# Thread: Another rate of change question

1. ## Another rate of change question

A dog instantly spots a squirrel sitting on the ground 15m away. She approaches the squirrel at a constant rate of 2m/s and the squirrel darts straight up the tree at a constant rate of 4m/s. Assuming the ground is smooth and flat and the tree is perfectly vertical:
a) At what rate is the distance between the dog and the squirrel changing at the moment when they are the closest together? (i.e. at the minimum distance apart?)
b) Determin the time at which the distance is at its minimum
c) Calculate the minimum distance between the dog and the squirrel.

I know that this problem involves the dog and squirrel forming a triangle, but there doesnt seem to be enough information given to solve the question, please help.

2. The thing to do first is to find the time when they are closest.

Since the dog is moving at 2 m/s, her distance is (15-2t).

The squirrels distance is 4t

Therefore the time they're closest together is by ol' Pythagoras.

$\displaystyle D(t)=(15-2t)^{2}+(4t)^{2}$

$\displaystyle D'(t)=40t-60$

$\displaystyle t=\frac{3}{2} \;\ sec$

After 3/2 seconds they are closest together.

If the dog is moving at 2 m/s for 3/2 seconds then she moves 3 feet.

Therefore, she is 15-2(3/2)=12 feet from the tree.

The squirrel is 4(3/2)=6 feet up the tree,

The distance between them along the hypoteneuse is

$\displaystyle \sqrt{12^{2}+6^{2}}=6\sqrt{5}\approx{13.42} \;\ meters$

3. I just have one question, when you used the pythagorean theorem for the D(t) equation, why dont you have to take the squareroot of the right side?

4. Hello, sugar_babee!

When you used the pythagorean theorem for the D(t) equation,
why don't you have to take the squareroot of the right side?
Galactus used a very handy "trick" . . .

You are correct . . . the distance is: .$\displaystyle d \;=\;\sqrt{(15-2t)^2 + (4t)^2}$

. . Square both sides: .$\displaystyle d^2 \;=\;(15-2t)^2 + (4t)^2$

Now let $\displaystyle D = d^2\!:\;\;D \;=\;(15-2t)^2 + (4t)^2$

Can you see that if we minimize $\displaystyle D$, we are also minimizing $\displaystyle d$ ?

So we can solve: $\displaystyle D' = 0$ and avoid that one-half power.

5. okay so how would I find the rate for part a?

6. You have all the info now, use related rates to solve.

Differentiate implicitly.

$\displaystyle D^{2}=x^{2}+y^{2}$

$\displaystyle D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$