When the base is 3 feet from the wall, how fast is the bottom of the ladder moving away from the wall?

Question

A 5 foot ladder is sliding down a wall with the top of the ladder moving down the wall at the rate of 1/2 foot per second. When the base is 3 feet from the wall, how fast is the bottom of the ladder moving away from the wall?

I got x²+y²=z² for my model equation and for my variables:

The base, x: 3 feet

The ladder, z: 5 feet

The wall, y: 4 feet (because of the 3:4:5 ratio)

The rate of the ladder, z': 1/2 feet

Since I'm looking for x', I got my equation to x'=[z(z')-y(y')]/x.

And then I began plugging in my numbers, [(5)(1/2)-4(y')]/3, to which I got x'=(5-8y')/6. I then got y'=5/8 and plugged that into the x' equation and got x'=35/42 feet per second.

My teacher told me that I put 'too much effort' and crossed out most of what I did but didn't explain why. Can someone help me explain what I did wrong/the right way to do it?

Answer 1

z is the length of the ladder, which is CONSTANT.

x² + y² = 25

dy/dt = -1/2 when x = 3, y = 4.

Find dx/dt:

2x(dx/dt) + 2y(dy/dt) = 0

6(dx/dt) - 4 = 0

dx/dt = 2/3 ft/sec