# Calculus

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A man 2m tall walks away from a lamppost whose light is 5m above the ground. If he walks at a speed of 1.5m/s, at what rate is his shadow growing when he is 10m from the lamppost?

I tried to draw a diagram, but I don't understand where the 5m is, the height of the lamppost? Is the question asking for the derivative of the speed?

• Calculus - ,

draw a lamppost, vertical line, and label it 5 m high (we assusme the light is at the top of the lamppost)
draw a horizontal line, the sidewalk? , and draw the man , vertical line, somewhere on the sidwalk.
Join the top of the lamppost to the man's head and continue until you hit the sidewalk.
label the distance between the post and the man as y, label the length of his shadow x
we are given dy/dt as 1.5 m/s, we are to find dx/dt when y = 10 m

I see to right-angled triangles, with a smaller inside a larger.
By similar triangles:
5/(x+y) = 2/x
cross-multiply, ...
5x = 2x + 2y
3x = 2y

differentiate with respect to t
3dx/dt = 2dy/dt
dx/dt = 2(1.5)/3 = 1 m/s

Notice we did not need the 10 m, there was no place to sub it in.

We have shown that the man's shadow is growing at a constant rate of 1 m/s, no matter where he is.

Be careful with this question.
Had it asked "how fast is the man's shadow moving", we would have to add the man's speed to the 1 m/s for a speed of 2.5 m/s.

• Calculus - ,

My goodness!
I just noticed that I typed "to" instead of "two" in "I see to right-angled triangles ..."

• Calculus - ,

Thank you so much for clearing it up showing the steps crystal clear. How do you know that 1.5m/s is dy/dt?

• Calculus - ,

2.5 km/h

• Calculus - ,

thanks a lot Reiny.

• Calculus - ,

The y distance is based on how fast the man is walking. In respect to time, dy/dt is 1.5m/s.