Calculus Minimization Problem - Pythagorean Theorem ~ Learn Calculus | Daily Calculus Problems

Problem #49 from Section 4.7 Stewart Calculus 5th Edition:

We have a point p, somewhere on line AD. We need to minimize the length of the black lines

and express the length of these lines as a function of x, L(x).

Now what data points do we need? We definitely need the lengths of the line segments because the length function L(x) will be the sum of them (AP + PB + PC). We know that AD = 5. Remember now that AP isn't fixed, so we can say that AP = x. So PD is 5-x.

What about PC and PC? Well now since PD, is 5-x we can use the Pythagorean theorem.

PB = sqrt(2^2 + (5-x)^2)
PC = sqrt(3^2 + (5-x)^2)

They both share the (5-x) line segment PD recall, and we just squared the lengths that were given in the problem, BD and DC.

So L(x) is simply:

x + sqrt(4 + (5-x)^2) + sqrt(9 + (5-x)^2)

Let's differentiate L(x)

So - L '(x) = 1 + 1/2(4 + (5-x)^2)^-1/2 + 1/2(9 + (5-x)^2)^-1/2

still going with the chain rule - we had to take the derivative of the inner function (5-x)^2.

L '(x) = 1 + 1/2(4 + (5-x)^2)^-1/2 * 2(5 - x)-1 + 1/2(9 + (5-x)^2)^-1/2 * 2(5 - x)-1

This simplifies to: L ' (x) = 1 + [(x-5)/sqrt(4 + (5-x)^2)] + [(5-x)/sqrt(9 + (x-5)^2)]

Well - that's a mess and we have to find the roots of the equation so we can evaluate the critical points. Let's use a graph instead:

We see that L(x) experiences a minimum as L '(x), it's derivative crosses the x axis. When any derivative changes from negative to positiv

e, the curve experiences a minimum. So what's the value of x that minimizes the function?

Let's take a closer look:

It seems that the derivative dL/dx is crossing the x axis at 3.593. Plugging in this x value into the original function:

x + sqrt(4 + (5-x)^2) + sqrt(9 + (5-x)^2)

= 9.35 meters or so.

You can also see this in the first graph as the Y value where the curve experiences a minimum.

That's pretty much it - you needed to define the line segments, find a function which expresses the sum of the line segments in terms of x and differentiate. The function was minimized as it's derivative crossed the x axis. The graphs were made in Mathematica, by Wolfram.

For those of you may still be hung up on the chain rule - here's a great video on YouTube which clearly describes the operation taking place.