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Math  /  Calculus

QuestionA ladder 30 ft long-rests against a vertical wall. If the top of the ladder is being pulled up the wall at a rate of 4ft/s4 \mathrm{ft} / \mathrm{s}, at what rate is the bottom of the ladder moving towards the wall when the top of the ladder is 6 ft from the ground?
Do not include units in your answer. Your answer can be exact or approximate. If it is approximate, round to three decimal places.

Studdy Solution

STEP 1

1. The ladder forms a right triangle with the wall and the ground.
2. The length of the ladder is constant at 30 ft.
3. The rate at which the top of the ladder moves up the wall is 4 ft/s.
4. We need to find the rate at which the bottom of the ladder moves towards the wall when the top is 6 ft from the ground.

STEP 2

1. Set up the relationship using the Pythagorean theorem.
2. Differentiate the relationship with respect to time.
3. Solve for the rate at which the bottom of the ladder moves towards the wall.

STEP 3

Let x x be the distance from the wall to the bottom of the ladder, and y y be the height of the top of the ladder from the ground. According to the Pythagorean theorem, we have:
x2+y2=302 x^2 + y^2 = 30^2

STEP 4

Differentiate both sides of the equation x2+y2=900 x^2 + y^2 = 900 with respect to time t t :
2xdxdt+2ydydt=0 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0

STEP 5

Substitute the known values into the differentiated equation. When y=6 y = 6 , we find x x using the Pythagorean theorem:
x2+62=900 x^2 + 6^2 = 900
x2=864 x^2 = 864
x=864=126 x = \sqrt{864} = 12\sqrt{6}
Now substitute x=126 x = 12\sqrt{6} , y=6 y = 6 , and dydt=4 \frac{dy}{dt} = 4 into the differentiated equation:
2(126)dxdt+2(6)(4)=0 2(12\sqrt{6}) \frac{dx}{dt} + 2(6)(4) = 0

STEP 6

Solve for dxdt \frac{dx}{dt} :
246dxdt+48=0 24\sqrt{6} \frac{dx}{dt} + 48 = 0
246dxdt=−48 24\sqrt{6} \frac{dx}{dt} = -48
dxdt=−48246 \frac{dx}{dt} = -\frac{48}{24\sqrt{6}}
dxdt=−26 \frac{dx}{dt} = -\frac{2}{\sqrt{6}}
Rationalize the denominator:
dxdt=−266 \frac{dx}{dt} = -\frac{2\sqrt{6}}{6}
dxdt=−63 \frac{dx}{dt} = -\frac{\sqrt{6}}{3}
The rate at which the bottom of the ladder is moving towards the wall is:
−63 \boxed{-\frac{\sqrt{6}}{3}}

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