Applied Calculus

Start by finding the Profit function, then find the implicit time derivative.

\(\displaystyle \frac{dP}{dt}=12000\) dollars per day.

We begin by creating the profit function

Then take the implicit time derivative of the profit function as an implicit equation

Then plug in the values that are given, \(x=40\) and \(\displaystyle \frac{dx}{dt}=30:\)

1. Remember that we need to apply the chain rule to the right side of the equation to differentiate \(r\) with respect to \(t\text{.}\)

2. Look at the units.

3. Use your answer to part (b) to help fill in all missing information. Only one variable should be left.

4. Which radius results in a larger rate? Note that this question asks about the rate of change of the radius, not of the volume.

1. \(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.\)

2. \(\frac{dV}{dt}=20\) cubic inches per second.

3. \(20=4\pi (6)^2 \frac{dr}{dt},\) so \(\frac{dr}{dt}=\frac{5}{36\pi}\) inches per second when the diameter is 12 inches.

4. \(\frac{dr}{dt}=\frac{5}{64\pi}\) inches per second when the diameter is 16 inches. Thus, the radius is changing more rapidly when the diameter is 12 inches.

1. Starting with the equation \(V=\frac{4}{3}\pi r^3\text{,}\) we differentiate both sides with respect to \(t\text{.}\) We need to use the chain rule to differentiate the right side of the equation since \(r\) is really \(r(t)\text{,}\) a function of \(t\text{.}\) So, the equation we have is really \(V(t)=\frac{4}{3}\pi [r(t)]^3\text{.}\) Differentiating with respect to \(t\) we have

   \begin{equation*} \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}. \end{equation*}

2. Since the units are cubic inches per second, this is a volume rate. So, \(\frac{dV}{dt}=20\) cubic inches per second.

3. Since \(\frac{dV}{dt}=20\) and \(r=6\) when \(d=12\text{,}\) we have \(20=4\pi(6)^2\frac{dr}{dt}\text{.}\) Solving this equation for \(\frac{dr}{dt}\text{,}\) we have \(\frac{dr}{dt}=\frac{5}{36\pi}\) inches per second.

4. Doing the same thing as in part (c), we find that \(\frac{dr}{dt}=\frac{5}{64\pi}\) inches per second when the diameter is \(16\) inches (the only difference in our equation here is that \(r=8\)). This number is smaller than the one calculated in part (c), so the radius is changing more rapidly when the diameter is \(12\) inches. Intuitively, this makes sense since the balloon's volume is growing at a constant rate: as the balloon grows, a small change in the radius will have a larger impact on the change in volume; equivalently, the same change in volume corresponds to a smaller change in the radius when the balloon is large.

1. As the base of the ladder moves away from the wall at a constant rate, the top of the ladder slides down the wall with a constant rate.

2. \(\frac{dh}{dt}\) is negative.

3. The base is moving away from the wall at a rate of \(\frac{dx}{dt}=\frac{24}{\sqrt{112}}\) feet per second.

1. As the base of the ladder moves away from the wall at a constant rate, the top of the ladder slides down the wall with a constant rate.

2. The rate of change of the height of the ladder on the wall would be negative, that is, \(\frac{dh}{dt} \lt 0\text{.}\) This is negative since the distance from the bottom of the wall to the top of the ladder, \(h \text{,}\) is decreasing.

3. Here we want \(\frac{dx}{dt}\text{,}\) how fast the base of the ladder is moving away from the wall. We know \(h=12\) and \(\frac{dh}{dt}=-4\text{.}\)

   We will use the Pythagorean Theorem, which says \(x^2+h^2=16^2\text{,}\) to compute the length of the missing side \(x\) by plugging in \(h=12:\)

   \begin{equation*} x^2+12^2=16^2 \implies x^2=16^2-12^2 \implies x=\sqrt{16^2-12^2}=\sqrt{112} \end{equation*}

   Then we take the implicit time derivative

   \begin{equation*} \frac{d}{dt}\left[x^2+h^2=16^2\right] \end{equation*}
   \begin{equation*} 2x\frac{dx}{dt}+2h\frac{dh}{dt}=0. \end{equation*}

   Finally, substitute in what is known and solve for \(\displaystyle \frac{dx}{dt}\)

   \begin{equation*} 2(\sqrt{112})\frac{dx}{dt}+2(12)(-4)=0 \implies \frac{dx}{dt}=\frac{24}{\sqrt{112}}. \end{equation*}

   Thus the base is moving away from the wall at a rate of \(\displaystyle \frac{dx}{dt}=\frac{24}{\sqrt{112}}\) feet per second.

