Some Applications of the Chain Rule

Now we can examine applications which involve more complicated functions.

Example 4: A ball at the end of a rubber band (Fig. 1) is oscillating up and down, and its height 

(in feet) above the floor at time $t$ seconds is $h(t)=5+2 \sin (t / 2)$. (t is in radians)

(a) How fast is the ball travelling after 2 seconds? after 4 seconds? after 60 seconds?

(b) Is the ball moving up or down after 2 seconds? after 4 seconds? after 60 seconds?

(c) Is the vertical velocity of the ball ever $0$?

Solution: (a) $v(t)=D(h(t))=D(5+2 \sin (t / 2))$

                $=2 \cos (t / 2) \mathbf{D}(t / 2)=\cos (t / 2)$ feet/second so

                $v(2)=\cos (2 / 2) \approx 0.540 \mathrm{ft} / \mathrm{s}, \mathrm{v}(4)=\cos (4 / 2) \approx-0.416 \mathrm{ft} / \mathrm{s}$, and

                $v(60)=\cos (60 / 2) \approx 0.154 \mathrm{ft} / \mathrm{s}$

(b) The ball is moving upward when $\mathrm{t}=2$ and 60 seconds, downward when $\mathrm{t}=4$.

(c) $v(t)=\cos (t / 2)$ and $\cos (t / 2)=0$ when $t=\pi \pm n \cdot 2 \pi \quad(n=1,2, \ldots)$.

Example 5: If 2400 people now have a disease, and the number of people with the disease appears to double every 3 years, then the number of people expected to have the disease in $\mathrm{t}$ years is $\mathrm{y}=2400 \cdot 2^{\mathrm{t} / 3}$.

(a) How many people are expected to have the disease in 2 years?

(b) When are 50,000 people expected to have the disease?

(c) How fast is the number of people with the disease expected to grow now and 2 years from now?

Solution: (a) In 2 years, $y=2400 \cdot 2^{2 / 3} \approx 3,810$ people.

(b) We know $\mathrm{y}=50,000$, and we need to solve $50,000=2400 \cdot 2^{\mathrm{t} / 3}$ for $\mathrm{t}$. Taking logarithms of each side of the equation, $\ln (50,000)=\ln \left(2400 \cdot 2^{t / 3}\right)=\ln (2400)+(t / 3) \cdot \ln (2)$ so $10.819=7.783+.231 \mathrm{t} \quad$ and $\mathrm{t} \approx$ 13.14 years. We expect 50,000 people to have the disease about 13.14 years from now.

(c) This is asking for $\mathrm{dy} / \mathrm{dt}$ when $\mathrm{t}=0$ and 2 years. $\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}\left(2400 \cdot 2^{\mathrm{t} / 3}\right)}{\mathrm{dt}}=2400 \cdot 2^{\mathrm{t} / 3} \cdot \ln (2) \cdot(1 / 3)$ $\approx 554.5 \cdot 2^{\mathrm{t} / 3}$. Now, at $\mathrm{t}=0$, the rate of growth of the disease is approximately $554.5 \cdot 2^{0} \approx 554.5$ people/year. In 2 years the rate of growth will be approximately $554.5 \cdot 2^{2 / 3} \approx 880$ people/year.

Example 6: You are riding in a balloon, and at time $\mathrm{t}$ (in minutes) you are $\mathrm{h}(\mathrm{t})=\mathrm{t}+\sin (\mathrm{t})$ feet high. If the temperature at an elevation $\mathrm{h}$ is $\mathrm{T}(\mathrm{h})=\frac{72}{1+\mathrm{h}} \quad$ degrees Fahrenheit, then how fast is your temperature changing when $\mathrm{t}=5$ minutes? (Fig. 2)

Solution: As $\mathrm{t}$ changes, your elevation will change, and, as your elevation changes, so will your temperature. It is not difficult to write the temperature as a function of time, and then we could calculate

