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A man stars walking north at $ 4 ft/s $ from a po…

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Problem 18 Medium Difficulty

A spotlight on the ground shines on a wall $ 12 m $ away. If a man $ 2 m $ tall from the spotlight towards the building at a speed of $ 1.6 m/s, $ how fast is the length of his shadow on the building decreasing when he is $ 4 m $ from the building?

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Video Transcript

We're told that a spotlight on the ground shines in a wall 12 m away. A man two m tall walks So so was really from the spotlight towards the building at a speed of 1.6 m per second. Were asked to determine how fast the length of his shadow is decreasing when he is four m from the building. So to understand this problem, I want to draw a diagram. That's how he well, intuitively, we know that the man's shadow will be shrinking as he walks away from the light. Now draw a diagram. We have our light here. Then we have our man who has a height of two m. And then we have our wall. Yeah. We don't know how tall the wall is be too, huh? So the man is walking from the light towards the wall? Yes. Now the shadow is going to be as tall as this Hypotenuse is on the back wall. Say yes. Now we know that the total length from the light to the wall. Yes, well, we're told this is 12 and this is in meters and the distance from the light to the man. Let's call this X she's and she's and the distance from the man's the wall. Therefore well, at this instant we're told that he's four m from the building like, yeah, so long and I'll call the height of the shadow on the wall. Yes. Why doesn't Yeah, to understand this problem, you should use similar triangles joint Some man forms a right triangle with respect to the floor and the light. The shadow also forms a right triangle with respect to the floor and the light been like nice And the angle of the light is the same. So by similar triangles what it follows that and literally I'm x over two is equal to 12 over Why, In other words, we have x times y equals 24. It's well, I'm like now we want to find dy DT. To do this will differentiate with respect to t buy the product rule. We have X dy DT plus dx DT times y equals zero solving for dy DT dy DT equals negative y over x times dx DT if you guys are you right right to the And we know that the man is four m from the building, so Okay, X is equal to yeah eight and DX DT, we're told, is 1.6 meters per second six. No idea. Now we want to find why right? We can use our original equation again. Similar triangles. So if X equals eight, then why is equal to 24/8 or three living and therefore Dy DT is equal to negative three over eight times dx DT which is 1.6 which is negative 0.6. The reason he started in the units are in meters per seconds, just one And so the length of his shadow on the building is decreasing at a rate of 0.6 m per second. Mhm. I was right, right? Get into It's decreased because we have a negative make work.

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Catherine Ross

Kristen Karbon

Samuel Hannah

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Kristen Karbon

Samuel Hannah

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Catherine Ross

Kristen Karbon

Samuel Hannah

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals