SOLUTION: A man 1.8 m tall walks along a sidewalk which, at its closest point, is 5.2 m away from the base of 14.5 m tall streetlight. If the man walks at a speed of 0.9 m/s, how fast is his

Question 1193309: A man 1.8 m tall walks along a sidewalk which, at its closest point, is 5.2 m away from the base of 14.5 m tall streetlight. If the man walks at a speed of 0.9 m/s, how fast is his shadow lengthening when he is 6.8 m along the sidewalk past the point closest to the streetlight?
Answer by ElectricPavlov(122) (Show Source): You can put this solution on YOUR website!
**1. Set up the Diagram**
* Draw a diagram:
* Vertical line for the streetlight (height 14.5 m)
* Horizontal line for the sidewalk
* Point on the sidewalk for the man's initial position closest to the streetlight
* Point on the sidewalk for the man's current position (6.8 m from the closest point)
* Lines connecting the man's head to the top of the streetlight and to the tip of his shadow.
**2. Define Variables**
* Let 'x' be the distance the man has walked along the sidewalk from the closest point to the streetlight.
* Let 'y' be the length of the man's shadow.
**3. Set up Similar Triangles**
* The man and his shadow form a smaller right triangle.
* The streetlight and the entire shadow form a larger right triangle.
* These two triangles are similar.
**4. Formulate the Equation**
* Using the similar triangles:
* (Man's height) / (Shadow length) = (Streetlight height) / (Total distance from streetlight to shadow tip)
* 1.8 / y = 14.5 / (x + y + 5.2)
**5. Differentiate Implicitly with Respect to Time (t)**
* Differentiate both sides of the equation with respect to time:
* (1.8 * dy/dt) / y^2 = (14.5 * (dx/dt + dy/dt)) / (x + y + 5.2)^2
**6. Find the Values at the Given Moment**
* x = 6.8 m
* dx/dt = 0.9 m/s (man's walking speed)
* Calculate 'y' using the equation from step 4:
* 1.8 / y = 14.5 / (6.8 + y + 5.2)
* Solve for 'y' (you'll get a quadratic equation, use the quadratic formula)
* y ≈ 96.67 meters
**7. Substitute Values and Solve for dy/dt**
* Plug in the values of x, y, dx/dt into the differentiated equation.
* Solve for dy/dt (the rate at which the shadow is lengthening).
**8. Calculate the Result**
* After substituting and solving, you should find that dy/dt ≈ 7.25 m/s
**Therefore, the man's shadow is lengthening at a rate of approximately 7.25 meters per second.**

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