a 20 ft ladder is leaning against a wall. The bottom is being pulled out at a constant rate of 2.5 ft/sec. Will the top of the ladder clide sown thr wall at a constant rate? What will the rate be? In relation to time?
It is not constant.
Use pythagorus: x^2 + y^2 = l^2, where l is the length of the ladder, and x is the x (horizontal) component, and y is vertical component. Then take the derivatives with respect to time, and solve for dy/dx.
I hope this is enough of a start for you...
Let x by horizontal, y be vertical.
20^2=x^2 + y^2
O=2x dx/dt + 2y dy/dy
You know dx/dt, which is a constant. Solve for dy/dt? is x, y constant (NO).
To find the rate at which the top of the ladder slides down the wall, you can use the derivative of the equation given by the Pythagorean theorem: x^2 + y^2 = l^2.
Differentiate both sides of the equation with respect to time (t) using the chain rule:
0 = 2x(dx/dt) + 2y(dy/dt).
Here, dx/dt represents the rate at which the bottom of the ladder is being pulled out, which is given as 2.5 ft/sec.
Now, you need to solve for dy/dt, which represents the rate at which the top of the ladder is sliding down the wall. However, you do not have enough information about how x and y change with respect to time to determine whether dy/dt is constant or not. More information, such as the specific relationship between x and y, or the rate at which they change, would be needed to find the exact value of dy/dt in relation to time.