1. ## [SOLVED] Related rate

Hi guys !

There's a related rate problem I really can't figure out. I think I have an idea, but I'm not too sure about it. Heres the situation :
A ball is dropped at t=0 at 15 m high, 9 m from a lamp. This lamp has a height of 15 m too. Knowing that the height of the ball can be known with the formula y(t)=15-4.9t², at what speed goes the shadow of the ball at t=0.5 (ie. 0.5s after being dropped)

Here is my drawing of the situation.

I called x the distance between under the ball and the shadow of it. I first calculated what height the ball would be after 0.5s, which gave me :
y(0.5) = 15-4.9(0.5)² = 13.775 m

Then, I decided to calculate its speed after at this moment. This would logically be the rate dy/dt, I thought. It gave me :
dy/dt = (13.775-15)/(0.5) = -2.45 m/s

I tried to figure out how I could relate y (height of the ball) and x, and the only thing I thought of was : Let's use similar triangle rules ! The biggest triangle was 15 m*(9+x) m and the smallest, y*x. I used the following formula :
x/(x+9) = y/15

I isolated x :
x = (9y)/(15-y)

And then, I derivated it in function of the time.
dx/dt = [135(dy/dt)]/[(15-y)²]

I knew that dy/dt = -2.45 m/s and that y = 13.775. So the final answer was :
dx/dt = -220.4 m/s

So, at t=0.5, the shadow of the ball moves toward the lamp at the speed of 220.4 m/s.

Did I do anything wrong ? Is the answer correct or logical/plausible ?

Thanks a lot !

(Sorry, I don't know anything about LaTEX)

2. ## Re: Related rate

All good except for

Originally Posted by Zaide
Then, I decided to calculate its speed after at this moment. This would logically be the rate dy/dt, I thought. It gave me :
dy/dt = (13.775-15)/(0.5) = -2.45 m/s
I guess you're reasoning geometrically on the assumption of constant speed - your straight line sketch? ... but that's against x not t!

Why not differentiate y(t), or use v = u + at.

3. ## Re: Related rate

Originally Posted by tom@ballooncalculus
All good except for

I guess you're reasoning geometrically on the assumption of constant speed - your straight line sketch? ... but that's against x not t!

Why not differentiate y(t), or use v = u + at.
Well, I was considering that at t=0.5, the speed of the ball was -2.45 m/s. I know it's not constant. If it would have been at t=1, i would've use -4.9 m/s.
Si you would just differentiate y(t) = -4.9t²+15 ? Don't really get it. How do I put the distance of the shadow (x) in this ?!

4. ## Re: Related rate

Originally Posted by Zaide
Well, I was considering that at t=0.5, the speed of the ball was -2.45 m/s.
Why?

I know it's not constant.
But it would have to have been, to justify using distance/time to calculate it.

So you would just differentiate y(t) = -4.9t²+15 ? Don't really get it.
I said do it! So do it, do it, do it! The Simpsons - Look at all those idiots - YouTube

Originally Posted by Zaide
How do I put the distance of the shadow (x) in this ?!
Like I said, you did everything right apart from the velocity at half a second.

5. ## Re: Related rate

I don't really get what you're trying to say. How would you do it ? I just started doing these related rates, and there are a few things I don't understand.
Do I derivate y to get dy/dt = -9.8t, and then replace the dy/dt in the equation I had ? Like that :
dx/dt = [135(dy/dt)]/[(15-y)²]
dx/dt = [135(-9.8t)]/[(15-y)²]
dx/dt = -440.8 m/s ?

6. ## Re: Related rate

Looks good to me.

7. ## Re: Related rate

Well, thanks a lot !