Answer to Question #175429 in Calculus for kienth bryan doroin

Question

The top of a 23 feet ladder, leaning against a vertical wall is slipping down the wall at a rate of 3 feet per second. How fast is the bottom of the ladder slipping along the ground when the bottom of the ladder is 10 feet away from the base of the wall?

Answer 1

Let for any time t the horizontal distance from the base of the wall to the bottom of the ladder be x and the vertical distance from the base of the wall to the top of the ladder be y.

For all time t, the hypotenuse i.e. length of ladder remains the same i.e. 23 foot.

Thus:

$$x^2+y^2=23^2=529$$

Differentiation with respect to time t,

$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$

Thus we need to find $\frac{dx}{dt}$

$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 \implies \frac{dx}{dt}=-\frac{2y}{2x}\cdot\frac{dy}{dt}=-\frac{y}{x}\cdot\frac{dy}{dt}$$ $$x^2+y^2=23^2=529 \implies y=\sqrt{529-x^2}$$

where $\frac{dy}{dt}=−3$ feet/second when $x = 10$ feet.

$y=\sqrt{529-x^2}=\sqrt{529-100}=\sqrt{429}$

$\frac{dx}{dt}=-\frac{y}{x}\cdot\frac{dy}{dt}=\frac{\sqrt{429}}{10}\cdot3=6.21$

Answer: 6.21 feet per second