# Math Help - Related Rates (Story Problems)

1. ## Related Rates (Story Problems)

1) A ball is dropped from a height of 20 meters, 12 meters away from the top of a 20-meter lamppost. The ball's shadow, caused by the light at the top of the lamppost, is moving along the level ground. How fast is the shadow moving 1 second after the ball is released?

2) A swimming pool is 12 meters long, 6 meters wide, 1 meter deep at the shallow end, and 3 meters deep at the deep end. Water is being pumped into the pool at 1/4 cubic meter per minute, and there is 1 meter of water at the deep end.
a) What percent of the pool is filled?
b) At what rate is the water level rising?

Hopefully, you can understand what these questions are asking without the accompanying diagram. I found these two particularly hard and wasn't sure how to solve them, so help me out with the reasoning and try to let me figure out the answer myself.

Thanks!

2. Originally Posted by seuzy13
1) A ball is dropped from a height of 20 meters, 12 meters away from the top of a 20-meter lamppost. The ball's shadow, caused by the light at the top of the lamppost, is moving along the level ground. How fast is the shadow moving 1 second after the ball is released?
hopefully you have sketched a picture.

let h = height of the ball above the ground

x = horizontal distance from the ball's shadow to the point on the ground directly beneath it

using proportions formed by similar triangles ...

$\frac{20}{12+x} = \frac{h}{x}$

$20x = h(12+x)$

$20 \frac{dx}{dt} = h \frac{dx}{dt} + (12+x) \frac{dh}{dt}$

finally, note that $h = 20-5t^2$ (using a "nice" approximation $g = 10 \, m/s^2$)

or

$h = 20-4.9t^2$ (using the more accurate $g = 9.8 \, m/s^2$)

3. Thank you so much. I think I understand how to do the first problem now, but I'm still totally lost on the second. Could I please get some help with that from someone?

4. ## Thank you

Originally Posted by skeeter
hopefully you have sketched a picture.

let h = height of the ball above the ground

x = horizontal distance from the ball's shadow to the point on the ground directly beneath it

using proportions formed by similar triangles ...

$\frac{20}{12+x} = \frac{h}{x}$

$20x = h(12+x)$

$20 \frac{dx}{dt} = h \frac{dx}{dt} + (12+x) \frac{dh}{dt}$

finally, note that $h = 20-5t^2$ (using a "nice" approximation $g = 10 \, m/s^2$)

or

$h = 20-4.9t^2$ (using the more accurate $g = 9.8 \, m/s^2$)

thanks for opening my eyes! i'm reluctant sometimes because I think the solution isn't supposed to be hard to get or that all the information needed to solve the problem should be provided. (Horrible by product of high school)

thanks again!