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Math  /  Calculus

QuestionA ladder 23 ft long rests against a vertical wall. If the top of the ladder is being pulled up the wall at a rate of 19ft/s19 \mathrm{ft} / \mathrm{s}, at what rate is the bottom of the ladder moving towards the wall when the top of the ladder is 7 ft from the ground?

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 23 ft.
3. The rate at which the top of the ladder moves up the wall is 19 ft/s.
4. We need to find the rate at which the bottom of the ladder moves towards the wall when the top is 7 ft above the ground.

STEP 2

1. Set up the relationship using the Pythagorean theorem.
2. Differentiate with respect to time to relate the rates of change.
3. Solve for the desired rate.

STEP 3

Set up the Pythagorean theorem for the right triangle formed by the ladder, the wall, and the ground. Let x x be the distance from the wall to the bottom of the ladder, and y y be the height of the ladder on the wall. The equation is: x2+y2=232 x^2 + y^2 = 23^2

STEP 4

Differentiate both sides of the equation 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. We know dydt=19 ft/s \frac{dy}{dt} = 19 \, \text{ft/s} and y=7 ft y = 7 \, \text{ft} .
First, solve for x x using the Pythagorean theorem: x2+72=232 x^2 + 7^2 = 23^2 x2+49=529 x^2 + 49 = 529 x2=480 x^2 = 480 x=480=16×30=430 x = \sqrt{480} = \sqrt{16 \times 30} = 4\sqrt{30}
Now substitute into the differentiated equation: 2(430)dxdt+2(7)(19)=0 2(4\sqrt{30}) \frac{dx}{dt} + 2(7)(19) = 0

STEP 6

Solve for dxdt \frac{dx}{dt} : 830dxdt+266=0 8\sqrt{30} \frac{dx}{dt} + 266 = 0 830dxdt=−266 8\sqrt{30} \frac{dx}{dt} = -266 dxdt=−266830 \frac{dx}{dt} = -\frac{266}{8\sqrt{30}}
Simplify the expression: dxdt=−133430 \frac{dx}{dt} = -\frac{133}{4\sqrt{30}}
Rationalize the denominator: dxdt=−13330120 \frac{dx}{dt} = -\frac{133 \sqrt{30}}{120}
The rate at which the bottom of the ladder is moving towards the wall is:
−13330120 ft/s \boxed{-\frac{133 \sqrt{30}}{120} \, \text{ft/s}}

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