Related Rate Again..

• Mar 7th 2009, 09:20 PM
Kayla_N
Related Rate Again..
1. Let http://webwork1.math.utah.edu/webwor...0fcdd79cd1.png be the area of a circle with radius http://webwork1.math.utah.edu/webwor...79cb7891c1.png. If http://webwork1.math.utah.edu/webwor...985d565591.png, find http://webwork1.math.utah.edu/webwor...375c713261.png when http://webwork1.math.utah.edu/webwor...c95730d941.png.

2. The radius of a right circular cone is increasing at a rate of 4 inches per second and its height is decreasing at a rate of 3 inches per second. At what rate is the volume of the cone changing when the radius is 50 inches and the height is 20 inches?

3. At noon, ship A is 20 nautical miles due west of ship B. Ship A is sailing west at 23 knots and ship B is sailing north at 22 knots. How fast (in knots) is the distance between the ships changing at 6 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

4. The top of a 28 foot ladder, leaning against a vertical wall, is slipping down the wall at the rate of 4 feet per second. How fast is the bottom of the ladder sliding along the ground when the bottom of the ladder is 7 feet away from the base of the wall?
• Mar 7th 2009, 09:47 PM
Mentia
Lets take a look at #1 and #2

1)

Area of a circle:

$\displaystyle A = \pi r^{2}$

Then, by implicit differentiation,

$\displaystyle \frac{dA }{dt } = 2 \pi r \frac{dr }{dt }$

And we know $\displaystyle \frac{ dr}{dt } = 4$ and r = 1, so we can get $\displaystyle \frac{dA }{dt }$

2)

Volume of a right circular cone:

$\displaystyle V = \frac{ \pi }{3 }r^{2}h$

Then, again by implicit differentiation,

$\displaystyle \frac{dV }{dt } = \frac{2 \pi }{3 }rh\frac{dr }{dt } + \frac{ \pi }{3 }r^{2}\frac{dh }{dt }$

and this time we know $\displaystyle \frac{dr }{dt } = 4$, $\displaystyle \frac{dh }{dt } = -3$, r = 50, h = 20
• Mar 7th 2009, 09:48 PM
Kayla_N
Sorry but just dont do number 4..i got the answer already :)