Question

A 393939-meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at 101010 meters per minute. At a certain instant, the bottom of the ladder is 363636 meters from the wall. What is the rate of change of the distance between the top of the ladder and the ground at that instant (in meters per minute)

Answer 1

The rate at which the distance between the top of the ladder and the ground at that instant is decreasing at 42 meters per minute.

Given length of ladder 39 m and the rate at which the wall is increasing is 101, the bottom of the ladder is 36 m from the wall.

Let bottom is x and the length of wall be y.

We have to find the rate at which the distance between the top of the ladder and the ground at that instant.

we have to find dy/dt.

We have to apply pythagoras theorem first.

$x^{2} +y^{2} =39^{2}$---------------1

$x^{2} +y^{2} =1521$

put the value of x=36 in the above equation

$36^{2} +y^{2} =1521$

$y^{2} =225$

y=15 m

We have to find the derivative of equation 1 with respect to t.

2x*dx/dt+2y*dy/dt=0

Because it is given that bottom is increasing at 101 meters per minute so dx/dt=101

2x*101+2y*dy/dt=0

put the value of y=15

2x*101+2*15*dy/dt=0

2x*101+2*15*dy/dt=0

dy/dt=-3030/2*36

dy/dt=-42.08

Hence the rate at which the rate at which the top of the ladder and the ground at that instant is decreasing at 42 meters per minute.

Learn more about pythagoras theorem at brainly.com/question/343682

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Question is wrong so it includes ladder length be 39 meters and the rate of increasing of bottom of ladder from bottom of wall is 101 meters per minute and the bottomis 36 m from the wall.