Related Rates: The Plan to Ambush the Wild Mathemon

After spotting the injured man in the forest and hearing about the Mathemon down the road, Jane and Derivee decide that they should try and capture that Mathemon. To make sure they get to the Mathemon quickly, Jane devises a plan for which pathway they should take that will lead them to the Mathemon the quickest in order to hurry back and help the injured man.

A wild Mathemon is 20 miles down the road. Currently, Jane and Derivee are 4 miles away from the nearest point on the road. Derivee can only travel 10 mph along the road and only 5 mph through the forest. What is the least amount of time it would take to get to the wild Mathemon?

First draw a diagram to get a better understanding.

We are trying to minimize the time it takes for Deravee to get to the other Mathemon

Looking at the diagram, we can make up an equation for time for both the road and forest

Time is distance over rate

2 16

5

x

20

10

x

Forest time =

Road time =

Using the equations made from the previous slide so we can find total time

It is just the sum of the road time and the forest time

2 16 20( )

5 10

x xT x

The next step in calculating the least amount of time for Derivee to get to the other Mathemon is to find the rate of change of time

To make it easier to find the derivative, some algebraic massaging is needed

21( ) 16 2

5 10

xT x x

Power Rule Take the exponent and multiply to the coefficient Subtract one from the exponent and that’ll become the

new exponent

Chain Rule Derivative of the outer composed of inner Multiplied by the derivative of the inner

Constant Multiple Rule Derivative of the function multiplid by the constant

12 2

1 1'( ) ( 16) (2 )

10 10T x x x

Applying Derivative Rules:

12 2

1 1'( ) ( 16)

5 10T x x x

12 2

2 1'( ) ( 16)

10 10T x x x

Simply multiply the 2x to the 1/10

To simplify further we need to find a common denominator, which is 10

All that is needed is to rewrite the function under the denominator of 10 and to rewrite the exponent of -1/2 by moving the root down to the denominator

2

2 1'( )

10 16

xT x

x

Now that the function is simplified we can easily find a root

The denominator will never equal zero because of the squared

That means we’ll only need to look at the numerator to find the zero, which is the critical point

'( ) 0T x at1

2x

Using the First Derivative Test:

To see if at x = ½ there is a min on the time function we’ll need to do a line analysis of the derivative

If it is a min then it’ll be negative to the left and positive to the right of the critical point

We can do this by choosing any point in between negative infinite and ½ and positive infinite

Therefore by the first derivative test at ½ is a min because to the left the derivative is negative and to the right it is positive

1

2

'( )T x

Now that we know at ½ is a minimum, we can plug it into the total time equation to find the value

1( ) 2.762

T hrs

It takes approximately 3 hours to reach the wild Mathemon at the end of the road

through the forest.

Shortest time solved! The plan of attack is complete!