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# 1 Chapter 3

## 1.1 Rates of Change

The Example $x$ is time, $y(x)$ is the position.

1. Horizontal Motion
Motion along straight line

$\frac{dy}{dx}=f^{\prime}(x)$ is the velocity.
2. Vertical (altitude)
1. Typically, know the position (from odometer), find velocity.
2. Also, know the velocity (from speedometer), find the position.

### 1.1.1 Examples

$f^{\prime}(x) = 2x$, what is $f$? \begin{aligned} f(x) &= x^{2} \\ f(x) &= x^{2} + C \end{aligned} (antidifferentiation).

Other Examples:

• Exponential growth/decay: Different but $x$ is still time.
• Linear Density

Given a metal rod.

Uniform Pure Metal Bar

Define: $\text{linear density} = \frac{\text{mass}}{\text{length}}$.

Technically, the average density. What if the rod isn't uniform, like an alloy? For example, suppose we have two pieces of different metals, the we melt them together (no stir!)

Non-Uniform Metal Bar
Suppose the densities are 1 lbs/in and 2 lbs/in. Then the density of this alloy will gradually vary from 1 to 2.

Instead, consider the mass function where $x$ is the location. $y=m(x)$ is the mass of the rod from $O$ to $x$. Consider how $m$ is changing:

Mass function with respect to $x$.

Take a small piece of the rod, $\Delta x$ long. Then $$\text{Average Density} = \frac{\text{Mass}}{\text{Length}} = \frac{\Delta m}{\Delta x} = \frac{m(x + \Delta x) - m(x)}{\Delta x}$$ So, $$\text{linear density} = \lim_{\Delta x \to 0} \frac{\Delta m}{\Delta x} = m^{\prime}(x)$$

### 1.1.2 Bar of Non-Uniform Density

1 lbs/in & 2 lbs/in. What is the total mass?

Density Distribution of Bar

$$\text{Mass} = \text{Average Density} \cdot \text{Length} = 1.5 \cdot 2 = 3$$

## 1.2 Exponential Models

Suppose we have a population of bacteria that doubles every day. For example, each divied in half everyday. Let $y$ be the number of bacteria, $x$ is time in days. Then $$y = f(x)$$ Re-write the above $$\underbrace{\text{population at time } x+1}_{f(x + 1)} = \underbrace{2f(x)}_{\text{time, } x}$$ To know $f(x)$ for all $x$, we need to know $f(0)$. But not good for calculus with incremental increase. So we create a continuous model: the rate of growth is proportional to the size of population.

\begin{alignat}{3} & \text{time, } x &\quad &\text{time, } x+1 & & \\ y & = 10 & \quad y &= 20 &\quad \Delta y &= 10 \\ y & =100 & \quad y&= 200 &\quad \Delta y &= 100 \end{alignat} If triples: \begin{alignat}{3} & \text{time, } x &\quad &\text{time, } x+1 & & \\ y & = 10 & \quad y &= 30 &\quad \Delta y &= 20 \\ y & =100 & \quad y&= 300 &\quad \Delta y &= 200 \end{alignat} We can restate this in calculus terms: The derivative of $y$ is proportional to $y$: $$\frac{dy}{dx} = ky$$ this is called a differential equation, where $k$ is the growth rate. This is the solution: $$y = Ce^{kx}$$ for any $C$.

Verify: \begin{aligned} \frac{dy}{dx} &= (Ce^{kx})^{\prime} \\ \text{CMR} &= C(e^{kx})^{\prime} \end{aligned}

Right Hand Side \begin{aligned} \text{CR} & = C \cdot e^{kx} \cdot (kx)^{\prime} \\ & = Ce^{kx}\cdot k \end{aligned}

Left Hand Side $$ky = k \cdot e^{kx}$$

What is $C$? Given $$y = Ce^{kx}$$ subsitute $x=0$. Then $$y(0) = Ce^{k\cdot 0} = ce^{0} = C$$ So, $C$ is the initial population. Re-phrase $$y(x) = y(0)e^{kx}$$

Growth
Know if $k > 0$, then this function is increasing and $\lim\limits_{x \to \infty} y(x) = \infty$.
Decay
if $k < 0$, the this function is decreasing and $\lim\limits_{x \to \infty} y(x) = 0$.
Graphs showing exponential growth and decay rates.

Until this point, this is all algebra, not calculus.

