NOTE: You may view the original PDFs of the notes written in class by clicking on the link for the 1:30 Section or 2:30 Section, respectively.

Lecture 14: Related Rates (Perimeter/Area/Volume Relations)

Goal: If a quantity changes over time, explain how its derivative changes over time.

1 Introduction (Circles)

Problem 1.

A circle's area and radius are related by the equation:

\[ A = \pi r^2 \]

If \( A \) are \( r \) are changing as time advances, is there any relation between \( \frac{dA}{dt} \) and \( \frac{dr}{dt} \)?

That is, suppose we have a circle of radius \( r \):

Hand-drawn circle with a radius segment from center labeled r
Figure 1
A hand-drawn circle. A dot marks the center, and a straight line segment is drawn from the center outward to the circle, angled up and to the right. The segment is labeled with the letter r, indicating the radius of the circle.

Then if we let a little time pass, there will be a change in the radius. As a result, there will be a change to the area.

Circle of radius r surrounded by a shaded annulus labeled dr/dt and dA/dt
Figure 2
A hand-drawn diagram showing a black circle with a dot at its center and a line segment from the center to the edge labeled r. Surrounding this black circle is a slightly larger irregular blue outline, and the ring-shaped region between the black circle and the blue outline is filled with blue hatching lines. A short red segment spans the gap from the black circle to the blue outline on the upper right, labeled in red with \( \frac{dr}{dt} \). To the right of the hatched ring, a blue label \( \frac{dA}{dt} \) points to the shaded annular region, indicating the rate of change of area corresponding to the rate of change of radius.

To describe \( \frac{dA}{dt} \), we recognize \( A \) and \( r \) are functions of time and differentiate both sides of the equation \( A = \pi r^2 \)

\[ \begin{aligned} A &= \pi r^2 \\ A_{(t)} &= \pi \left[ r_{(t)} \right]^2 \\ \frac{d}{dt}\left( A_{(t)} \right) &= \frac{d}{dt}\left( \pi \left[ r_{(t)} \right]^2 \right) \\ \frac{dA}{dt} &= \pi \left( 2 \left[ r_{(t)} \right] \frac{dr}{dt} \right) \\ \frac{dA}{dt} &= 2\pi r \frac{dr}{dt} \end{aligned} \]

Notice that \( \frac{dA}{dt} \) is a function of both \( \frac{dr}{dt} \) (the change in the radius) and \( r \) (the radius itself).

Small circle of radius r inside a much larger shaded region showing a big change in area
Figure 3
A hand-drawn sketch in black and red ink. A small black circle with its center marked by a dot and a short radius segment labeled r sits near the middle. Surrounding it is a much larger, roughly circular red outline, and the wide ring-shaped region between the small black circle and the large red outline is filled with red hatching lines. The sketch illustrates that when \( \frac{dr}{dt} \) is large, the resulting change in area is large. A large bracket to the right of the drawing groups it with the accompanying note.

\( \frac{dr}{dt} \) being large causes a large change in area

Large circle of radius r with a thin shaded ring around it
Figure 4
A hand-drawn sketch in black and red ink. A large black circle is drawn with its center marked by a dot and a long radius segment labeled r extending to the upper right of the circle. Just outside the black circle is a slightly larger red outline, forming a thin ring that is filled with short red hatch marks all the way around. The thin ring represents a small change in radius, yet because the radius itself is large the added area is substantial. A large bracket to the right of the drawing groups it with the accompanying note.

Even if \( \frac{dr}{dt} \) is small, if the radius is large, it will cause a large change in area.

Problem 2.

(a) In the previous example, if \( r = 2 \) and \( \frac{dr}{dt} = 3 \), then what is the value of \( \frac{dA}{dt} \)? Interpret.

Given

\[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \]

we plug in the appropriate values

\[ \begin{gathered} \left. \frac{dA}{dt} \right|_{\substack{r = 2 \\ r' = 3}} = 2\pi (2)(3) \\[0.5em] = 12\pi \end{gathered} \]

Therefore, when the radius is 2 units long, the area is increasing at a rate of \( 12\pi \) units\( ^2 \)/sec.

Circle of radius 2 surrounded by a large shaded region labeled 3
Figure 5
A hand-drawn diagram showing a black circle with a dot at its center and a radius segment labeled 2. Surrounding the circle is a much larger, roughly circular red outline, and the ring-shaped region between the black circle and the red outline is filled with red diagonal hatching. The width of this ring is labeled 3, indicating the rate of change of the radius. The large hatched area illustrates the resulting change in area of \( 12\pi \) square units per second.

(b) In the previous example, if \( r = 1 \) and \( \frac{dA}{dt} = 2\pi \), then what is the value of \( \frac{dr}{dt} \)?

