A changing circle

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\tkzTabSlope }[1]{\foreach \x /\y /\z in {#1}{\node [left = 3pt] at (Z\x 1) {\scriptsize $\y $}; \node [right = 3pt] at (Z\x 1) {\scriptsize $\z $}; }} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ddt }[0]{\frac {d}{\d t}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\descriptionlabel }[1]{\hspace {\labelsep }\textbf {#1:}}$

Two young mathematicians discuss a circle that is changing.

Check out this dialogue between two calculus students (based on a true story):

Devyn: Riley, I’ve been thinking about calculus.
Riley: YOLO.
Devyn: Consider a circle of some radius $r$ .
Riley: Ha! What else would we ever call the “radius?”
Devyn: Exactly. Now the formula for the perimeter of a circle is?
Riley: $P=2\cdot \pi \cdot r$ baby.
Devyn: And its area?
Riley: You know it’s $A=\pi \cdot r^2$ .
Devyn: Right, but here’s what’s bugging me: If I know $r'$ , what is $P'$ ? What’s $A'$ ?
Riley: Oooh. Ouch. Hmmm. I wanna say it’s $P' = 2\cdot \pi \cdot r' \qquad \text {and}\qquad A' = \pi (r')^2$ but I’m not sure that is right.
Devyn: Yeah…me too. But I’m not sure that’s right either. Are we forgetting something?

Do you think our young mathematicians above are correct?

Yes. $P' = 2\cdot \pi \cdot r'$ and $A' = \pi (r')^2$ . No. While $P' = 2\cdot \pi \cdot r'$ , $A' \ne \pi (r')^2$ . No. While $A' = \pi (r')^2$ , $P' \ne 2\cdot \pi \cdot r'$ . No. $P' \ne 2\cdot \pi \cdot r'$ and $A' \ne \pi (r')^2$ . There is no way to tell.

Set $r(t)=3\cdot t$ . What is $r'(t)$ when $r=15$ ? $r'(t)=\answer {3}$

Set $r=3\cdot t$ . Now $P(t) = 2\cdot \pi \cdot 3\cdot t$ . What is $P'(t)$ when $r=15$ ? $P'(t)=\answer {2\cdot \pi \cdot 3}$

Describe what $P'$ means in this context.

It describes how fast the perimeter is changing at a particular instant in time.

Set $r=3\cdot t$ . Now $A(t) = \pi \cdot (3\cdot t)^2$ . What is $A'(t)$ when $r=15$ ? $A'(t)=\answer {2\cdot \pi \cdot 15\cdot 3}$

Describe what $A'$ means in this context. Does it make sense that $A'$ is positive?

It describes how fast the area is changing. It does make sense that $A'$ is positive, since as the radius gets larger, the area should get larger, too.

What, if anything, did our two young mathematicians forget about above?

They forgot the chain rule.