### 1.2.1 Population Growth

Population grows by 10% a year. How long will it take to double? $$y = Ce^{kx}, \quad x \text{ time }, \quad k=?$$ look at the algebra: assume \begin{aligned} y(0) & = 1 \\ y(1) & = 1.1 \end{aligned} Substitute these: \begin{aligned} Ce^{k\cdot 0} & = 1 \to C = 1 \\ Ce^{k\cdot 1} & = 1.1 \to e^{k} = 1\\ \to k &= \ln 1.1 \end{aligned}

To answer this question, we are looking for $x$, time, satisfying \begin{aligned} y(x) & = 2, \quad \text{ substitute} \\ e^{kx} & = 2, \quad \text{ solve } \\ kx & = \ln 2 \\ x &= \frac{\ln 2}{k} = \frac{\ln 2}{\ln 1.1} \end{aligned}

Carbon decays, loses half of its mass over a certain period of time. Then we have exponential decay here.

Percentage of this elements, $^{14}\text{C}$

Idea:

1. know the element's decay constant, $k$.
2. Measure % $\Rightarrow$ know time when tree was cut.

#### 1.2.2.1 Carbon Dating

Half life is 5730 years (Time take to go from 100% to 50%). Parchment has 74% of $^{14}\text{C}$ left. How old is it?

Estimate assumes that decay is linear.

Close to $\frac{1}{4}\text{-life} = \frac{1}{2}\frac{1}{2}\text{-life} = 2865$ Actual age is younger.

Given $y = e^{kx}$, find $k$.

Set \begin{aligned} y(0) = 1, y(5730) & = \frac{1}{2}, \text{ substitute } \\ e^{k\cdot 5730} & = \frac{1}{2}, \text{ Solve for } k \\ k \cdot 5730 & = \ln \frac{1}{2} \\ k &= \frac{\ln \frac{1}{2}}{5730} \end{aligned}

Find $x$ such that, $y(x) = 0.74$, or \begin{aligned} e^{kx} & = 0.74 \\ kx &= \ln 0.74 \\ x & = \frac{ln 0.74}{k} \\ &= \frac{ln 0.74}{\ln 0.5}\cdot 0.5730 = 2455 \end{aligned}

### 1.2.3 Newton's Law of Cooling

The rate of cooling of an object is proportional to the difference between its temperature and the temperature of the atmosphere. (Assumed $T > T_{0}$). Previsou discussion suggests the differential equation: $$\frac{dT}{dx} = k( T - T_{0})$$ where $T_{0}$ is the ambient temperature. What about warming? Does this still apply? Yes. But first,
Question: What is the sign of $k$?
Answer: $k$ is negative.

Verify:

1. Warmer object: $k< 0, T - t_{0} > 0$, so $\frac{dT}{dx} < 0$, so $T \searrow$: cooling.
2. Cooler object: $k< 0, T - t_{0} < 0$, so $\frac{dT}{dx} > 0$, so $T \nearrow$: warming.

Next, what is $T$? The solution to this new differential equation? Try this. What if $T_{0} = 0$, then $$T = Ce^{kx}$$ (know from before)

What if $T \neq 0$, then $$T = T_{0} + Ce^{kx}$$

We just shift the graph up by $T_{0}$.

Exercise: Subsitute to confirm.
Fact: Horizontal asymptote is $y = T_{0}$.

### 1.2.4 Review Exercise

Differentiate $y = \sin(\cos(\tan x ))$.

Flowchart: \begin{alignat}{4} y &= \sin (v) & \gets v & = \cos(u) & \gets u &= \tan(x) &\gets x \\ \frac{dy}{dv} & = \cos(v) & \frac{dv}{du} &= -\sin(u) & \frac{du}{dx} &= \sec^{2}x & \\ \end{alignat} Multiply these (Chain Rule - substitute $v = \cos u$ and $u = \tan x$) \begin{aligned} \frac{dy}{dx} & = \cos(v)\left(-\sin(u)\right)\sec^{2}(x) \\ &= \cos(\cos(u)\cdot(-\sin(u)\cdot\sec^{2}x \\ &= -\cos\cos\tan x \cdot \sin\tan x \cdot \sec{2}x \end{aligned}

