# Thread: Relating rates - completely lost. HW due tomorrow.

1. ## Relating rates - completely lost. HW due tomorrow.

We have a ton of problems due tomorrow and I can't even fathom where to begin. Here are a few problems that I think if I understand how to do, I may be able to get the rest of the problems:

1) A spherical balloon is inflated wiht helium at the rate of 100pift^3/min. How fast is the baloon's radius increasing at the instant the radius is 5 ft? How fast is the surface area increasing at that instant?

2) A balloon is rising vertically above a level, straight road at a constant rate of 1 ft./sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance between the bicycle and balloon increasing 3 sec later?
(The book gives a triangle like diagram. The baloon has a line underneath it with an up arrow that says y(t) next to it, the bicycle going right with x (t) underneath it and then the hypotenuse is s(t).

2. Originally Posted by Medicatedsoap

1) If a and b are the lengths of two sides of a triangle, and θ the measure of the included angle, the area A of the triangle is A=(1/2)absinθ. How is dA/dt related to da/dt, db/dt, and dθ/dt?
I'm assuming they are all supposed to be variables. d/dx (ABC) = A'BC+AB'C+ABC' in case you've never seen the extended product rule.

$\frac{dA}{dt}=\frac{1}{2} \left( \frac{da}{dt}b\sin(\theta)+\frac{db}{dt}a\sin(\the ta)+\frac{d \theta}{dt}ab\cos( \theta) \right)$

You deleted this one! Did you solve it?

3. I did delete that one because I finally figured it out. XD But what you've done confirms my answer. Thank you!

Still working on the other two.

4. Originally Posted by Medicatedsoap

1) A spherical balloon is inflated wiht helium at the rate of 100pift^3/min. How fast is the baloon's radius increasing at the instant the radius is 5 ft? How fast is the surface area increasing at that instant?
$V=\frac{4}{3} \pi r^3$

$\frac{dV}{dt}=4 \pi r^2 \frac{dr}{dr}$

$100=4 \pi r^2 \frac{dr}{dt}$

$100=4 \pi (5^2) \frac{dr}{dt}$ Solve for dr/dt.
S.A. of a sphere is $4 \pi r^2$
$\frac{dSA}{dt}= \ 8 \pi r \frac{dr}{dt}$ r=5 and plug in dr/dt once you solve it above.