1. The rate of change of the distance distance from the person to the boat is positive, i.e. \(\frac{dy}{dt} > 0\text{.}\)

2. The distance from the person to the boat is increasing at a rate of \(\frac{dy}{dt}=\frac{12}{\sqrt{13}}\) feet per minute.

1. The rate of change of the distance from the person to the boat is positive, that is, \(\frac{dy}{dt}\gt 0\text{,}\) since the distance is increasing as the boat sails away.

2. To find how fast the distance \(y \) is changing, we need to compute \(\frac{dy}{dt}\text{.}\) We know \(y=130\) and \(\frac{dx}{dt}=10\) feet per minute.

   We will use the Pythagorean Theorem, which tells us that \(x^2+50^2=y^2\text{.}\) First we find length of the missing side \(x\) by plugging in \(y=12\text{:}\)

   \begin{equation*} x^2+50^2=130^2 \implies x^2=130^2-50^2 \implies x=\sqrt{130^2-50^2}=120. \end{equation*}

   Then we take the implicit time derivative, noting that both \(x \) and \(y \) are changing with time:

   \begin{equation*} \frac{d}{dt}\left[x^2+50^2=y^2\right] \end{equation*}
   \begin{equation*} 2x\frac{dx}{dt}+0=2y\frac{dy}{dt}. \end{equation*}

   Finally, we substitute in what is known and solve for \(\frac{dy}{dt}\text{:}\)

   \begin{equation*} 2(120)(10)=2(130)\frac{dy}{dt} \implies \frac{dy}{dt}=\frac{120}{\sqrt{130}}. \end{equation*}

   Therefore, the distance from the person to the boat is increasing at a rate of \(\frac{dy}{dt}=\frac{12}{\sqrt{13}}\) feet per minute.

Draw some pictures! Label everything you know with a constant, and everything that is changing with a variable and a corresponding rate of change (with respect to time, \(t\)). For the instant in question, you will also want to write down the values of the variables at that instant and identify what quantities you need to find. Note that the basepaths meet at right angles.

Let \(x\) denote the position of the ball (as a distance from home plate) at time \(t\) and \(z\) the distance from the ball to first base, as pictured below.

Let \(r\) be the runner's position (as a distance from home plate) at time \(t\) and let \(s\) be the distance between the runner and the ball, as pictured.

We let \(x\) denote the position of the ball (as a distance from home plate) at time \(t\) and \(z\) the distance from the ball to first base, as pictured below.

By the Pythagorean Theorem, we know that \(x^2 + 90^2 = z^2\text{;}\) differentiating with respect to \(t\text{,}\) we have

At the instant the ball is halfway to third base, we know \(x = 45\) and \(\left. \frac{dx}{dt} \right|_{x = 45} = 100\text{.}\) Moreover, by the Pythagorean Theorem, \(z^2 = 90^2 + 45^2\text{,}\) so \(z = 45\sqrt{5}\text{.}\) We can solve for \(\frac{dz}{dt} \) using the equation above and then plug in all of the known variables:

Therefore, the distance between the ball and first base is at this instant growing at a rate of about \(44.72\) feet per second.

For the second question, we still let \(x\) represent the ball's position at time \(t\text{,}\) but now we introduce \(r\) as the runner's position (also as a distance from home plate) at time \(t\) and let \(s\) be the distance between the runner and the ball. In this setting, as seen in the diagram below, \(x\text{,}\) \(r\text{,}\) and \(s\) form the sides of a right triangle, so that

by the Pythagorean Theorem.

Differentiating each side of Equation(4.3) with respect to \(t\text{,}\) it follows that the three rates of change are related by the equation

We are given that at the instant \(x = 45\text{,}\) \(r = \frac{90}{8}=\frac{45}4\text{,}\) so solving for \(s \) in the equation above, we have \(s = \frac{45}{4}\sqrt{17}\text{.}\) In addition, at this same instant we know that \(\left. \frac{dx}{dt} \right|_{x = 45} = 100\) and \(\left. \frac{dr}{dt} \right|_{x = 45} = 30\text{.}\) Solving for \(\frac{ds}{dt} \) and applying this information, we end up with

Therefore, the distance between the ball and the runner is at this instant increasing at a rate of about \(104.29\) feet per second.