Thread: Related Rates

I need help understanding the steps to solving these problems:

1. A spherical balloon is expanding. If the radius is increasing at the rate of 2 inches per minute, at what rate is the volume increasing when the radius is 5 inches?

I know to start off, I have to take the derivative of V, which is 4piR^2 * dr/dt, which equals dr/dt=dV/dt/(4piR^2). I'm stuck after that.

2. A particle moves in a circular orbit x^2 + y^2 = 16. As it passes through the point (2, 2sqrt(3)), its y-coordinate decreases at a rate of 3 units per second. At what rate is the x-coordinate changing?

3. A triangle, whose h = 2b, has an area A = 1/2bh=1/2b(2b)=b^2 is expanding with time. If dA/dt = 8 cm2/s, what is db/dt when b = 2?

4. A heap of garbage in the shape of a cube is being compacted. Given that the volume decreases at a rate of 2 cubic meters per minute, find the rate of change of an edge of the cube when the volume is exactly 27 cubic meters.

There are a lot more, but I think if I knew how to do these, I could figure out the others.

Thanks!

for #1

dr/dt=dV/dt/(4piR^2).

you want dV/dt = 4pi r^2 dr/dt

you know dr/dt and r don't you ?

For #2

x^2 + y^2 = 16.

differentiate 2xdx/dt + 2ydy/dt = 0

you know dy/dt, x , and y don't you?

once you have #1 and # 2 then the others should be no problem

Mmm...Thank you, but I'm still having a little trouble understanding :/

For number one, I have another similar problem including the solution, but I'm having trouble seeing the steps:
A sphere is expanding at the rate of 6 cm^3/sec, the formula for volume of a sphere is: V=4/3pir^3, what is dr/dt when r = 2?
Here is the solution:
V=4/3pir^3
dV/dt=4pir^2*dr/dt
dr/dt=dV/dt/(4piR^2)
=3/8cm/s

I don't understand how the answer became 3/8cm/s from the third step.


For number two, this is what I did but I'm not sure if it's correct.
2xdx/dt + 2ydy/dt = 0
2(2)*dx/dt+2(2sqrt(3))(-3)=0 [-3 since it's decreasing?]
4*dx/dt-20.78=0
dx/dt=5.2

Did I solve it correctly?

I tried to solve number one and this is what I got:

dV/dt = 4pi r^2 dr/dt
4pi(5^2)(2)=200pi

And I understand the problem and the solution on the previous post. Sorry, my posts are very confusing.

dx/dt=5.2

is correct for #1

dV/dt = 4pi r^2 dr/dt = 200pi is correct

dr/dt=dV/dt/(4piR^2)
=3/8cm/s

dV/dt = 6 r= 2

dr/dt=dV/dt/(4piR^2) = 6/(4pi *4) = 3/(8pi)

Alright, thanks so much. You've helped me immensely on this lesson.

I think you have the ability but I think you need to slow down and realize the problems are more straightforward than you think.

Calculus always takes more patience than intelligence.


If interested I have several related rates problems with animations and solutions on my web site at Optimization and Related Rates.

I think you're completely correct about the patients.
Thank you, and again for the link - I started Optimization today!