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a 13-ft ladder is leaning against a house when its base starts to slide away. by the time the base is 12 ft from the house, the base is moving at the rate of 5 ft/sec. at what rate is the angle between the ladder and the ground changing then

User Olivier P
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Final answer:

To determine the rate at which the angle between the ladder and the ground is changing, we can use trigonometry to calculate the height of the ladder. By finding the derivative of the angle with respect to time, we can determine the rate of change of the angle, which is approximately 0.208 radians/sec.

Step-by-step explanation:

To find the rate at which the angle between the ladder and the ground is changing, we can use trigonometry. Let's consider the ladder as the hypotenuse of a right triangle, with the base of the ladder as the adjacent side and the height of the ladder against the house as the opposite side.

Given that the ladder is 13 ft long, the base is moving at a rate of 5 ft/sec, and the base is currently 12 ft from the house, we can find the height of the ladder using the Pythagorean theorem.

Using the equation a^2 + b^2 = c^2, where a is the height, b is the base, and c is the length of the ladder, we have:

12^2 + b^2 = 13^2

Simplifying the equation, we get: b^2 = 169 - 144 = 25

So, b = 5 ft.

Now, since tan(theta) = opposite/adjacent, we can find the angle theta using the equation tan(theta) = 5/12. Taking the inverse tangent of both sides, we get:

theta = tan^(-1)(5/12)

Now, to find the rate at which theta is changing, we can take the derivative with respect to time. Using the chain rule, we have:

d(theta)/dt = (1/((1+(5/12)^2)*12)) * (0+5)

Simplifying the equation, we get: d(theta)/dt = 0.208 radians/sec.

Therefore, the angle between the ladder and the ground is changing at a rate of approximately 0.208 radians/sec

User Jake Jones
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