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Applications of the derivative

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This is a part of "Related rates".

Air balloon

Suppose we have: an air balloon, spherical in shape. Air is pumped in it. The rate: $5 {}^{\text{in }^{3}}/_{\text{sec}}$ of air.

Question: That's the volume's rate of growth, what is the rate of growth of the radius?

Observation: The rate of growth will slow down.

Solution: Step 1 in word problems: introduce variables. Let $x$ be time, $V$ volume.

The volume depends on $x$: $\frac{dv}{dx} = 5$, from above. This is a sphere, so $V = \frac{4}{3}\pi r^{3}$, $r$ is the radius.

Dependence: $x \to V \gets r$

Question: Find $\frac{dr}{dx}$.

We could reverse the arrow: to find this, $r = \sqrt[3]{\frac{3}{4\pi}V}$. OK, but the differentiation will be HARD!

Instead, we differentiate the equation itself. Thus, if two variables are related (via an equation), then so are their derivatives == "rates of change". (Hence, "related rates".)

Relation : $V= \frac{4}{3} \pi r^{3}.$

Here both V and r are functions of x. Differentiate them now.

LHS, very simple: $$\frac{d}{dx}V=\frac{dV}{dx}.$$

RHS: $r^{3}$ is a composition! So use Chain Rule: $$\frac{d}{dx}(\frac{4}{3} \pi r^{3}) = \frac{4}{3} \pi \cdot 3r^{2} \frac{dr}{dx}.$$ Thus $$\frac{dV}{dx} = \frac{4}{3} \pi \cdot 3r^{2} \frac{dr}{dx}.$$

Recall the rate of change of $V$ is 5, so: $$5 = 4\pi r^{2} \frac{dr}{dx},$$ or $$\frac{dr}{dx} = \frac{5}{4\pi r^{2}}.$$

Next, what's the rate of growth of $r$ when $r = 1, r = 2, r = 3$?

Answer: For $$\begin{alignat}{2} r & = 1 &\quad \frac{dr}{dx} & = \frac{5}{4\pi} \\ r & = 2 &\quad \frac{dr}{dx} & = \frac{5}{16\pi} \\ r & = 3 &\quad \frac{dr}{dx} & = \frac{5}{36\pi} \end{alignat} $$ So, indeed there is a slow-down.

The key is implicit differentiation, via the Chain Rule. This is the dependence: $$ x \to r \to y = r^{3}.$$ Derivatives: $\frac{dr}{dx}$ and $\frac{dy}{dr}=3r^{3}$. Using CR $$ \frac{dy}{dx} = \frac{dr}{dx}\cdot 3r^{2}, $$ or $$\frac{d}{dx} (r^{3}) = \frac{dr}{dx} 3r^{2}.$$

Use this idea to find derivatives of inverses. $$f(x) = e^{x}, f^{-1}(y) = \ln y,$$ satisfy the "canceling rule": $$ x \to e^{x} \overset{y}{\to} \ln y \overset{x}{\to} $$ or $$e^{\ln x} = x $$ Differentiate this (implicitly) to find the derivative of the logarithm.

Rules of differentiation are directly associated with algebraic operations:

Infection Spread

This models spread of number or infection $$\lim_{t \to \infty} \frac{1}{1 + a e^{–kt}}, \quad a,k > 0$$

Exponential Growth.

Simplifying $$\begin{aligned} \lim_{t \to \infty} \frac{1}{1 + a e^{–kt}} & = \frac{1}{\lim_{t \to \infty}(1 + ae^{–kt})}, \quad \text{QR} \\ & = \frac{1}{1 + \lim_{t \to \infty} a e^{-kt}} \quad \text{SR} \\ & = \frac{1}{1 + a \lim_{t \to \infty} e^{-kt}} \quad \text{CMR} \\ & = \frac{1}{1 + a\cdot 0} \\ & = 1 \end{aligned}$$ Here we used a limit $\lim\limits_{t \to \infty} e^{-kt}$ of a specific function: as $t \to \infty, -kt \to \infty,, x = -kt \to -\infty$.

Conclusion: eventually 100% are infected.

Next, find $$\left(\frac{1}{1 + ae^{–kt}}\right)^{\prime}$$ We need $(e^{-kt})^{\prime} = -ke^{–kt}$, then use the rules: $$\begin{aligned} \overset{\text{QR}}{=} \frac{0 \cdot (~~) - 1 \cdot (1 + ae^{-kt})^{\prime}}{(1 + ae^{-kt})^{2}} & \overset{\text{SR}}{=} - \frac{(1)^{\prime} + (ae^{-kt})^{\prime}}{(a + ae^{-kt})^{2}} \\ & \overset{\text{CMR}}{=} - \frac{a(e^{-kt})^{\prime}}{(1 + ae^{-kt})^{2}} \\ & = - \frac{a(-ke^{-kt})}{(1 + ae^{-kt})^{2}} \\ & = \frac{ake^{-kt}}{(1 + ae^{-kt})^{2}} \to 0 \end{aligned}$$ as $t \to \infty$. We used the limit from part (a) to compute the limits of the numerator and denominator.

Conclusion: the graph flattens.

Indeed: Infection spread.png