$\frac{\mathrm{d} \mathrm{T}(\mathrm{t})}{\mathrm{dt}}=\mathrm{T}^{\prime}(\mathrm{t})$ and evaluate $\mathrm{T}^{\prime}(5)$, or we could use the Chain Rule:

$\frac{\mathbf{d} \mathrm{T}(\mathrm{t})}{\mathbf{d} \mathbf{t}}=\frac{\mathbf{d} \mathrm{T}(\mathrm{h}(\mathrm{t}))}{\mathbf{d} \mathrm{h}(\mathrm{t})} \cdot \frac{\mathrm{d} \mathrm{h}(\mathrm{t})}{\mathrm{dt}}=\frac{\mathbf{d} \mathrm{T}(\mathrm{h})}{\mathbf{d} \mathrm{h}} \cdot \frac{\mathrm{d} \mathrm{h}(\mathrm{t})}{\mathrm{dt}}=\frac{-72}{(1+\mathrm{h})^{2}} \cdot(1+\cos (\mathrm{t}))$

When $t=5$, then $h(t)=5+\sin (5) \approx 4.04$ so $T^{\prime}(5) \approx \frac{-72}{(1+4.04)^{2}} \cdot(1+.284) \approx-3.64^{\circ} /$ minute.

Practice 4: Write the temperature $\mathrm{T}$ in the previous example as a function of the variable $\mathrm{t}$ alone and then differentiate $\mathrm{T}$ to determine the value of $\mathrm{dT} / \mathrm{dt}$ when $\mathrm{t}=5$ minutes.

Example 7: A scientist has determined that, under optimum conditions, an initial population of 40 bacteria will grow "exponentially" to $\mathrm{f}(\mathrm{t})=40 \cdot \mathrm{e}^{\mathrm{t} / 5}$ bacteria after $\mathrm{t}$ hours.

(a) Graph $\mathrm{y}=\mathrm{f}(\mathrm{t})$ for $0 \leq \mathrm{t} \leq 15$. Calculate $\mathrm{f}(0), \mathrm{f}(5), \mathrm{f}(10)$.

(b) How fast is the population increasing at time $t$ ? (Find $f^{\prime}(t)$).

(c) Show that the rate of population increase, $f^{\prime}(t)$, is proportional to the population, $f(t)$, at any time t. $\quad$ (Show $\mathrm{f}^{\prime}(\mathrm{t})=\mathrm{K} \cdot \mathrm{f}(\mathrm{t})$ for some constant $\mathrm{K}$).

Solution: (a) The graph of $\mathrm{y}=\mathrm{f}(\mathrm{t})$ is given in Fig. 3. $\mathrm{f}(0)=40 \cdot \mathrm{e}^{0 / 5}=40$ bacteria. $\mathrm{f}(5)=40 \cdot \mathrm{e}^{5 / 5} \approx 109$ bacteria and $f(10)=40 \cdot \mathrm{e}^{10 / 5} \approx 296$ bacteria.

(b) $f^{\prime}(t)=\frac{\mathbf{d}}{d t}(f(t))=\frac{\mathbf{d}}{d t}\left(40 \cdot e^{\mathrm{t} / 5}\right)=40 \cdot \mathrm{e}^{\mathrm{t} / 5} \frac{\mathbf{d}}{\mathbf{d} t}(\mathrm{t} / 5)$

$=40 \cdot \mathrm{e}^{\mathrm{t} / 5}(1 / 5)=8 \cdot \mathrm{e}^{\mathrm{t} / 5}$ bacteria/hour.

(c) $\quad \mathrm{f}^{\prime}(\mathrm{t})=8 \cdot \mathrm{e}^{\mathrm{t} / 5}=\frac{1}{5} \cdot\left(40 \cdot \mathrm{e}^{\mathrm{t} / 5}\right)=\frac{1}{5} \mathrm{f}(\mathrm{t})$ so $\mathrm{f}^{\prime}(\mathrm{t})=\mathrm{K} \cdot \mathrm{f}(\mathrm{t})$ with $\mathrm{K}=1 / 5$.