Suppose we have money in the bank at APR 10% cpmpounded annually. Then after a year, given $\$1,000initial deposit, you have \begin{aligned} 1000 + 1000\cdot 0.10 &= 1000(1 + 0.1) \\ &= 1000 \cdot 1.1 \end{aligned} Same every year. Aftert$years, it$1000\cdot1.1^{t}$. What if it is compounded semi=annually, same APR? After$\frac{1}{2}$year,$1000\cdot 0.05$, or total $$1000 + 1000\cdot 0.05 = 1000\cdot 1.05$$ after another$\frac{1}{2}$year $$\left(1000\cdot 1.05\right)\cdot 1.05 = 1000 \cdot 1.05^{2}$$ After$t$years $$1000\cdot (1.05^{2})^{t} = 1000\cdot 1.05^{2t}$$ Note:$1.05^{2} = 1.1025 > 1.1 $Try compound quarterly, $$1000\cdot 1.025^{4t}$$ In compounded$n$times, then $$1000 \cdot \left(1 + \frac{1}{n}\right)^{nt}$$ where$\frac{1}{n}$is the interest in one period. Generally, for APR$r$(decimal), for initial deposit$A_{0}$, the after$t$years, the current amount $$A(t) = A_{0}\left(1 + \frac{r}{n} \right)^{nt}$$ if compounded$n$times per year. What if we compounded more and more,$n \to \infty? \begin{aligned} \lim_{n \to \infty} A(t) & = \lim_{n \to \infty} A_{0} \left( 1 + \frac{r}{n} \right)^{nt} = \infty ? \\ \text{CMR} &= A_{0} \lim_{n \to \infty} \left( 1 + \frac{r}{n} \right)^{nt} \\ & = A_{0} \left( \lim_{n \to \infty} \left(1 + \frac{r}{n}\right)^{n}\right)^{t} \\ & = A_{0} (e^{r})^{t}) \end{aligned} Note:\lim\limits_{n\to\infty} \left(1 + \frac{1}{n}\right)^{n} = e$. Withe$\text{APR} = r$, initial deposit$A_{0}$, after$t$years you have $$A(t) = A_{0} e^{rt}$$ if intereste is compounded continuously Example: APR 10%,$A_{0} = 1000$,$t = 1$. $$A(1)=1000\cdot e^{1.1} = 1000\cdot e^{0.1} \approx \1,105$$ interest$ \$1,105 > \$100 $(annual). Example: APR 100% $$A(i) = 1000\cdot e = \2,718$$ annual:$ \$1000 < \$1,718 $. Example: How long does it take to triple your money with APR=5%, compounded continuously?$A_{0} = 1$. Solve for$t. \begin{aligned} 3 & = 1\cdot e^{0.05t} \\ \ln 3 &= 0.05t \\ t &= \frac{\ln e}{0.05} = 22 \text{ years.} \end{aligned} Related Rates Recall how we define inverses: \begin{alignat}{2} &\text{implicit} & &\text{explicit} \\ x & = \sin y &\to y&= \sin^{-1}x \\ x &= e^{y} & \to y &= \ln x \end{alignat} So these are just solutions of equations, but the relation betweenx$and$yos present in both. Except, it's implicit in the first column. ### 1.2.6 Cop with Radar Gun Cop with a radar gun. Your speed is 80 mph, what is the reading? Assume the cop car is stationary. Signal Measurements made by cop Signal is sent, then it comes back, the time is measured. \begin{aligned} d_{1} &= \underbrace{\text{speed}}_{\text{know}} \cdot \underbrace{\text{time}}_{\text{measure}} \\ d_{2} & = \text{speed} \cdot \text{time} \end{aligned} Reading: \frac{d_{2} - d_{1}}{\text{Time between signals}} \neq 80 \text{m/h} No reading as car passes cop's position Reading = 0 when just passing the cop's car... \begin{aligned} D & = \text{distance from cop car to the road. FIXED!