Question

A ladder 24ft long leans against a vertical wall the lower end is moving away at the rate of 3ft/sec ., find the rate at which the top of the ladder is moving downward ,if the foot is 8 ft from the wall

Answer 1

Answer:

Given:  $\frac{dx}{dt} = 3ft/sec$ and we have to calculate  $\frac{dy}{dt}$

From the figure 1;

x and y are related by the Pythagoras theorem.

Pythagoras theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

therefore,

$x^2+y^2=(24)^2$ or

$x^2+y^2=576$

Differentiate both sides of this equation with respect to t to get;

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

$\frac{dy}{dt}=-\frac{x}{y} \frac{dx}{dt}$

When x= 8ft from the wall, we have

$y=\sqrt{576-64}= \sqrt{512} =16\sqrt{2}$

Therefore,

$\frac{dy}{dt}=-\frac{8}{16\sqrt{2}} \cdot 3 =  \frac{-3}{2\sqrt{2}}=-\frac{3\sqrt{2}}{4}$

As, the top of the ladder is moving downward (because of the negative sign in the result) at a rate of $\frac{3\sqrt{2}}{4}$ ft/sec

Answer 2

Step-by-step explanation:

here's the answer :)......