# 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...

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 casts the woman's shadow on the ground. How fast is the length of her shadow changing when the woman is 12ft from the point on the path closest to the lamppost?

embizze | High School Teacher | (Level 1) Educator Emeritus

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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.