Rates of change question

**auonline** · September 6th 2009, 02:30 AM

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

**HallsofIvy** · September 6th 2009, 03:34 AM

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, $x^2+ y^2= (5.1)^2$ . Change that "static" equation into a "dynamic" equation by differentiating each side with respect to x: $2x\frac{dx}{dt}+ 2y\frac{dy}{dt}= 0$ . You are told that x= 2.4 and $\frac{dx}{dt}= -1.5$ . You can solve $(2.4)^2+ y^2= (5.1)^2$ for y and then solve $2(2.4)(-1.4)+ 2y\frac{dy}{dt}= 0$ for $\frac{dy}{dt}$ .

**auonline** · September 6th 2009, 04:25 AM

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

+ rep and thank'd.

Thanks

Lachlan =D

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