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

In summary, the problem involves a man walking away from a light source hanging 15ft above a straight horizontal path. The man's height is 6ft and he is moving at a rate of 5ft/sec. The question asks for the rate at which his shadow is lengthening and the rate at which the tip of his shadow is moving. By drawing a diagram and using similar triangles, an equation can be obtained for the length of the man's shadow in terms of time. Differentiating this equation and substituting the given rates, an expression for the shadow's rate of change can be found. This leads to a final answer of 25/3 ft/sec for the rate at which the tip of the man's shadow is moving
  • #1
sarah22
26
0

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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  • #2
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).
 

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

1. What is a related rates problem?

A related rates problem is a type of mathematical problem where the rates of change of two or more related quantities are known, and the goal is to find the rate of change of another quantity that is dependent on the first quantities.

2. How do you approach solving a related rates problem?

The first step is to identify the given rates and the quantity that needs to be solved for. Then, draw a diagram and set up an equation using the given rates and the related quantities. Finally, use the chain rule to find the derivative of the dependent quantity with respect to time and solve for the unknown rate.

3. What is the chain rule and how is it used in related rates problem solving?

The chain rule is a calculus rule that allows us to find the derivative of a composite function. In related rates problem solving, the chain rule is used to find the derivative of the dependent quantity with respect to time, by taking the derivative of each related quantity separately and multiplying them together.

4. Can you give an example of a related rates problem?

For example, a ladder is leaning against a wall and sliding down the wall at a constant rate of 2 ft/s. The base of the ladder is also sliding away from the wall at a rate of 1 ft/s. How fast is the top of the ladder sliding down the wall when the base is 3 feet from the wall?

5. What are some common mistakes to avoid when solving related rates problems?

Some common mistakes to avoid include forgetting to use the chain rule, not labeling the diagram correctly, and not setting up the equation correctly. It is also important to check the units of the given rates and the units of the quantity that needs to be solved for, to ensure they are consistent.

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