# How would I approach this question?A woman 6ft tall is walking at 2ft/s along a straight path on level ground. There is a lamppost 5ft to the side is the path. A light 16ft high on the lamppost...

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### 1 Answer

Let x represent the distance the woman is from the point on the path closest to the lamppost. Let y be the length of the shadow. Then there are two right triangles to consider.

(1) Along the ground there is a right triangle with legs of 5 and x. The hypotenuse of this right triangle is `sqrt(x^2+25)` .

(2) The other right triangle has legs of 16 and `sqrt(x^2+25)+y` (The lamppost and the distance from the foot of the post to the tip of the shadow.)

(3) Consider the right triangle from (2). The woman's height is 6 feet, and this creates a similar triangle with `16/6=(sqrt(x^2+25)+y)/y` .

Cross-multiplying yields `8y=3sqrt(x^2+25)+3y`

(4) We want to find `(dy)/(dt)=(dy)/(dx)*(dx)/(dt)` . We know `x=12,(dx)/(dt)=2` .

Differentiating the equation from (3) yields:

`8(dy)/(dt)=((3/2)(2x)(dx)/(dt))/(sqrt(x^2+25))+3(dy)/(dt)`

`5(dy)/(dt)=(3x(dx)/(dt))/(sqrt(x^2+25))` Substituting for x and `(dx)/(dt)` we get:

`(dy)/(dt)=(` `(3(12)(2))/(5sqrt(12^2+25))=72/65`

So the change in the length of the shadow is `72/65` feet per second.