Expert: Paul Klarreich - 11/30/2007

Question
I'm having a really tough time understanding problems where the angle is changing. I thought I understood them, but then when my answer came out, it just did not seem correct.

Here is the problem verbatim: "A balloon is rising vertically above a level, straight road at a constant velocity of 1 foot / sec. Just when the balloon is 65 foot above the ground, a bicycle passes under it going 17 feet / sec. A) How fast is the distance between the bicycle and the balloon increasing in 3 seconds and B) how fast is the angle of elevation changing in 3 seconds?"

Anyways, here is what I have done so far:
1) Given information:
Original altitude = 65
Change in altitude per second = 1
speed of biker = 17

2) Define Variables:
dy/dt = 1
y = 65 + 3 * dy/dt = 68
dx/dt = 17
x = 3 * dx/dt = 51
z = sqrt(68^2+51^2) = 85

3) Part A
d/dt(x^2+y^2=z^2)
2x * dx/dt + 2y * dy/dt = 2z * dz/dt
51 * 17 + 68 * 1 = 85 * dz/dt
867 + 68 = 85 * dz/dt
dz/dt = 935/85
dz/dt = 11 ft. / sec.

4) Part B
sin(theta) = y/z
cos(d(theta)/dt) = (z * dy/dt - y * dz/dt) / z^2
cos(d(theta)/dt) = (85 * 1 - 68 * 11) / 8562
cos(d(theta)/dt) = (85 - 748) / 7225
cos(d(theta)/dt) = -663 / 7225
d(theta)/dt = arccos(-663 / 7225)
d(theta)/dt = 1.6627 Radians / Second = 95.265 Degrees / Second

The change in angle, as you can see, appears to be much, much too high. Have I done something wrong?

Thank you for your time.
-Mike

Answer
Questioner:   Mike
Category:  Calculus
Private:  No
 
Subject:  Related Rates: Angle Problems
Question:  I'm having a really tough time understanding problems where the angle is changing. I thought I understood them, but then when my answer came out, it just did not seem correct.

Here is the problem verbatim: "A balloon is rising vertically above a level, straight road at a constant velocity of 1 foot / sec. Just when the balloon is 65 foot above the ground, a bicycle passes under it going 17 feet / sec. A) How fast is the distance between the bicycle and the balloon increasing in 3 seconds and B) how fast is the angle of elevation changing in 3 seconds?"

Anyways, here is what I have done so far:
1) Given information:
Original altitude = 65
Change in altitude per second = 1
speed of biker = 17

2) Define Variables:
dy/dt = 1
y = 65 + 3 * dy/dt = 68
dx/dt = 17
x = 3 * dx/dt = 51
z = sqrt(68^2+51^2) = 85

3) Part A
d/dt(x^2+y^2=z^2)
2x * dx/dt + 2y * dy/dt = 2z * dz/dt
51 * 17 + 68 * 1 = 85 * dz/dt
867 + 68 = 85 * dz/dt
dz/dt = 935/85
dz/dt = 11 ft. / sec.

4) Part B
sin(theta) = y/z
cos(d(theta)/dt) = (z * dy/dt - y * dz/dt) / z^2<<<< NO, THAT'S WRONG!!!!!!!!!

>>>>>>>> Should be  cos theta  d(theta)/dt.

cos(d(theta)/dt) = (85 * 1 - 68 * 11) / 8562
cos(d(theta)/dt) = (85 - 748) / 7225
cos(d(theta)/dt) = -663 / 7225
d(theta)/dt = arccos(-663 / 7225)
d(theta)/dt = 1.6627 Radians / Second = 95.265 Degrees / Second

The change in angle, as you can see, appears to be much, much too high. Have I done something wrong?

Thank you for your time.
-Mike
 

...................................
Hi, Mike,

It is good to see what you did.  Obviously, you worked hard to organize things. And nearly all of it was right.  I'll make a couple of suggestions that might help -- maybe you won't have to work so hard.

Variables (first -- actual values don't come in until the end)

y = height of balloon.
x = distance from 'base' of balloon to biker.
r = distance from bike to balloon.
@ = angle of elevation.  [@ = theta]

Rates:
dy/dt = rate of rise of balloon, GIVEN AS 1 ft/sec
dx/dt = speed of biker, GIVEN as 17 ft/sec
dr/dt = rate of change of distance bike-balloon, TO BE FOUND
d@/dt = rate of change of angle, TO BE FOUND.

Relation between variables.

(One at a time.)

x^2 + y^2 = r^2

Diff. w.r.t. time:

2x dx/dt + 2y dy/dt = 2r dr/dt

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

NOW put in some values.  We want x,y,r, dx/dt, dy/dt, all at t = 3:

x = 17(3) = 51,  y = 65 + 3 = 68 = 17(4) by coincidence, so r = 17(5) = 85.

[So far, you are exactly right.]

Now we have:

x dx/dt + y dy/dt = r dr/dt  (again)

17(3)(17) + 17(4)(1) = 17(5) dr/dt

(3)(17) + (4)(1) = (5) dr/dt

51 + 4 = 5 dr/dt

dr/dt = 11  (RIGHT!!)

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

(Next one)

tan @ = y/x << OK, you used cosine. Let's see what happens.

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

(copied from above)
NOW put in some values.  We want x,y,r, dx/dt, dy/dt, all at t = 3:

x = 17(3) = 51,  y = 65 + 3 = 68 = 17(4) by coincidence, so r = 17(5) = 85.

sec @ = 1/cos @ = r/x = 85/51 = 5/3

25            17(3) (1) - 17(4)(17)
(--)  d@/dt = ---------------------
 9                 (17)(3)(17)(3)


25            (3) (1) - (4)(17)
(--)  d@/dt = ---------------------
 9                 (3)(17)(3)


25            3 - 68
(--)  d@/dt = ---------
 9            (9)(17)


25             - 65
(--)  d@/dt = ---------
 9            (9)(17)


5             - 13
(--)  d@/dt = ---------
1              (17)

        - 13
 d@/dt = --- = - 0.153 radians/sec
          85

Do you see where you went wrong?