Question

A 15-ft ladder leans against a wall. The lower end of the ladder is being pulled away from the wall at the rate of 1.5 ft/sec. Let x be the distance from the bottom of the ladder to the wall, y be the distance from the top of the ladder to the ground and l be the length of the ladder. How fast is the top of the ladder moving along the wall at the instant it is 9 feet above the ground

Answer 1

Answer:

The top of the ladder is sliding down at a rate of 2 feet per second.

Step-by-step explanation:

Refer the image for the diagram. Consider $\Delta ABC$ as right angle triangle. Values of length of one side and hypotenuse is given. Value of another side is not known. So applying Pythagoras theorem,

$\left ( AB \right )^{2}+\left ( BC \right )^{2}=\left ( AC \right )^{2}$

From the given data, $L=15\:ft=AC$, $y=9\:ft=AB$ and $x=BC$

Substituting the values,  

$\therefore \left ( 9 \right )^{2}+\left ( x \right )^{2}=\left ( 15 \right )^{2}$

$\therefore 81+x^{2}=225$

$\therefore x^{2}=225-81$

$\therefore x^{2}=144$

$\therefore \sqrt{x^{2}}=\sqrt{144}$

$\therefore x=\pm 12$

Since length can never be negative, so $x= 12$.

Now to calculate $\dfrac{dy}{dt}$ again consider following equation,  

$\left ( y \right )^{2}+\left ( x \right )^{2}=\left ( l \right )^{2}$

Differentiate both sides of the equation with respect to t,  

$\dfrac{d}{dt}\left(y^2+x^2\right)=\dfrac{d}{dt}\left(l^2\right)$

Applying sum rule of derivative,

$\dfrac{d}{dt}\left(y^2\right)+\dfrac{d}{dt}\left(x^2\right)=\dfrac{d}{dt}\left(l^2\right)$

$\dfrac{d}{dt}\left(y^2\right)+\dfrac{d}{dt}\left(x^2\right)=\dfrac{d}{dt}\left(225\right)$

Applying power rule of derivative,  

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

Simplifying,  

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

Substituting the values,  

$9\dfrac{dy}{dt}+12\times1.5=0$

$9\dfrac{dy}{dt}+18=0$

Subtracting both sides by 18,

$9\dfrac{dy}{dt}=-18$

Dividing both sides by 9,

$\dfrac{dy}{dt}= - 2$

Here, negative indicates that the ladder is sliding in downward direction.  

$\therefore \dfrac{dy}{dt}= 2\:\dfrac{ft}{sec}$