} \\ S & = \text{distance between cop car to your car. VARY} \\ P &= \text{Distance between your car to the point on the road closest to the cop car. VARY} \\ t &= \text{time, the independent variable} \end{aligned} 80 m/h is your speed, so\frac{dP}{dt} = 80$. That's what the cop wnats to know. But what does the radar measure?$\frac{dS}{dt}$. How do we connect the two,$\frac{dS}{dt}$,$\frac{dP}{dt}$? We use geometry, Pythagoras Theorem: $$P^{2} + D^{2} = S^{2} \gets \text{Not numbers, variabes, functions.}$$ This connects$P$and$S$, not$\frac{dP}{dt}$and$\frac{dS}{dt}$! We differentiate Pythagoras Theorem with respect to$t: \begin{aligned} \frac{d}{dt}\left(P^{2} + D^{2}\right) & = \frac{d}{dt}\left(S^{2}\right) \\ \frac{d}{dt}\left(P^{2}\right) + \frac{d}{dt}\left(D^{2}\right) & = \frac{d}{dt}\left(S^{2}\right) \\ 2P\cdot \frac{dP}{dt} + 2D\underbrace{\frac{dD}{dt}}_{=0} &= 2S\cdot \frac{dS}{dt} \\ p\cdot\frac{dP}{dt} &= S\cdot \frac{dS}{dt} \\ \therefore \frac{dS}{dt} &= \frac{P}{S}\frac{dP}{dt} \end{aligned} 4 functions oft$. Analyze this. Cosine Rule This is what the cop sees $$\frac{dS}{dt} = \frac{P}{s} 80$$ Simplyfy, what's$\frac{P}{S}$$$\cos \alpha = \frac{P}{S}$$ depends on$t$. $$\frac{dS}{dt} = 80 \cos \alpha$$ How does$\alpha$change as you drive? Early:$\alpha$is small and close to 0,$\cos \alpha$lose to 1, so \frac{dS}{dt} is close to 80. The angle,$\alpha$, is small in the beginning. Then as$\alpha$increases,$\cos \alpha$decreases, then $$\left| \frac{dS}{dt} \right| < 80$$ When$\alpha = \frac{\pi}{2} (90^{\circ})$,$\cos \alpha = 0$,$\frac{dS}{dt} = 0$. The angle,$\alpha$, increases and approaches zero as$\alpha \to 90$Then$\alpha$increases,$\cos \alpha$decreases as$\cos \alpha < 0$, $$\frac{dS}{dt} > 0, \left|\frac{dS}{dt}\right| < 80$$ Angle$\alpha$decreases as$90 < \alpha < 180$. Later$\alpha$appraches$\pi$,$\cos \alpha$approaches 1. Measured Speed vs. Time Conclusion: The radar gun always underestimates your speed. What can you do to "improve" the reading? What do you want$\alpha$to be? As large as possible! Take the right lane if the cop in the incoming lane Advice: Take the right lane if the cop in the incoming lane ### 1.2.7 Cylinder Tank 1. Radius 5 2. Water$3 \frac{\text{m}^{3}}{\text{min}}$coming in, 3. How fast is the level increasing? 1. Name things:$r$radius,$V$volume,$A$Area of the bottom,$H$Height of water. Rephrase the data in terms of$t$time. 2. Collect information about them: 1.$r=5$2.$\frac{dV}{dt}=3$3.$\frac{dH}{dt} = ?3. Connect these variables: \left.\begin{aligned} A & = \pi r^{2} \\ V &= AH \end{aligned} \right\} \to V = \pi r^{2} H = 25\pi H 4. Differentiate the equation in (3). \begin{aligned} \frac{d}{dt}\left(V\right) &= \frac{d}{dt}\left( 25 \pi H\right) \\ \frac{dV}{dt} & = 25 \pi \frac{dH}{dt} \end{aligned} 5. Substitute data on the rates we know. \begin{aligned} 3 &= 25 \pi \frac{dH}{dt} \\ \text{So } \frac{dH}{dt} &= \frac{3}{25\pi} \frac{\text{m}}{\text{min}}\end{aligned} ## 1.3 Ladder Against Wall A ladder against the wall 1. 10 ft. ladder 2. Bottom moves 1 ft/sec. 3. Question: How fast is the top moving when the bottom is 6 ft from the wall? 4. Diagram of falling ladder leaning against wall. Introduce variablesx$,$y(h)$, functions of time,$t$. Translate the variables into equations. Translate the geometry too. 1.