A man 2m tall is watching a street light 6m high as he jogs toward it at a rate of 2m/s. How fast is the angle of elevation of the man's line ofsight increasing at the instant that he is 3m from...

A man 2m tall is watching a street light 6m high as he jogs toward it at a rate of 2m/s. How fast is the angle of elevation of the man's line of

sight increasing at the instant that he is 3m from the base of light?

Asked on by islnds

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nathanshields | High School Teacher | (Level 1) Associate Educator

Posted on

Draw a little diagram, with the dude, the light, and label the heights.  Now get the triangle in there, with leg lengths a=4 (the difference between the man and the light), and b=3 (the distance between the man and the light).

Call the angle of elevation of line of sight A. We want to now how fast this angle changes, eh?

How would you find that angle, anyway?  Inverse tangent?  Dang right. Remember, "a" is constant but "b" varies:

tan A = a/b = 4/b

A = arctan 4/b

`A = arctan(4b^-1)`

Now let's do a derivative, since we're interested in finding dA/dt.

`(dA)/dt=(-4b^-2)(1/(1+(4b^-1)^2))(db)/dt`

`=(-4/(b^2))(1/(1+16b^-2))(db)/dt`

`=-4/(b^2+16)(db)/dt`

We're interested in dA/dt right when b = 3, but what would db/dt equal?  Think about how fast b is changing... 2 m/s... shrinking!  So db/dt = -2 m/s
`=-4/(9+16)*-2=8/25`

What would the units of dA/dt be?

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