# Thread: Rates of change question

1. ## Rates of change question

Hello

I have a maths exam tomorrow, this is one question I cannot get

Any help is GREATLY appreciated!!!

A ladder 5.1m long is pushed on level ground against a vertical wall. The foot of the ladder is pushed towards the wall at 1.5m/s. At what rate is the top of the ladder rising when its bottom is 2.4m from the wall?

Thanks a lot!!

Lachlan

2. Originally Posted by auonline
Hello

I have a maths exam tomorrow, this is one question I cannot get

Any help is GREATLY appreciated!!!

A ladder 5.1m long is pushed on level ground against a vertical wall. The foot of the ladder is pushed towards the wall at 1.5m/s. At what rate is the top of the ladder rising when its bottom is 2.4m from the wall?

Thanks a lot!!

Lachlan
First draw a picture. You should see a right triangle with the ladder, of length 5.1 m, as hypotenuse. If you use x and y to me the "base" and "height" of that triangle, $\displaystyle x^2+ y^2= (5.1)^2$. Change that "static" equation into a "dynamic" equation by differentiating each side with respect to x: $\displaystyle 2x\frac{dx}{dt}+ 2y\frac{dy}{dt}= 0$. You are told that x= 2.4 and $\displaystyle \frac{dx}{dt}= -1.5$. You can solve $\displaystyle (2.4)^2+ y^2= (5.1)^2$ for y and then solve $\displaystyle 2(2.4)(-1.4)+ 2y\frac{dy}{dt}= 0$ for $\displaystyle \frac{dy}{dt}$.

3. I can't believe I didn't think of doing that!!

+ rep and thank'd.

Thanks

Lachlan =D