$h = 10$2.$\frac{dx}{dt} = 1$3.$\frac{dy}{dt} = ?$when$x = 6$. 4.$x^{2} + y^{2} = h^{2}This is all we need. (No need for the words or the picture anymore.) Simplify: Given: \begin{aligned} x^{2} + y^{2} & = 100, \\ \frac{dx}{dt} &= 1 \end{aligned} Find\frac{dy}{dt}$for$x=6. Differentiate the equation with respect to the independent variables: \begin{aligned} \frac{d}{dt}\left( x^{2} + y^{2} \right) & = \frac{d}{dt}\left(100\right) \\ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} & = 0, \text{ Solve for } \frac{dy}{dt} \\ \frac{dy}{dt} &= - \frac{x}{y}\frac{dx}{dt}, \text{ Substitute } \\ &= -\frac{6}{y} 1, \\ & \text{ Now } y \text{ comes from } x^{2} + y^{2} = 100 \\ &= -\frac{6}{8} \\ & = -\frac{3}{4} \end{aligned} ## 1.4 Linear Approximations The tangent line approximates the graph ofy=f(x)$around$x=a$. in fact, it's the best one among all other lines! Assume$f$is differentiable at$x=a$. The tangent line will (but any other line won't) merge with the graph. This is the geometric meaning of best approximation. What about algebra? What is the equation of the tangent line? $$\text{slope } = f^{\prime}(a), \text{ passes through }(a,f(a))$$ Point-slope form: $$y - y_{0} = m(x - x_{0})$$ or $$y - f(a) = f^{\prime}(a)(x - a)$$ This is in fact a function. We solve for$y$to make explicit: $$y = f(a) + f^{\prime}(a)(x - a)$$ What kind of function is this? Linear (because the power of$x$is 1). The best linear approximation at$x = a$(or around$a$) of the (differential) function$y = f(x)$is $$f(x) = f(a) + f^{\prime}(a)(x - a)$$ How good is it? $$\text{error } = | f(x) - \ell(x) | = \text{ lengths of these segments.}$$ It's 0 at$x = a$, and may grow as you move away from$a$. Example: Approxmate$\sqrt{4.1}$. Cann't compute$\sqrt{x}$by hand. So$f(x) = \sqrt{x}$is approximated at$x = a = 4. $$\sqrt{4} = 2, \text{ exactly.}$$ What else do we know about 4? $$f(4) = 2$$ Answer: the dervative: \begin{aligned} f^{\prime}(x) &= \frac{1}{2\sqrt{x}} \\ \Rightarrow f^{\prime}(4) &= \frac{1}{2\sqrt{4}} = \frac{1}{4} \end{aligned} So to compute\sqrt{4.1}, use the approximation: \begin{aligned} \ell(x) &= f(a) + f^{\prime}(a) (x - a) \\ & = 2 + \frac{1}{4} (x - 4) \end{aligned} Good Enough. \begin{aligned} \ell(4.1) &= 2 + \frac{1}{4}(4.1 - 4) \\ & = 2 + \frac{1}{4} \cdot 1 \\ & = 2 + 0.025 \\ & = 2.025 \end{aligned} This is a linear approximation (1st degree). Compare to the "dumb" approximation: 4.1 is close to 4, then\sqrt{4.1}$is close to$\sqrt{4}$. So $$\sqrt{4.1} \approx 2$$ (0 degree) Compare to the true value $$\sqrt{4.1} \approx 2.0248$$ So, the linear approximation is better. (Taylor polynomials provide higher degree approximations, see Cale 2). ## 1.5 Differentials Question: Can we interpret$\frac{dy}{dx}$as a fraction? No. Answer: Yes, if you create new variables,$dx$and$dy$. Suppose$y = f(x)$,$f^{\prime}(a) = 3$or $$\frac{dy}{dx} = 3$$ So $$dy = 3 dx$$ These are new variables. Slope of the secant line is$m = \frac{\Delta y}{\Delta x}, or $$\Delta y = m\cdot\Delta x$$

