Two Calculus problems I need help with!?

2013-02-28 8:45 am
Im trying to figure out how to solve these two problems, would anyone mind helping me out please?

1. A cube of ice is melting so that each edge is decreasing at the rate of 4 inches per hour. Find how fast the volume of the ice is decreasing at the moment when each edge is 5 inches long.

2. A rocket fired straight up in the air is being tracked by a radar station 3 miles from the launching pad. If the rocket is traveling at 2 miles per second, how fast is the distance between the rocket and the radar station changing at the moment when the rocket is 4 miles up?
(I got 5 miles?)

Thank you for your help!

回答 (3)

2013-02-28 9:51 am
✔ 最佳答案
1)
V = x^3
dV/dt = dV/dx * dx/dt (chain rule)
= 3x^2 * (-4)
= -12x^2 cubic inch(es)/hour

When each edge is 5 inches long,
-12(5)^2
= -300 cubic inches/hour, meaning that the volume is decreasing by 300 cubic inches per hour.

2)
Let x, y and z, respectively, be the distance between the pad and the station, the rocket and the pad, and the rocket and the station.

x^2 + y^2 = z^2 (Pythagorean theorem)
9 + y^2 = z^2
d(y^2)/dt = d(z^2)/dt
2y * dy/dt = 2z * dz/dt (chain rule)
dz/dt = 2y/z mile(s)/second

When the rocket is 4 miles up, and the distance between the rocket and the station is 5 miles,
The distance between the rocket and the station is increasing by
2 * 4/5 = 8/5 = 1.6 miles/second.
2016-10-24 2:48 pm
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2013-02-28 10:38 am
1. If the edges have length x inches and the volume is v inches^3 then
v=x^3 so dv/dt = (dv/dx)(dx/dt)=3x^2dx/dt. Given that dx/dt=-4 and x=5
you can now find dv/dt.

2. When the rocket R has moved y miles its distance from the
launching pad L is y. The distance from The radar station S to R is z
when angle SRL=Θ. dy/dt = 2 mps .
When y=4 from triangle RSL SR=5 and cosΘ=4/5.
dz/dt = dy/dt(cosΘ)=2(4/5)=1.6 miles per second.
Your 5 miles is the distance from S to R.
You can get dz/dt using z=√(y^2+9) and dz/dt=(dz/dy)(dy/dt)
=[y/√(y^2+9)]dy/dt and sub y=4, dy/dt=2


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