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6. ( 6 points) A ladder 10 feet long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of \( 1 \mathrm{ft} / \mathrm{s} \). Be sure to include units in your answer. a. ( 5 points) How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall? b. ( 1 point) Briefly explain why the sign of your answer to part a makes sense. Solution

Ask by Mitchell Ryan. in the United States
Dec 03,2024

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a. The top of the ladder is sliding down the wall at a rate of \( -\frac{3}{4} \, \text{ft/s} \) when the bottom is 6 ft from the wall. b. The negative sign indicates that the top of the ladder is moving downward as the bottom moves away from the wall, which is expected to maintain the ladder's length.

Solution

To solve this problem, we can use related rates and the Pythagorean theorem. Let: - \( x \) be the distance from the bottom of the ladder to the wall (in feet). - \( y \) be the height of the top of the ladder on the wall (in feet). - The length of the ladder is constant at \( L = 10 \) feet. According to the Pythagorean theorem, we have: \[ x^2 + y^2 = L^2 \] Substituting \( L = 10 \): \[ x^2 + y^2 = 10^2 \] \[ x^2 + y^2 = 100 \] ### Part a: Finding the rate at which the top of the ladder is sliding down the wall We are given that the bottom of the ladder is sliding away from the wall at a rate of \( \frac{dx}{dt} = 1 \, \text{ft/s} \). We want to find \( \frac{dy}{dt} \) when \( x = 6 \, \text{ft} \). First, we differentiate the equation \( x^2 + y^2 = 100 \) with respect to time \( t \): \[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(100) \] Using the chain rule, we get: \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] We can simplify this to: \[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \] Now, we need to find \( y \) when \( x = 6 \): \[ 6^2 + y^2 = 100 \] \[ 36 + y^2 = 100 \] \[ y^2 = 64 \] \[ y = 8 \, \text{ft} \] Now we can substitute \( x = 6 \), \( y = 8 \), and \( \frac{dx}{dt} = 1 \, \text{ft/s} \) into the differentiated equation: \[ 6(1) + 8 \frac{dy}{dt} = 0 \] \[ 6 + 8 \frac{dy}{dt} = 0 \] \[ 8 \frac{dy}{dt} = -6 \] \[ \frac{dy}{dt} = -\frac{6}{8} = -\frac{3}{4} \, \text{ft/s} \] Thus, the top of the ladder is sliding down the wall at a rate of \( \frac{dy}{dt} = -\frac{3}{4} \, \text{ft/s} \). ### Part b: Explanation of the sign of the answer The negative sign of \( \frac{dy}{dt} \) indicates that the height \( y \) is decreasing, which makes sense in this context. As the bottom of the ladder slides away from the wall, the top of the ladder must slide down the wall to maintain the length of the ladder at 10 feet. Therefore, the negative rate of change reflects the fact that the top of the ladder is moving downward as the bottom moves away from the wall.

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To solve this problem, we can use the Pythagorean theorem and related rates. The ladder forms a right triangle with the wall and the ground. Let \( x \) be the distance from the wall to the bottom of the ladder, and \( y \) be the height of the ladder on the wall. According to the Pythagorean theorem, we have: \[ x^2 + y^2 = 10^2. \] Given that the bottom of the ladder is sliding away from the wall at a rate of \( \frac{dx}{dt} = 1 \mathrm{ft/s} \), we want to find \( \frac{dy}{dt} \) when \( x = 6 \) ft. 1. Differentiate both sides of the equation with respect to time \( t \): \[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0. \] 2. Rearranging gives: \[ y\frac{dy}{dt} = -x\frac{dx}{dt}. \] 3. Now, substituting \( x = 6 \) ft into the Pythagorean equation to find \( y \): \[ 6^2 + y^2 = 10^2 \implies 36 + y^2 = 100 \implies y^2 = 64 \implies y = 8 \text{ ft}. \] 4. Substitute \( x = 6 \), \( y = 8 \), and \( \frac{dx}{dt} = 1 \): \[ 8\frac{dy}{dt} = -6(1) \implies 8\frac{dy}{dt} = -6 \implies \frac{dy}{dt} = -\frac{6}{8} = -\frac{3}{4} \mathrm{ft/s}. \] **Answer for part a:** The top of the ladder is sliding down the wall at a rate of \( -\frac{3}{4} \mathrm{ft/s} \). For part b: The negative sign of the answer indicates that the height of the ladder, \( y \), is decreasing over time. This makes sense because as the bottom of the ladder is sliding away from the wall, the top of the ladder must naturally move downward along the wall to maintain the length of the ladder at 10 feet.

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