A 13 foot ladder is resting against the wall. The bottom is initially 10 feet away from the wall and is being pushed towards the wall at a rate of 0.25 ft/sec. How fast is the top of the ladder moving up the wall 20 seconds after we start pushing? Hint: the hypotenuse (ladder) is a constant since the length of the ladder never changes.

This is what I think I start with:

x^2+y^2=13^2

Yes, you're on the right track! To solve this problem, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the ladder in this case) is equal to the sum of the squares of the lengths of the other two sides.

Let's assign variables to the different quantities involved:
- Let x be the distance between the bottom of the ladder and the wall.
- Let y be the distance between the top of the ladder and the ground.
- The length of the ladder is given as 13 feet.

So, the equation we start with is: x^2 + y^2 = 13^2

Now, let's differentiate both sides of the equation with respect to time (t) to find the rates at which x and y are changing:

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

Since we are given that dx/dt = 0.25 ft/sec, we can substitute this value into the equation:

2x(0.25) + 2y(dy/dt) = 0

Simplifying, we have:

0.5x + y(dy/dt) = 0

Now, we need to find the value of y when x = 10 (after 20 seconds), which will help us solve for dy/dt:

From the original equation, x^2 + y^2 = 13^2, substitute x = 10:

10^2 + y^2 = 13^2

Solving for y, we get:

y^2 = 13^2 - 10^2
= 169 - 100
= 69

y = √69 (taking the positive square root since y represents a distance)

Now, substitute this value of y into the equation 0.5x + y(dy/dt) = 0:

0.5(10) + √69(dy/dt) = 0

5 + √69(dy/dt) = 0

Solving for dy/dt, we have:

√69(dy/dt) = -5

dy/dt = -5/√69

Therefore, the top of the ladder is moving down the wall at a rate of -5/√69 ft/sec, 20 seconds after we start pushing.