How Fast Does a Shadow Lengthen as a Man Walks Away from a Light?

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SUMMARY

The problem involves a man walking away from a light source 15 feet above a horizontal path, with a height of 6 feet and a walking speed of 5 feet per second. The lengthening rate of his shadow is derived using similar triangles, leading to the conclusion that the tip of the man's shadow moves at a rate of 25/3 feet per second. The key equations utilized include the relationship between the heights and distances of the triangles formed by the light, the man, and his shadow.

PREREQUISITES
  • Understanding of similar triangles and their properties
  • Basic knowledge of calculus, specifically derivatives
  • Familiarity with related rates problems in physics or mathematics
  • Ability to set up and solve equations involving geometric relationships
NEXT STEPS
  • Study the concept of related rates in calculus
  • Learn how to apply similar triangles in geometric problems
  • Practice solving problems involving rates of change
  • Explore real-world applications of similar triangles in physics
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Students studying calculus, particularly those focusing on related rates, as well as educators seeking to enhance their teaching methods for geometric applications in real-world scenarios.

sarah22
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Homework Statement


A light is hung 15ft above a straight horizontal path. If a man 6 ft tall is walking away from the light at the rate of 5ft/sec. How fast is his shadow lengthening? And at what rate is the tip of the man's shadow moving?

y = 6? or 15?
dx/dt = 6


Homework Equations


z^2 = x^2 + y^2 ?


The Attempt at a Solution


I tried looking for angle of it or whatever that came up on my mind. I really don't know where to start after getting the given.

My professor didn't show a same problem just like the above. It just shows us some spherical, cone, two cars moving from different location, a helicopter above and a satellite dish below that is rotating. But this one is so different.

It's not an assignment by the way. I'm just practicing and came up with this problem on the book. The answer at the back for the rate of the tip of the man's shadow moving is 25/3 ft/sec. How the heck did he get that?
 
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Draw a picture if you haven't already done so.
Let x(t) = the man's horizontal distance from the light at time t. His height is 6', so don't use a variable for this constant. Let s(t) = the length of the man's shadow at time t.
You should have right two triangles: a small one whose altitude is 6' and whose base is s(t), and a large one whose altitude is 15' and whose base is x(t) + s(t). These triangles are similar, so you should be able to get an equation for s(t).

After that, get a new equation with the derivatives (the rates) and substitute the given rates to find an expression for s'(t).
 

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