2.9 Related Rates Example 04 (Man walking with his shadow)

http://www.rootmath.org | Calculus 1

A more difficult problem involving the rate of change of a man's shadow as he walks away from a lamp post.

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[xCelsius451x]

Does the sign of the rates matter? For example, the man is walking toward the light, while the shadow will stretch away from it. Should one rate be + and the other - ?

[rootmath]

@xCelsius451x The sign does matter! But the man is walking away from the lamp and his shadow is also stretching away from the lamp so they are both positive. Think if you had a lamp right over your head, your shadow would be a little circle around your feet, as you walk away from the lamp your shadow grows

[sh3aty]

I'm kind of confused. It says that a man is walking away from the lap at 5 ft/s. If the man is walking away, wouldn't the 5 ft/s be associated with dl/dt? As if the man moves, X is changing as well as Y.

So, I guess my question is why is 5 ft/s associated as dx/dt and not dl/dt?

[rootmath]

@sh3aty The man's position is given by x (forget about L and y for a moment). So the derivative of his position (with respect to time) is his velocity. So dx/dt = 5 ft/s.

On the other hand, dL/dt refers to how fast the man AND his shadow are moving. It's related to how fast he is walking (hence "related rates") but it won't be the same because we have to consider how fast his shadow is growing also.

See your inbox for more detail and message me back if you have more questions.

[w4tch0rb3w4tch3d]

Hey, wouldn't it be easier to just find the derivative of y=(2/3)x?

Since dx/dt is already a known value.

Good video, helped!

[rootmath]

@w4tch0rb3w4tch3d You're absolutely correct! For part b. it would have been easier to do as you suggested, it was an oversight on my part. So now we have 2 different ways to solve the same problem. Thanks for the comment!

[babyimananarchistx]

Thank you so much for posting this! This same problem is in my textbook, but it is also given that x=40ft. With that information, how would you solve the problem differently?

[qthu1hu]

Thank you for making this video, it helped me out quite a bit.

[kickit246]

This video helped me... barely. You never show your work and I have to keep rewinding to see where you pulled numbers from. It was a bit frustrating, but helped regardless I suppose.

[animal872]

Thank you. this helped a lot.

[RoXSaNgnOiRz]

lol 2/3+3/2 is not 5/2 @ 8:25

[odoc01]

Thanks! totally saved my life here!