Expert: Alon Mandes - 11/9/2008

Question
I have these practive problems, and I seriously don't understand them at all. Please help!  1. Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 6 mi/hr. how fast is the radius of the spill icnreasing when the area is 9 mi?  2. An aircraft is climbing at a 30* angle to the horizontal. How fast is the aircraft gaining altitude is the speed is 500 mi/hr?  3. A spherical balloon is to be deflated so the the radius decreases at a constant rate of 15 cm/min. At what rate must air be removed when the redius is 9 cm?  4. Sand pouring from a chute forms a conical pile whoseheight is always equal to its diameter. If the height increases at a constatn rate of 5 ft/min, at what rate is sand pouring from the chute when the pile is 10 feet high?  5. A particle is moving along the curve whose equation is (xy^2/(1+y^2))=8/5. Assume that the x-coordinate is increasing at a rate of 6 units/sec when the particle is at the point (1,2). a.) At what rate is the -coordinate of the point changing at the isntant? b.) Is the particle risig or falling at that instant? Explain.  6. a 13-foot ladder is leaning agasint a wal.. If the top of the ladder slips down the wall at a rate of 2 ft/sec, how fast will the foot be moving away from the wall when the top is 5 feet above the ground?  7. A point P is moving along the line whose equation is y=2x. How fast is the distance between P and the point (3,0) changing at the instant when P is at (3,6) if x is desreasing at the rate of 2 units/sec at that instant?  8. Two baots leave the same port at the same time with one boat traveling north at 15 knots per housr and the other boat traveling west at 12 knots per hour. how fast is the distance between the two boats changing after 2 hours?  9. A metal cube contracts when it is cooled. If the edge of the cube is descreasing at a rate of 0.2 cm/hr, how fast is the volume changing when the edge is 60 cm.?  10. A balloon rises at the rate of 8 feet per second from a point on the ground 60 feet from an observer. Find the rate of the angle of elevation when the balloon is 25 feet above the ground.

Answer
Hello Arnold, This a very long exercises & too much. But, I am willing to help you.
I want 1st to explain to you the method where such kind of exercises
are solved, then I'm going to solve 1 or two problems from the above
list. I also want you to try to follow my leads & solve the rest.
You are also invited to propose 1 or 2 difficult chosen exercise from the list above to help you with.

In problems concerning calculation of " rate of change " or
" speed of change " The method is simple :
1st, you must build a function with 1 variable, usually this
variable is the amount of each that affects the change of the
desired parameter in the exercise. This function will connect all
the parameters of the problem in 1 equation.
2nd, you have to derive this function in order to find the ration.
3rd, you will calculate the derivative at the point given to you.

Let's solve problem no 1 :
We have been given the rate of change of the area of the split,
& this is the derivative of the area as function of time. We know
that A'(t)=6, so A(t)=6t.
Now we must build our function : we are asked about the radius..
We already know that A=πR^2 That means R=√(A/π) & as function of
time : R(t)=√(6t/π). Ok, we have built the function. Now we want
to calculate the " rate of change " which is the derivative. So,
let's derive : R'(t)=(3/π)/√(6t/π). & we want to calculate R'(t)
at time to. Time to is the time when the area A=9. So, we claim
6to=9 -> to=3/2. & we will plug this value into our derivative :
R'(3/2)=(3/π)/√(6*(3/2)/π)=1/√π. & that is the answer !!

Let's solve problem no 2 :
The volume of a spherical balloon is (4/3)πR^3. It's given that
the rate of change of the radius is R'(t)=-15, so that means
that R(t)=C-15t. Where C is the initial radius before deflating,
& it's constant ("We don't have to know him, it will be dropped
out during the calculations"). Note that also - sign because it
is decreasing. So, we have R as function of time, what else ?
We want to know what the rate of change in the volume will be
when the radius is 9. Let's build the equation of the volume :
V(t)=(4/3)πR^3=(4/3)π(C-15t)^3. Now let's derive :
V'(t)=-60π(C-15t)^2. & we are interested in calculating this
derivative at to, which resemble the time the radius is 9:
So, we claim : 9=C-15to -> C-9=15to -> to=(C-9)/15. Let's plug
this value into V'(t) : V'(to)=-60π(C-15[C-9]/15)^2 ->
V'(to)=-60π(C-C+9)^2=-4860π. & this is the answer !

You are welcome to proceed solving the rest. I'm here if you need
me.

Alon.