Tangent line instead $$dy = 3 dx$$ Slope of tangent = 3. Where are $dx$, $dy$?

They are run, $x$ and rise, $y$, of the tangent line.

Why? Answer: What if the Universe is curved?

Negatively Curved Universe

(Einstein: It is).

## 1.6 Implicit Differentiation

Find $\frac{dy}{dx}:$ of $$2xy \sin y = y \cos x$$ Mathematics is not a spectator sport

\begin{aligned} \frac{d}{dx}(2xy) & = 2\frac{d}{dx}(xy) \\ & = 2 \left(\frac{d}{dx}(x) \cdot y + \frac{d}{dx}(y) \cdot x \right) \\ & = 2 \left( 1 \cdot y + \frac{dy}{dx} x \right) \\ \frac{d}{dx}(\sin y) &= \frac{d}{dx} \underbrace{(\sin(y(x)))}_{\text{Diff: } \sin \text{ and } y} = \cos y\cdot \frac{dy}{dx} \end{aligned}

## 1.7 Math 229 Calculus 1

### 1.7.1 Question 1

Calculate the derivative of $$f(x) = x^{e} + e^{x} + x + e$$

\begin{aligned} f^{\prime} & = ( x^{e} + e^{x} + x + e)^{\prime} \\ & = (x^{e})^{\prime} + (e^{x})^{\prime} + (x)^{\prime} + (e)^{\prime} \\ & = ex^{e-1} + e^{x} + 1 + 0 \end{aligned}

### 1.7.2 Question 2

Differentiate $$g(x) = \sqrt{x} \cos (x)$$

\begin{aligned} g^{\prime} & = (\sqrt{x} \cos x)^{\prime} \\ & = (\sqrt{x})^{\prime} \cos x + \sqrt{x} (\cos x)^{\prime} \\ &= \frac{1}{2\sqrt{x}} \cos x + \sqrt{x}(-\sin x) \end{aligned}

### 1.7.3 Question 3

Evaluate $\frac{dy}{dx}$ for $$y = \sqrt{e^{x}}$$

\begin{alignat}{2} y & = \sqrt{u} & \quad u &= e^{x} \\ \frac{dy}{du} &= \frac{1}{2\sqrt{u}} & \quad \frac{du}{dx} &= e^{x} \end{alignat} Therefore $$\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = \frac{1}{2\sqrt{u}}e^{x} = \frac{1}{2\sqrt{e^{x}}}\cdot e^{x}$$

### 1.7.4 Question 4

Evaluate $\frac{dy}{dx}$ for $$xy = \cos y + x$$

\begin{aligned} \frac{d}{dx}(xy) &= \frac{d}{dx}(\cos y + x ) \\ \frac{d}{dx}(x) y + x\frac{d}{dx}(y) &= \frac{d}{dx}(\cos y) + \frac{d}{dx}(x) \\ 1\cdot y + x \cdot \frac{dy}{dx} & = -\sin y \frac{dy}{dx} + 1 \\ \frac{dy}{dx} & = \frac{1-y}{x + \sin y} \end{aligned}

### 1.7.5 Question 5

Suppose the altitude, in $m$, of an object is given by the function $$y = t^{2} + t, \quad t \geq 0$$ where $t$ is time, in sec. What is the velocity when the altitide is 12 meters?

Altitude $$f(t) = t^{2} + t$$ Velocity $$f^{\prime}(t) = 2t + 1$$ Altitude is 12 $$\therefore t^{2} + t = 12$$ Solve for $x$ $$x= 3$$ Velocity $$f^{\prime}(3) = 2\cdot 3 + 1 = 7$$

### 1.7.6 Question 6

The population of a city declines by 10% every year. How long will it take to drop by 50% of the current population?

Population $$f(t) = Ce^{kt}$$ Declines by 10% in a year $$0.9 = 1 e^{k\cdot 1}$$ Solve for $k$ $$k = \ln 0.9$$ Drops to 50% in $t$ years $$0.5 = 1e^{k\cdot t}$$ Solve for $t$ $$kt = \ln 0.5$$ So $$t = \frac{\ln 0.5}{k} = \frac{\ln 0.5}{\ln 0.9}$$

### 1.7.7 Question 7

The area of a circle is increasing at a rate of 5 cm2/sec. At what rate is the radius of the circle increasing when the area is 2cm?

$r$ is the radius, $A$ is the area, $t$ is time. $$A = \pi r^{2}$$ Differentiate with respect to $t$ $$\frac{dA}{dt} = \pi r \frac{dr}{dt}$$ Substitute $$5 = \pi\cdot 2 \frac{dr}{dt}$$ So $$\frac{dr}{dt} = \frac{5}{2\pi}$$

### 1.7.8 Question 8

Find the linear approximation of $f(x) = 3\sin(x)$ at $a = 0$. Use it to estimate $\sin(0.02)$.

\begin{aligned} f^{\prime}(x) &= 3\cos x \\ f^{\prime}(0) &= 3\cos 0 = 3 \end{aligned} Then \begin{aligned} \ell(x) &= f(0) + f^{\prime}(0)( x - 0) \\ & = 0 + 3( x-0 ) &= 3x \end{aligned} So $$\ell(-0.02) = 3\cdot (-0.02) = -0.06$$