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woman walks directly tow

woman walks directly tow

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 94E
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### Problem Description A wall lamp is positioned 7 feet up a wall. A 6-foot tall woman walks directly towards the light at a speed of 5 feet per second. Draw and label a diagram of the situation. Then, showing all your work as instructed in class, find the rate at which the tip of her shadow is moving when she is 20 feet away from the wall. If necessary, round your answer to 3 decimal places and give proper units. ### Solution Explanation #### Step-by-Step Instructions: 1. **Diagram Setup**: - Draw the wall and indicate the position of the lamp 7 feet above the ground. - Draw the woman who is 6 feet tall and represent her walking towards the wall. - Indicate her speed of 5 feet per second. - Draw the shadow cast by the woman towards the wall. 2. **Using Similar Triangles**: - Note that two similar triangles are formed: one with the lamp's position to the tip of the shadow and the other with the height of the woman and the length of her shadow. - Let `x(t)` be the distance of the woman from the wall, and `y(t)` be the length of her shadow on the ground. 3. **Formulate Equations**: - From the properties of similar triangles: $$ \frac{7}{y(t)} = \frac{6}{y(t) - x(t)} $$ - Rearrange this to: $$ 7(y(t) - x(t)) = 6y(t) $$ $$ 7y(t) - 7x(t) = 6y(t) $$ $$ y(t) = 7x(t) $$ 4. **Derive the Rate**: - Differentiate both sides with respect to `t`: $$ \frac{dy(t)}{dt} = 7 \frac{dx(t)}{dt} $$ - Given that the woman walks towards the wall at 5 ft/sec (` \frac{dx(t)}{dt} = -5 ft/sec `): $$ \frac{dy(t)}{dt} = 7 \times (-5) $$ $$ \frac{dy(t)}{dt} = -35 ft/sec $$ 5. **Interpret the Result**: - The negative sign indicates that the length of the shadow is
Transcribed Image Text:### Problem Description A wall lamp is positioned 7 feet up a wall. A 6-foot tall woman walks directly towards the light at a speed of 5 feet per second. Draw and label a diagram of the situation. Then, showing all your work as instructed in class, find the rate at which the tip of her shadow is moving when she is 20 feet away from the wall. If necessary, round your answer to 3 decimal places and give proper units. ### Solution Explanation #### Step-by-Step Instructions: 1. **Diagram Setup**: - Draw the wall and indicate the position of the lamp 7 feet above the ground. - Draw the woman who is 6 feet tall and represent her walking towards the wall. - Indicate her speed of 5 feet per second. - Draw the shadow cast by the woman towards the wall. 2. **Using Similar Triangles**: - Note that two similar triangles are formed: one with the lamp's position to the tip of the shadow and the other with the height of the woman and the length of her shadow. - Let `x(t)` be the distance of the woman from the wall, and `y(t)` be the length of her shadow on the ground. 3. **Formulate Equations**: - From the properties of similar triangles: $$ \frac{7}{y(t)} = \frac{6}{y(t) - x(t)} $$ - Rearrange this to: $$ 7(y(t) - x(t)) = 6y(t) $$ $$ 7y(t) - 7x(t) = 6y(t) $$ $$ y(t) = 7x(t) $$ 4. **Derive the Rate**: - Differentiate both sides with respect to `t`: $$ \frac{dy(t)}{dt} = 7 \frac{dx(t)}{dt} $$ - Given that the woman walks towards the wall at 5 ft/sec (` \frac{dx(t)}{dt} = -5 ft/sec `): $$ \frac{dy(t)}{dt} = 7 \times (-5) $$ $$ \frac{dy(t)}{dt} = -35 ft/sec $$ 5. **Interpret the Result**: - The negative sign indicates that the length of the shadow is
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