A 15-ft ladder rests against a vertical wall. If the top of the ladder slides down the wall at a rate of 0.33 ft/sec, how fast, in ft/sec, is the bottom of the ladder sliding away from the wall, at the instant when the bottom of the ladder is 9 ft from the wall? Answer with 2 decimal places. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

Question

A 15-ft ladder rests against a vertical wall. If the top of the ladder slides down the wall at a rate of 0.33 ft/sec, how fast, in ft/sec, is the bottom of the ladder sliding away from the wall, at the instant when the bottom of the ladder is 9 ft from the wall? Answer with 2 decimal places. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

Answer 1

x^2+y^2 = 225

so,
x dx/dt + y dy/dt = 0

Now plug in your numbers and solve for dx/dt

Answer 2

To solve this problem, we can apply the concept of related rates. Let's denote the distance between the bottom of the ladder and the wall as x and the distance between the top of the ladder and the ground as y. We are given that dy/dt = -0.33 ft/sec, where the negative sign indicates that y is decreasing.

Using the Pythagorean theorem, we have:

x^2 + y^2 = 15^2

Differentiating both sides of the equation with respect to time (t):

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

We want to find dx/dt, the rate at which x is changing when x = 9 ft. We can rearrange the equation to solve for dx/dt:

2x(dx/dt) = -2y(dy/dt)
dx/dt = -y(dy/dt) / x

Now let's substitute the given values:

dx/dt = -9 * (-0.33) / 15

Simplifying the expression:

dx/dt = 0.198 ft/sec

Therefore, the bottom of the ladder is sliding away from the wall at a rate of 0.20 ft/sec (rounded to 2 decimal places).

Answer 3

To solve this problem, we can use related rates. Let's assign some variables:

Let x be the distance between the bottom of the ladder and the wall (in ft).
Let y be the distance between the top of the ladder and the ground (in ft).

We are given the following information:

dx/dt = -0.33 ft/sec (rate at which the top of the ladder is sliding down the wall)
x = 9 ft (distance between the bottom of the ladder and the wall)

We need to find dy/dt (rate at which the bottom of the ladder is sliding away from the wall) when x = 9 ft.

We can use the Pythagorean theorem to relate x and y:

x^2 + y^2 = 15^2

Differentiating both sides of this equation with respect to time (t), we get:

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

Plugging in the given values:

2(9)(-0.33) + 2y(dy/dt) = 0

-5.94 + 2y(dy/dt) = 0

2y(dy/dt) = 5.94

dy/dt = 5.94 / (2y)

Since x^2 + y^2 = 15^2, we can solve for y:

9^2 + y^2 = 15^2

81 + y^2 = 225

y^2 = 225 - 81

y^2 = 144

y = 12 ft

Now we can substitute this value of y back into the equation for dy/dt:

dy/dt = 5.94 / (2(12))

dy/dt = 5.94 / 24

dy/dt = 0.2475 ft/sec

Therefore, the bottom of the ladder is sliding away from the wall at a rate of 0.25 ft/sec (rounded to 2 decimal places) at the instant when the bottom of the ladder is 9 ft from the wall.