Question

. A 25-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially 20ft away from the wall, how fast does the ladder move up the wall 5sec after we start pushing?

Answer 1

Answer:

0.75 feet per second.

Step-by-step explanation:

Please find the attachment.

We have been given that a 25-ft ladder is leaning against a wall. We can see from the attachment that ladder forms a right triangle with respect to wall and ground.

So we can set a Pythagoras theorem as:

$x^2+y^2=25^2$

$x^2+y^2=625$

Now, we need to find the derivative of above equation with respect to time.

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

Since the adder is moving toward the wall at a rate of 1 ft/sec for 5 sec, so x after 5 seconds would be: $20-1(5)=20-5=15$

Let us solve for y using Pythagoras theorem.

$y^2+15^2=25^2$

$y^2+225=625$

$y^2=625-225$

$y^2=400$

Take positive square root:

$\sqrt{y^2}=\sqrt{400}$

$y=20$

Upon substituting our given values in derivative equation, we will get:

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

$2(15)\cdot(-1)+2(20)\cdot \frac{dy}{dt}=0$

$-30+40\cdot \frac{dy}{dt}=0$

$40\cdot \frac{dy}{dt}=30$

$\frac{dy}{dt}=\frac{30}{40}$

$\frac{dy}{dt}=0.75$

Therefore, the ladder is moving up at a rate of 0.75 feet per second.