Related Rates Applications
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# Related Rates Applications

One great application of calculus is being able to solve problems regarding related rates. What we intend to do is compute the rate of change of some quantity with a rate of change of another quantity, both of which rates are with respect to time $t$. Solving problems like these can be difficult, so we will provide a strategy.

• 1. Be sure to read the entire problem carefully and identify any important information in the problem. Write down important information that is given to you, and what important information you intended to find to answer the problem while introducing appropriate notation.
• 2. Illustrate the problem and what you are given with a clear and concise diagram. This will help you visualize the problem and have your verify the problem.
• 3. Find an equation that relates both quantities together, and then differentiate both sides of the equation.
• 4. Substitute any relevant given information into the new equation and solve for the unknown rate. Do NOT substitute relevant information too early!

## Example 1

A ladder whose height is $10m$ rests against a vertical wall and is slowly sliding down the way. The bottom of the ladder slides away from the wall at a rate of $1m/s$. Determine how fast the top of the ladder is sliding down when the ladder is $4m$ away from the wall.

Let $y$ denote the distance from the top of the ladder to the ground, and let $x$ denote the distance from the bottom of the ladder to the wall. The following diagram illustrates our problem:

We are given that the rate that the bottom of the ladder slides away from the wall is $1m/s$, and therefore, $\frac{dx}{dt} = 1m/s$. We now must find a relationship between $x$ and $y$ that is applicable to our question… the best being that $x^2 + y^2 = (10)^2$. If we implicitly differentiate this and isolate for what we need to find, $\frac{dy}{dt}$, then:

(1)
\begin{align} 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \\ 2x (1) + 2y \frac{dy}{dt} = 0 \\ \frac{dy}{dt} = \frac{- x}{y} \end{align}

Now when the ladder is $x = 4m$ away from the wall, we use Pythagoras' theorem again to get that $y = \sqrt{84}$ away from the wall, and thus, $\frac{dy}{dt} = \frac{-4}{\sqrt{84}} \approx -0.436 m/s$.

## Example 2

Two cars $A$ and $B$ on different roads are approaching each other at a common intersect. Car $A$ is approaching north at a rate of $70 \mathrm{kmph}$, and car $B$ is approaching east at a rate of $60 \mathrm{kmph}$. What rate are the cars approaching when car $A$ is $3\mathrm{km}$ from the intersection and car $B$ is $4 \mathrm{km}$ from the intersection?

We first draw a diagram to illustrate the problem:

Now let $x$ denote the distance car $A$ is from the intersection, and let $y$ denote the distance car $B$ is from the intersection. Let $z$ denote the distance between the cars. We note that we are given $\frac{dx}{dt} = -70\mathrm{kmph}$, $\frac{dy}{dt} = -60\mathrm{kmph}$. We want to figure out $\frac{dz}{dt}$. One such formula that relates all of this together is Pythagoras' theorem once again:

(2)
\begin{align} x^2 + y^2 = z^2 \\ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt} \\ x \frac{dx}{dt} + y \frac{dy}{dt} = z \frac{dz}{dt} \\ \frac{dz}{dt} = \frac{1}{z} \left ( x \frac{dx}{dt} + y \frac{dy}{dt} \right ) \end{align}

We note that from the question that $x = 3$ and $y = 4$, therefore $z = 5$. Substituting these values into our equation, we get:

(3)
\begin{align} \frac{dz}{dt} = \frac{1}{5} \left ( 3 (-70) + 4(-60) \right ) \\ \frac{dz}{dt} = -90 \mathrm{kmph} \end{align}

Thus, the cars are approaching each other at a rate of $90 \mathrm{kmph}$.

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