Example 2

  • 3.9.12.
    • A particle is moving along a hyperbola \(xy=8\). As it reaches the point \((4,2)\), the \(y\)-coordinate is decreasing at a rate of 3 cm/s. How fast is the \(x\)-coordinate of the point changing at that instant?
    • Here, the expression relating \(x\) and \(y\) is already given to us, so we'll have to derivate it implicitly.
    • What you need to look out for is that the variable with respect to which we're differentiating is \(t\), the time.
    • Therefore implicit differentiation yields the following. \[ x'y+xy'=0 \]
    • Substituting the known numerical values, we get the following. \[ x'\cdot2+4\cdot(-3)=0. \]
    • This yields \(x'=6\) cm/s.

Example 3

  • 3.9.14.
    • If a snowball melts so that its surface area decreases at a rate of \(1\,\mathrm{cm}^2/\mathrm{min}\), find the rate at which the diameter decreases when the diameter is 10 cm.
    • The area of a sphere is \(A=4r^2\pi\). But we need an equation with the area \(A\), and the diameter \(d=2r\).
    • Therefore, we want to use \[ A=4(d/2)^2\pi=d^2\pi. \]
    • Implicit differentiation gives \[ A'=2dd'\pi \]
    • Substituting the given numerical values, we get \[ -1=2\cdot10\cdot d'\cdot\pi \]
    • Therefore, we get \(d'=-\frac{1}{20\pi}\) cm/min.
  • Exercises. 3.9: 4, 6, 8, 16, 18, 22, 24, 30, 38