METHOD 1:

\[ \begin{gathered} \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \\[0.5em] \frac{dr}{dt} = \frac{1}{2\pi r} \frac{dA}{dt} \end{gathered} \]

Thus,

\[ \begin{gathered} \left. \frac{dr}{dt} \right|_{\substack{r = 1 \\ A' = 2\pi}} = \frac{1}{2\pi (1)} (2\pi) \\[0.5em] = 1 \end{gathered} \]

METHOD 2:

At this point, you can plug in the numbers and solve for \( \frac{dr}{dt} \).

\[ \begin{aligned} \frac{dA}{dt} &= 2\pi r \frac{dr}{dt} \\ (2\pi) &= 2\pi (1) \frac{dr}{dt} \\ 2\pi &= 2\pi \frac{dr}{dt} \\ \frac{dr}{dt} &= 1 \end{aligned} \]

Therefore, if the radius is 1 unit long, in order for the area to change at a rate of \( 2\pi \) units\( ^2 \)/sec, the radius will need to change at a rate of \( 1 \frac{\text{unit}}{\text{sec}} \).

Problem 3.

The radius of a circle r is increasing at a constant rate 3 cm/min.

(a) Find the rate of change of the area of the circle (\( A \)) when the radius is 5cm.

We want to translate from English to the language of math. To aid with this, it is a good idea to draw a picture if you can. So, in this case a circle of radius \( r \):

Circle with a line segment drawn from its center to the edge, labeled r
Figure 6
A hand-drawn circle with a dot at its center. A short line segment is drawn from the center dot outward to the upper-right portion of the circle's edge, and this segment is labeled with the letter r, indicating the radius of the circle.

Now, we need to determine the following from the word problem:

  1. What are the quantities in play?
    • We see the circle's radius and area mentioned. So, let \( r \) and \( A \) be the radius and area respectively.
  2. Is there a formula that relates all of our quantities?
    • Yes, namely, \( A = \pi r^2 \)
  1. What do we know?
    • "r is increasing at a constant rate 3 \( \frac{\text{cm}}{\text{min}} \)." Ie, \( \frac{dr}{dt} = 3 \).
  2. What do we need to know?
    • "the rate of change of the area... when the radius is 5." Ie, we need to know \( \frac{dA}{dt} \) when \( r = 5 \).

Now, we use the relation is (2) to find \( \frac{dA}{dt} \):

\[ \begin{gathered} A = \pi r^2 \\[0.5em] \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \end{gathered} \]

Now, when \( r = 5 \), \( \frac{dr}{dt} = 3 \), so we have enough information to find \( \frac{dA}{dt} \) when \( r = 5 \).

\[ \begin{aligned} \frac{dA}{dt} &= 2\pi(5)(3) \\ \frac{dA}{dt} &= 30\pi \end{aligned} \]

If we want to be prim and proper, our final step is to translate back into English. So, our final answer is something like:

"When the radius is 5 cm, the area of the circle is increasing at a rate of \( 30\pi \) cm\( ^2 \)/min."

(b) Find the rate of change of the circumference of the circle (\( C \)) when the radius is 5cm.

Circle with a segment drawn from its center to the edge, labeled r
Figure 7
A hand-drawn circle with a dot marking its center. A straight line segment is drawn from the center dot outward to a point on the circle, and the segment is labeled r, representing the radius.

Run through the checklist:

  1. The circle's radius and circumference are mentioned. Let \( r \) and \( C \) denote the radius and circumference respectively.
  2. Formula: \( C = 2\pi r \)
  3. What do we need to know? We NTK \( \frac{dC}{dt} \) when \( r = 5 \)
  4. What do we know? \( \frac{dr}{dt} = 3 \)

NOTE:

  • A common strategy is figuring out what we need to find (Ie, answer (4)), then work backwards. Hence, why the list is out of order
  • Your quantities may go in (3) or (4) depending on preference. For example, I say "we need to know \( \frac{dC}{dt} \) when \( r = 5 \)," but it is also valid to say "when we know \( r = 5 \), we need to know \( \frac{dC}{dt} \)."

Now,

\[ \begin{aligned} C &= 2\pi r \\ \frac{dC}{dt} &= 2\pi \frac{dr}{dt} \\ \left.\frac{dC}{dt}\right]_{r=5} &= 2\pi (3) \\ &= 6\pi \end{aligned} \]

When the radius is 5 cm, the circumference is increasing at a rate of \( 6\pi \) \( \frac{cm}{min} \).

2 Rectangular Prisms

Problem 4.

The edges of a cube are shrinking at a rate of \( 10 \) cm/s.

