Related Rates

A \(10ft\) tall ladder leans against a wall. The base of the ladder begins to slide away from the wall at a rate of \(2ft/s) As the base of the ladder slides away from the wall, the top of the ladder slides down the wall vertically. When the base of the ladder is \(9ft\) away from the wall, what is the rate at which the top of the ladder slides down the wall?

Step 1: Draw a diagram

Drawing a diagram of the problem will help us to better comprehend our known and unknown values.

Fig. 2. From the horizontal rate of change, we are tasked with finding the vertical rate of change.

Step 2: Identify known and unknown quantities

Before we can do any Calculus, we must fully understand the problem. We know that a \(10ft\) ladder slides away from a wall horizontally at a rate of \(2ft/s\). The problem wants to know at what rate the top of the ladder moves when the base of the ladder is \(9ft\) from the wall.

Using our diagram in step 1, we can organize known and unknown variable quantities:

\[\dfrac{dy}{dt}=?\]

\[y(t)=?\]

\[x(t)=9\]

\[\dfrac{dx}{dt}=2\]

\[z=10\]

\(x\) and \(y\) are functions of time in this problem, so they are written \(x(t)\) and \(y(t)\). However, the length of the ladder, \(z\), does not change with time, so it is not written with function notation.

Step 3: Use equations to relate information in the problem

Based on the information we have and the information we need, it should be obvious that the Pythagorean Theorem will be useful in this problem.

Looking at the diagram again, notice that the ladder and 2 walls make a right triangle. This is a perfect scenario to use the Pythagorean Theorem!

Remember, while the ladder moves horizontally and vertically, the hypotenuse of the triangle (length of the ladder) does not change.

\[(x(t))^2+ (y(t))^2=z^2\]

\[(x(t))^2+ (y(t))^2=10^2\]

\[(x(t))^2+ (y(t))^2=100\]

Notice that we are given the derivative of \(x\) with respect to time,

\[\dfrac{dx}{dt}\]

We are also asked to find the rate at which the ladder is moving vertically,

\[\dfrac{dy}{dt}\]

How can we make an equation with these variables? Implicit differentiation!

Step 4: Solve using implicit differentiation

Now that we have an equation, let's use implicit differentiation to get the equation in terms of two rates of change. We will take the derivative with respect to time.

\[\dfrac{d}{dt}[(x(t))^2+(y(t))^2]=\dfrac{d}{dt} 100\]

\[2(x(t))\dfrac{dx}{dt}+2(y(t))\dfrac{dy}{dt}=0\]

Step 5: Substitute in known values

Again, we want to find the rate at which the ladder slides down the wall vertically:

\[\dfrac{dy}{dt}\]

We know that \(x=9\) ft and

\[\dfrac{dx}{dt}=2ft/s\]

Plugging in our known values, we get

\[2 \cdot 9 \cdot 2 + 2(y(t))\dfrac{dy}{dt}=0\]

To solve for \(\dfrac{dy}{dt}\), we still need the value of \(y\) when \(x=9\). We can use the Pythagorean Theorem equation we set up earlier to find \(y\), subbing \(x(t)=9\).

\[(x(t))^2+ (y(t))^2=z^2\]

\[9^2+ (y(t))^2=10^2\]

\[ (y(t))^2=19\]

\[ (y(t))=\sqrt{19}\]

Plugging in \(y(t)\) and solving for \(\dfrac{dy}{dt}\).

\[36+2(\sqrt{19})\dfrac{dy}{dt}=0\]

\[\dfrac{dy}{dt}=-\dfrac{36}{2 \sqrt{19}}\]

\[\dfrac{dy}{dt}=-\dfrac{18}{\sqrt{19}}\]

\[\dfrac{dy}{dt}=-4.129ft/s\]

Considering a perfectly spherical balloon being filled with air. The balloon expands at a rate of \(3cm^2/s\). When the balloon's radius is \(4cm\), how fast is the radius increasing?

Step 1: Draw a diagram

Fig. 3. From the rate of change of the volume, we are tasked with finding the rate of change of the radius.

Based on our diagram, we are missing the rate of change of the radius. However, we do have the rate of change of the volume.

Step 2: Identify known and unknown quantities

We know that the volume of a spherical balloon increases at a rate of \(3cm^2/s\). We want to know the rate of change of the radius when the balloon has a radius of \(4cm\). Organizing into variables, we have

\[\dfrac{dV}{dt}=3cm^3/s\]

\[r(t)=4cm\]

\[\dfrac{dr}{dt}=?\]

Step 3: Use equations to relate information in the problem

Based on the information we need, and the shape of the balloon, the volume equation of a sphere will be useful in this problem.

\[V=\dfrac{4}{3} \pi \cdot r^3\]

Step 4: Solve using implicit differentiation

Now that we have an equation, let's use implicit differentiation to get the equation in terms of two rates of change. We will take the derivative with respect to time.

\[\dfrac{d}{dt}[V]=\dfrac{d}{dt}\left(\dfrac{4}{3} \pi \cdot r^3 \right)\]

\[\dfrac{dV}{dt}=4 \cdot \pi \cdot r^2 \dfrac{dr}{dt}\]

Step 5: Substitute in known values

We want to find the rate of change of the radius:

\[\dfrac{dr}{dt}\]

We know that \(r=4cm\) and:

\[\dfrac{dV}{dt}=3cm^3/s\]

Plugging in our known values, we get

\[3=4\pi \cdot 4^2 \dfrac{dr}{dt}\]

\[\dfrac{dr}{dt} \approx 0.01492 cm/s \]

\[\dfrac{dr}{dt} \approx 0.015 cm/s \]

The positive sign in our answer signifies that the radius is growing larger in the positive direction.

Therefore, the radius expands at a rate of about \(0.015cm/s\). Clearly, the radius grows at a very slow rate. However, this makes sense considering the volume also grows at a relatively slow rate.