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### Section 3.11 : Related Rates

3. For a certain rectangle the length of one side is always three times the length of the other side.

1. If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the longer side decreasing?
2. At what rate is the enclosed area decreasing when the shorter side is 6 inches long and is decreasing at a rate of 2 inches/minute?

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Hint : The equation needed here is a really simple equation. In fact, so simple it might be easy to miss…
a If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the longer side decreasing? Show All Steps Hide All Steps
Start Solution

Let’s call the shorter side $$x$$ and the longer side $$y$$. We know that $$x' = - 2$$ and want to find $$y'$$.

Now all we need is an equation that relates these two quantities and from the problem statement we know the longer side is three times shorter side and so the equation is,

$y = 3x$ Show Step 2

Next step is to simply differentiate the equation with respect to $$t$$.

$y' = 3x'$ Show Step 3

Finally, plug in the known quantity and solve for what we want : $$\require{bbox} \bbox[2pt,border:1px solid black]{{y' = - 6}}$$

Hint : Once we have the equation for the area we can either simplify the equation as we did in this section or we can use the result from the previous step and the equation directly.
b At what rate is the enclosed area decreasing when the shorter side is 6 inches long and is decreasing at a rate of 2 inches/minute? Show All Steps Hide All Steps
Start Solution

Again, we’ll call the shorter side $$x$$ and the longer side $$y$$ as with the last part. We know that $$x = 6$$, $$x' = - 2$$ and want to find $$A'$$.

The equation we’ll need is just the area formula for a rectangle : $$A = xy$$

At this point we can either leave the equation as is and differentiate it or we can plug in $$y = 3x$$ to simplify the equation down to a single variable then differentiate. Doing this gives,

$A\left( x \right) = 3{x^2}$ Show Step 2

Now we need to differentiate with respect to $$t$$.

If we use the equation in terms of only $$x$$, which is probably the easiest to use we get,

$A' = 6x\,x'$

If we use the equation in terms of both $$x$$ and $$y$$ we get,

$A' = x\,y' + x'y$ Show Step 3

Now all we need to do is plug in the known quantities and solve for$$A'$$.

Using the equation in terms of only $$x$$ is the “easiest” because we already have all the known quantities from the problem statement itself. Doing this gives,

$\require{bbox} \bbox[2pt,border:1px solid black]{{A' = 6\left( 6 \right)\left( { - 2} \right) = - 72}}$

Now let’s use the equation in terms of $$x$$ and $$y$$. We know that $$x = 6$$ and $$x' = - 2$$ from the problem statement. From part (a) we have $$y' = - 6$$ and we also know that $$y = 3\left( 6 \right) = 18$$. Using these gives,

$\require{bbox} \bbox[2pt,border:1px solid black]{{A' = \left( 6 \right)\left( { - 6} \right) + \left( { - 2} \right)\left( {18} \right) = - 72}}$

So, as we can see both gives the same result, but the second method is slightly more work, although not much more.