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Scilla [17]
3 years ago
11

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.0 ft/s,

how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 6 ft from the wall.)

Physics
2 answers:
kirill115 [55]3 years ago
7 0

Answer:

\frac{d\theta}{dt} = -0.125 rad/s

Explanation:

Let at any moment of time the foot of the ladder is at distance "x" from the wall and if the length of the ladder is L = 10 ft

so we have

\frac{x}{L} = cos\theta

now we have

x = L cos\theta

now differentiate the above equation with time

\frac{dx}{dt} = -L sin\theta\frac{d\theta}{dt}

so we have

\frac{d\theta}{dt} = \frac{-v_x}{L sin\theta}

\frac{d\theta}{dt} = \frac{-1 ft/s}{10 sin\theta}

as we know that

cos\theta = \frac{6}{10} = 0.6

\theta = 53 degree

\frac{d\theta}{dt} = \frac{-1 ft/s}{10 sin53}

\frac{d\theta}{dt} = \frac{-1 ft/s}{10 sin\theta}

\frac{d\theta}{dt} = -0.125 rad/s

natka813 [3]3 years ago
4 0

Answer:

The angle's rate of change is: -0.125 (degree/feet).

Explanation:

In this case of problem we need to find the angle's rate (α) of change when x=6 ft. First we need to relate (α) with x and y and the expression that do it is: \alpha =Arctan(\frac{y}{x}) where (α) is the angle between the ladder and the ground, x is the horizontal distance and y the vertical distance, now we need to have the variable y at function of x, so we can do it using the Pythagorean theorem and gets:y^{2} +x^{2} =10^{2} solving for y(x) we get:y(x)=\sqrt{100-x^{2}}. Replacing all we have got in the first equation: \alpha =Arctan(\frac{\sqrt{100-x^{2} } }{x}). Finally we derivate this equation at function of variable x and gets this result:\frac{d\alpha }{dx} =\frac{-1}{\sqrt{100-x^{2} } } evaluating at x=6 ft we get: -0.125(degree/feet). The negative signal means that the angle is decreasing.

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