Rate of how fast a shadow grows as you walk

daigo

Member

There is a lamp post 15 feet tall casting a shadow 'B' ft. long of a person that is 6 feet tall standing 'A' ft. from the lamp post. If the person moves away from the lamp post at 5 feet per second, how fast does the shadow lengthen?
So here I thought it might be something to formulate a triangle, or actually two:

So the problem is asking me to find the derivative of the shadow's length with respect to time: $$\frac{dB}{dt}$$

I think the triangles are similar:

$$\frac{15}{A+B} = \frac{6}{B} \\ \\ \cdots \\ B = \frac{2}{3}A$$

So we have B and I will attempt to take the derivative of this:

$$\frac{dB}{dt} = [\frac{2}{3}A]'$$

Constant/Chain rules:

$$\frac{dB}{dt} = \frac{2}{3} \cdot [A'] \\ = \frac{2}{3} \cdot \frac{dA}{dt}$$

Now how do I find the derivative of 'A'? It must have something to do with the movement velocity because it was given information and I haven't used it yet. Unless I am being tricked and it is extraneous/unnecessary information.

HallsofIvy

Well-known member
MHB Math Helper
You've done very well. Now, with your "A" and "B", the total distance from the light pole to the person is A+ B. You are told that the person is walking away from the light pole at 5 ft/s so that (A+ B)'= A'+ B'= 5 and therefore A'= 5- B'. Replace A' in your last equation with that and solve for B'.

Reckoner

Member
You've done very well. Now, with your "A" and "B", the total distance from the light pole to the person is A+ B. You are told that the person is walking away from the light pole at 5 ft/s so that (A+ B)'= A'+ B'= 5 and therefore A'= 5- B'. Replace A' in your last equation with that and solve for B'.
The distance from the pole to the person is actually $$A$$, according to the original poster's drawing.

So, daigo, if the person is moving 5 feet per second away from the post, then how fast is $$A$$ increasing?

daigo

Member
The distance from the pole to the person is actually $$A$$, according to the original poster's drawing.

So, daigo, if the person is moving 5 feet per second away from the post, then how fast is $$A$$ increasing?
By 5 ft./sec.? This is a constant velocity, so...this is just a slope of 0 on the y-axis since it's a horizontal line. If function 'A' is the distance travelled while going at a constant 5 ft./sec (f(A) = 5t), then the slope of a line is always going to be the same so taking the derivative of a line like '5t' would just be a constant 5, I think?

Reckoner

Member
By 5 ft./sec.?
Yes. $$A$$ is increasing by 5 ft every second. Therefore,

$\frac{dA}{dt} = 5\ \mathrm{ft}/\mathrm{s}.$

Now substitute this into your equation.