(a) How fast is the volume (\( V \)) shrinking when each side length is 9cm long?

Sketch of a cube in perspective with the bottom front edge labeled x
Figure 8
A freehand perspective sketch of a cube (rectangular prism with equal sides), drawn with visible front face and receding top and side faces, including interior edges showing the back corner. Below the bottom front edge a short segment is marked with a label \( x \), indicating the side length of the cube.

NOTE:

I will abbreviate "What do we know?" with "K" and "What do we need to know?" with "NTK" from now on.

Run through the check list:

  1. The box's side length and Volume
    • \( x = \) Length of the side
    • \( V = \) Volume
  2. \( V = x^3 \)
  3. K: \( \frac{dx}{dt} = -10 \)
  4. NTK: \( \frac{dV}{dt} \) when \( x = 9 \)

So,

\[ \begin{aligned} V &= x^3 \\ \frac{dV}{dt} &= 3x^2 \frac{dx}{dt} \end{aligned} \]

NOTE:

In general, you need to wait to plug in numbers until you are done taking derivatives. Otherwise, you may not get the correct expression for \( \frac{dV}{dt} \). Plugging in numbers too early is a very common mistake.

\[ \begin{aligned} \left.\frac{dV}{dt}\right]_{\substack{x=9 \\ x'=-10}} &= 3(9)^2(-10) \\ \left.\frac{dV}{dt}\right]_{\substack{x=9 \\ x'=-10}} &= -2430 \end{aligned} \]

Therefore, when the side lengths are 9 cm long, the volume is decreasing at a rate of \( 2430 \) \( \frac{cm^3}{s} \).

(b) How fast is the surface area (\( A \)) shrinking when each side length is 9cm long?

Run through the checklist:

  1. If \( x \) is as in Part (a), let \( A \) denote the surface area of the cube.
  2. Formula: \( A = 6x^2 \)
  3. K: \( \frac{dx}{dt} = -10 \)
  4. NTK: \( \frac{dA}{dt} \) when \( x = 9 \)
\[ \begin{aligned} A &= 6x^2 \\ \frac{dA}{dt} &= 12x \frac{dx}{dt} \\ \left.\frac{dA}{dt}\right]_{\substack{x=9 \\ x'=-10}} &= 12(9)(-10) \\ \left.\frac{dA}{dt}\right]_{\substack{x=9 \\ x'=-10}} &= -1080 \end{aligned} \]

When the side lengths of the cube are 9 cm long, the surface area of the cube is decreasing at a rate of \( 1080 \) \( \frac{cm^2}{s} \).

3 Spheres

Problem 5.

A balloon is (roughly) a sphere. The balloon deflates and its radius decreases at a rate of 2 cm/s.

(a) How fast is the volume (\( V \)) shrinking when the radius is 5cm long?

Sketch of a sphere with a dashed equatorial circle and a radius segment labeled r from the center.
Figure 9
A freehand drawing of a large circle representing a sphere. An ellipse drawn across the middle of the circle represents the equator; its back half is dashed and its front half is a solid curve, giving a three-dimensional appearance. A dot marks the center of the sphere, and a straight line segment with an arrowhead extends from that center dot up and to the right to the sphere surface. The segment is labeled with the letter r, denoting the radius of the balloon.

Run through the checklist:

  1. The balloon's radius and volume are mentioned. Denote them with \( r \) and \( V \) respectively.
  2. Formula: \( V = \frac{4}{3}\pi r^3 \)
  3. K: \( \frac{dr}{dt} = -2 \)
  4. NTK: \( \frac{dV}{dt} \) when \( r = 5 \)
\[ \begin{aligned} V &= \frac{4}{3}\pi r^3 \\ \frac{dV}{dt} &= 4\pi r^2 \frac{dr}{dt} \\ \left.\frac{dV}{dt}\right]_{\substack{r=5 \\ r'=-2}} &= 4\pi (5)^2 (-2) \\ \left.\frac{dV}{dt}\right]_{\substack{r=5 \\ r'=2}} &= -200\pi \end{aligned} \]

When its radius is 5 cm, the balloon's volume is decreasing at a rate of \( 200 \pi\) \( \frac{cm^3}{s} \).

(b) How fast is the surface area (\( A \)) shrinking when the radius is 5cm long?

