Question

A 10 foot ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 feet per second, how fast is the bottom of the ladder moving along the ground when the bottom of the ladder is 5 feet from the wall?

Answer 1

Answer:

  2√3 ft/s

Step-by-step explanation:

The relation between the height of the ladder on the wall and the distance along the ground is given by the Pythagorean theorem. If we let y represent the height and x represent the distance from the wall, we have ...

  x^2 + y^2 = 10^2

Differentiating with respect to t gives

  2x(dx/dt) +2y(dy/dt) = 0

Solving for dx/dt, we have ...

  dx/dt = (-y/x)(dy/dt)

When the bottom of the ladder is 5 ft from the wall, the top of the ladder is ...

  √(10² -5²) = 5√3 . . . . feet up the wall

Filling in the values for y, x, and dy/dt, we find ...

  dx/dt = (-5√3)/(5)·(-2 ft/s) = 2√3 ft/s

The bottom of the ladder is moving 2√3 ≈ 3.46 ft/s away from the wall.

Answer 2

The rate at which the bottom of the ladder is moving along the ground is 1.15ft/s

From the given equation, we can say that the relationship between the height of the ladder on the wall and the distance along the ground is given by the Pythagorean theorem shown below;

$c^2=a^2+b^2$

Given that;

c is the length of the ladder = 10feet

On substituting

$a^2+b^2=10^2\\a^2+b^2=100$

On differentiating the wall equation with respect to time "t"

$2a\frac{da}{dt}+2b\frac{db}{dt} = 0\\ a\frac{da}{dt} + b\frac{db}{dt}=0$

Given the following:

$\frac{db}{dt} = 2ft/s\\ a = 5$

Get the value of "b"

$5^2+b^2=100\\b^2=100-25\\b^2=75\\b=8.66ft$

Get how fast is the bottom of the ladder moving along the ground

$-5(2)+ 8.66\frac{db}{dt}=0\\-10 = -8.66\frac{db}{dt}\\\frac{db}{dt}=\frac{10}{8.66}\\\frac{db}{dt}= 1.15ft/s$

Hence the rate at which the bottom of the ladder is moving along the ground is 1.15ft/s

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