The top of a ladder slides down a vertical wall at a rate of 0.25 m/s. At the moment when the bottom of the ladder is 5 m from the wall, it slides away from the wall at a rate of 0.6 m/s. How long is the ladder?

Explanation:

L = length of ladder

Draw an xy axis. Focus on the upper right quadrant where x > 0 and y > 0

Let the origin (0,0) be the base of the wall.

Let y be the height of the top of the ladder and x be the horizontal distance the base of the ladder is from the base of the wall

Since the ladder is sliding down at a rate of 0.25 m/s, this means dy/dt = -0.25

At x = 5, we are given that dx/dt = 0.6

Use the pythagorean theorem to see that

a^2+b^2 = c^2

x^2 + y^2 = L^2

Now apply the derivative to both sides with respect to t. Don't forget to apply the chain rule while doing so.

d/dt[ x^2 + y^2 ] = d/dt[ L^2 ]

d/dt[ x^2 ] + d/dt[ y^2 ] =  d/dt[ L^2 ]

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

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

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

Note how the right hand side goes to 0 because the derivative of a constant is 0. The ladder length L doesn't change, so neither does L^2.

From here we plug in x = 5, dx/dt = 0.6, dy/dt = -0.25 and we solve for y

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

5*0.6 + y*(-0.25) = 0

0.3 - 0.25y = 0

3 = 0.25y

0.25y = 3

y = 3/0.25

y = 12

The top of the ladder is y = 12 meters off the ground and the base is x = 5 meters from the wall.

Therefore,

L^2 = x^2 + y^2

L = sqrt(x^2 + y^2)

L = sqrt(5^2 + 12^2)

L = 13

The ladder is 13 meters long.

answer: yes, since the line has a steeper rate of change than the exponential the y-values will be greater and they will only intersect one time.

answer is  

c.

36

step-by-step explanation: