MATH 111 Calculus I \(\quad\;\;\) Instructor: Difeng Cai

3.9 Related Rates

  1. Read the problem, identify and sketch the GEOMETRY.
  2. Set up an equation to connect quantities of interest. Depending on the geometry, for solids like cylinder/sphere/box, the equation could be the volume formula; for right-angled triangle/rectangle, the equation could be the area, the Pythagorean theorem \(x^2+y^2=z^2\), or trigs if angle is of interest.
  3. Differentiate (Chain Rule is often needed) both sides of the equation with respect to \(t\) and obtain an equation between the unknown rate and the given quantities
  4. Figure out and plug in the known quantities and solve for the unknown rate.

Example 1. A cylindrical tank with radius \(5\) \(m\) is being filled with water at a rate of \(2\) \(m^3/min\). How fast is the height of the water increasing?
Sol:

  1. Geometry: cylinder; Quantities: volume of water(cylinder): \(V,\;\) height of water: \(h,\;\) radius of cylinder: \(r=5\) \(m,\;\) time: \(t\).
  2. Find an equation to relate \(V\) and \(h\): \[V = \text{area of base} \times\text{height} = \pi r^2 h = 25\pi h\]
  3. Differentiate w.r.t. \(t\): \[\dfrac{dV}{dt}=25\pi \dfrac{dh}{dt}\]
  4. Given: \(\dfrac{dV}{dt}=2\) \(m^3/min\quad\) Unknown: \(\dfrac{dh}{dt}\)
  5. Plug in known rate and solve for the unknown rate: \[\dfrac{dh}{dt} = \frac{1}{25\pi}\dfrac{dV}{dt}=\frac{1}{25\pi}\cdot 2=\frac{2}{25\pi} \;m/min\]

Example 2. The radius of a sphere is increasing at a rate of \(4 \; mm/s\). How fast is the volume increasing when the diameter is \(40 \; mm\)?
Sol:

  1. Geometry: sphere; Quantities: radius \(r,\;\) volume of a sphere \(V,\;\) time \(t\)
  2. Find an equation to relate \(V\) and \(r\): \[V = \frac{4}{3}\pi r^3\]
  3. Differentiate w.r.t. \(t\) (chain rule is needed for \(r^3\)): \[\dfrac{dV}{dt}=\frac{4}{3}\pi \dfrac{d}{dt}(r^3) = \frac{4}{3}\pi \cdot 3r^2 \dfrac{dr}{dt} = 4\pi r^2\dfrac{dr}{dt}\]
  4. Given: \(\dfrac{dr}{dt}=4 \; mm/s\quad\) Need to find: \(\dfrac{dV}{dt}\) when \(r=\frac{40}{2}=20\)
  5. Plug in known values and solve for the unknown rate: \[\dfrac{dV}{dt}=4\pi \left(\frac{40}{2}\right)^2\cdot 4=6400\pi \; mm^3/s\]

Example 3. Two cars start moving from the same point. One travels south at \(60\) mi/h and the other travels west at \(25\) mi/h. At what rate is the distance between the cars increasing two hours later?
Sol:

  1. Geometry: right-angled triangle; Quantities: distance traveled by first car: \(x,\;\) distance traveled by second car: \(y,\;\) distance between two cars \(z\)
  2. Find an equation to relate \(x,y,z\): \[z^2 = x^2+y^2\]
  3. Differentiate w.r.t. \(t\): \[2z\cdot \dfrac{dz}{dt} = 2x\cdot \dfrac{dx}{dt} + 2y\cdot \dfrac{dy}{dt}\]
  4. Need to find: \(\dfrac{dz}{dt}\) two hours later (i.e., \(t=2\))
  5. Given: speed of first car: \(\dfrac{dx}{dt}=60,\;\) speed of second car: \(\dfrac{dy}{dt}=25\)
  6. Values of \(x,y,z\) ? Two hours later, \(x=60\cdot 2=120\), \(y=25\cdot 2=50\), \(z=\sqrt{x^2+y^2}=\sqrt{120^2+50^2}=130\)
  7. Plug in known values and solve for the unknown rate \(\dfrac{dz}{dt}\): \[ \dfrac{dz}{dt} = \frac{x}{z}\cdot \dfrac{dx}{dt} + \frac{y}{z}\cdot \dfrac{dy}{dt}\quad \text{according to the equation in 3.}\] so \[\dfrac{dz}{dt}= \frac{120}{130}\cdot 60+\frac{50}{130}\cdot 25=\frac{8450}{130}=65 \; mi/h\]

"You have now completed the Master It."