S) to the intersection.
z = the distance between the trucks.

Rates:

dx/dt = the speed of the first truck, GIVEN as  -k  km/hr  (minus, because the distance to the crash point is decreasing.

dy/dt = the speed of the second truck, also given as  -k  km/hr.
dz/dt = the rate at which the distance between them is decreasing.  TO BE FOUND.

Relationship:

Your diagram will show a right triangle with  x,y as the legs and z as the hypotenuse.

x^2 + y^2 = z^2

Differentiate implicitly:

2x dx/dt + 2y dy/dt = 2z dz/dt  or

x dx/dt + y dy/dt = z dz/dt

Put in known values.  We have  x = m and y = m, and can compute  z = m sqrt(2)

m (k) + m (k) = m sqrt(2) dz/dt

The m divides out.

2k = sqrt(2) dz/dt

Divide by  sqrt(2):

dz/dt = k sqrt(2)

So there is no problem using the symbolic values  k, m in the problem.  You don't have to pick numbers or anything, just deal with those values.  Yes, you have to be a little more careful doing that, but it works.
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