Run through the checklist:

  1. \( r \) is as in Part (a). Let \( A \) denote the surface area of the balloon.
  2. Formula: \( A = 4\pi r^2 \)
  3. K: \( \frac{dr}{dt} = -2 \)
  4. NTK: \( \frac{dA}{dt} \) when \( r = 5 \)
\[ \begin{aligned} A &= 4\pi r^2 \\ \frac{dA}{dt} &= 8\pi r \frac{dr}{dt} \\ \left.\frac{dA}{dt}\right]_{\substack{r=5 \\ r'=-2}} &= 8\pi (5)(-2) \\ \left.\frac{dA}{dt}\right]_{\substack{r=5 \\ r'=-2}} &= -80\pi \end{aligned} \]

When the radius of the balloon is 5 cm long, its surface area is decreasing at a rate of \( 80\pi \) \( \frac{\mathrm{cm}^2}{\mathrm{s}} \).

4 Cylinders

Problem 6.

A cylindrical tank with a radius and height of 100 cm stands upright. Water is being drained at a rate of \( 7cm^3/s \). How fast is the water level changing when the tank is half empty?

Sketch of an upright cylindrical tank, height and diameter labeled 100, with water surface and depth labeled h.
Figure 10
A hand-drawn sketch of an upright cylinder representing a water tank. A vertical double-headed arrow on the left side labeled 100 marks the full height of the tank. The bottom ellipse has a horizontal segment across it labeled 100, marking the diameter. Inside the cylinder, a wavy line drawn part way up represents the surface of the water. A vertical segment runs from the water surface down to the bottom of the tank and is labeled h, denoting the water level.

Run through the check list

  1. Let \( h \) denote the "water level" and \( V \) denote the amount of water in the tank.
  2. Formula: \( V = \pi (100)^2 h = 10000\pi h \)
  3. K: The radius and height are a constant 100 cm. We also know \( \frac{dV}{dt} = -7 \)
  4. NTK: \( \frac{dh}{dt} \) when \( h = \frac{1}{2}(100) = 50 \)

NOTE:

Take a moment to verify why we can plug in 100 for the tank's height and radius now as opposed to after we take derivatives. They are truly constant (not changing with time) as opposed to the height and volume of the water.

\[ \begin{aligned} V &= 1000\pi h \\ \frac{dV}{dt} &= 1000\pi \frac{dh}{dt} \\ \frac{dh}{dt} &= \frac{1}{1000\pi} \frac{dV}{dt} \end{aligned} \]

When \( h = 50 \), \( \frac{dV}{dt} = -7 \), so

\[ \begin{aligned} \left.\frac{dh}{dt}\right]_{h=50} &= \frac{1}{1000\pi}(-7) \\ \left.\frac{dh}{dt}\right]_{h=50} &= \frac{-7}{10000\pi} \ \frac{\mathrm{cm}}{\mathrm{s}} \\ \left.\frac{dh}{dt}\right]_{h=50} &\approx -2.2 \ \frac{\mathrm{mm}}{\mathrm{s}} \end{aligned} \]

So, when the tank is half way full, the water level is decreasing at a rate of roughly 2.2 \( \frac{\mathrm{mm}}{\mathrm{s}} \).

5 Cones

Problem 7.

Sand pours onto a surface at \( 15cm^3/s \), forming a conical pile with a base diameter that is always equal to the pile's altitude. How fast is the altitude of the pile increasing when the pile is 8cm high?

Sketch of a cone with dashed central altitude labeled h and base radius labeled r.

Run through the check list:

  1. Let \( V \) denote the volume of the pile, along with \( h \) and \( r \) being the height and radius respectively.
  2. Formula: \( V = \frac{\pi}{3} r^2 h \)
  3. K: \( \frac{dV}{dt} = 15 \), also \( 2r = h \).
  4. NTK: \( \frac{dh}{dt} \) when \( h = 8 \)

NOTE:

We know nothing about \( \frac{dr}{dt} \), so to solve the problem we need to rewrite \( V \) solely as a function of \( h \). This is why need the relation \( 2r = h \).

\[ \begin{aligned} V &= \frac{\pi}{3} \pi r^2 h \\ V &= \frac{\pi}{3} \pi \left(\frac{h}{2}\right)^2 h \end{aligned} \]\[ \begin{aligned} V &= \frac{\pi}{12} h^3 \\ \frac{dV}{dt} &= \frac{\pi}{4} h^2 \frac{dh}{dt} \\ \frac{dh}{dt} &= \frac{4}{\pi h^2} \frac{dV}{dt} \\ \left.\frac{dh}{dt}\right]_{h=8} &= \frac{4}{\pi (8)^2} (15) \\ \left.\frac{dh}{dt}\right]_{h=8} &= \frac{60}{64\pi} \\ \left.\frac{dh}{dt}\right]_{h=8} &= \frac{15}{16\pi} \end{aligned} \]

Therefore, when the pile is 8cm tall, the height is increasing at a rate of \( \frac{15}{16\pi} \frac{\mathrm{cm}}{\mathrm{s}} \).