242 lines
7.7 KiB
TeX
242 lines
7.7 KiB
TeX
\documentclass{report}
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\input{preamble}
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\input{macros}
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\input{letterfonts}
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\title{\Huge{Calculus III}}
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\author{\huge{Krishna Ayyalasomayajula}}
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\date{}
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\begin{document}
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\maketitle
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\newpage% or \cleardoublepage
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% \pdfbookmark[<level>]{<title>}{<dest>}
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\pdfbookmark[section]{\contentsname}{toc}
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\tableofcontents
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\pagebreak
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\chapter{Lab 7}
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\section{Work}
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\qs{}{
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\begin{align*}
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\vec{r} \in \mathbb{R} \therefore t \ge 0 \because e^{\sqrt{t}}\notin\mathbb{R} \; t|t<0 \\
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\text{This corresponds with answer choice D}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\forall t>4, \quad \vec{r}\cdot\hat{j}\notin\mathbb{R} \\
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\ln(t-1) \text{ is not defined } \forall t\le1 \\
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\text{This corresponds with answer choice D} \\
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\vec{r}\cdot\hat{j}\in[-4,4] \quad \land \quad \vec{r}\cdot\hat{i}\in[-3,3] \\
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}=\langle -3\sin t ,4\cos t \rangle \\
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \right |_{t=\tfrac{\pi}{2}}= \langle -3,0 \rangle \quad \vec{r}(0)=\langle 3,0 \rangle \\
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\text{This corresponds with answer choice B}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\text{This can be directly evaluated to: } \\
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\langle 28,-49 \rangle \\
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\text{This corresponds with answer choice D}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\vec{r}\cdot\hat{j} \text{ is not defined for } t=1 \text{ but the limit as $t\to6$ can be evaluated without affecting the proceess} \\
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\vec{r}\cdot\hat{i} \text{ is not defined for } t=1 \text{ but the limit as $t\to6$ can be evaluated without affecting the proceess} \\
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\text{Using directevaluation: } \\
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\langle \frac{5}{35},-\frac{36+12-3}{5} \rangle = \langle \tfrac{1}{7},-9 \rangle \\
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\text{This corresponds with answer choice B}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\implies \langle 0,-6 \rangle \\
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\text{This corresponds with answer choice B}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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e^{-\ln 6}=6^{-1}=\tfrac{1}{6} \\
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\implies \lim_{t\to\ln 6}{\langle 6e^{-t}, 3e^{-t} \rangle}=\langle 1,\tfrac{1}{2} \rangle \\
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\text{This corresponds witha nswer choice B}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle -14t,\tfrac{1}{3}t^2 \rangle \\
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\text{This corresponds with answer choice C}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle -\csc^2 t, -\cot t\csc t \rangle \\
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\text{This corresponds with answer choice A}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle 8te^{t^2}, -3, 2t \rangle \\
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\text{This corresponds with answer choice B}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle 18\frac{1}{6t},6t^2 \rangle \\
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\frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} = \langle \frac{-108}{36t^2}, 12t \rangle \langle \frac{-3}{t^2}, 12t \rangle \\
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\text{This corresponds with answer choice C}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle 15t^4,-60t^4,20t^4 \rangle \\
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\| \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \| \\
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&= \sqrt{15^2\cdot t^8 +60^2\cdot t^8 +20^2\cdot t^8} \\
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&= t^4 \sqrt{15^2+60^2+20^2} = 65t^4 \\
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\hat{T}=\langle \tfrac{15}{65},\tfrac{-60}{65},\tfrac{20}{65} \rangle \\
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\text{This corresponds with answer choice B}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\vec{r}(t) &= 6\sin^3(2t)\,\hat{i} + 6\cos^3(2t)\,\hat{j} \\[6pt]
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}
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&= \langle 36\sin^2(2t)\cos(2t),\; -36\cos^2(2t)\sin(2t) \rangle \\[6pt]
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&= 36\sin(2t)\cos(2t)\,(\sin(2t)\,\hat{i} - \cos(2t)\,\hat{j})
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\end{align*}
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\begin{align*}
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\left\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\right\|
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&= \sqrt{(36\sin^2(2t)\cos(2t))^2 + (-36\cos^2(2t)\sin(2t))^2} \\[6pt]
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&= 36\sqrt{\sin^4(2t)\cos^2(2t) + \cos^4(2t)\sin^2(2t)} \\[6pt]
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&= 36\sqrt{\sin^2(2t)\cos^2(2t)} \\[6pt]
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&= 36\,|\sin(2t)\cos(2t)| \\[6pt]
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&= 18\,|\sin(4t)|
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\end{align*}
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\begin{align*}
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\hat{T}(t)
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&= \frac{\dfrac{d\vec r}{dt}}{\left\|\dfrac{d\vec r}{dt}\right\|}
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= \frac{36\sin(2t)\cos(2t)(\sin(2t)\,\hat{i} - \cos(2t)\,\hat{j})}{36|\sin(2t)\cos(2t)|} \\[6pt]
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&= \sin(2t)\,\hat{i} - \cos(2t)\,\hat{j}
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\end{align*}
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\text{This corresponds with answer choice B.}
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}
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\qs{}{
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Let $\vec{C}$ encode the components of the constant of integration such that $\vec{C}\in\mathbb{R}^3$
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\begin{align*}
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle t^4+5t^2,4t \rangle \\
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\int\mathrm{d}\vec{r}= \int \langle t^4+5t^2,4t \rangle \mathrm{d}t \\
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\vec{r}(t)&=\langle \tfrac{1}{5}t^5+\tfrac{5}{3}t^3,2t^2 \rangle + \vec{C}\\
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\vec{C}&=\vec{r}(0)=\langle 5,-6 \rangle
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\end{align*}
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This corresponds with answer choice C.
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}
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\qs{}{
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\begin{align*}
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&\implies \int_0^3{\left \langle \frac{2}{\sqrt{1+t}},-9t^2,\frac{6t}{(1+t^2)^2}\right \rangle \mathrm{d}t }\\
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&\implies \int_0^3{\left \langle 2(1+t)^{-\tfrac{1}{2}},-9t^2,\frac{6t}{(1+t^2)^2}\right \rangle \mathrm{d}t }\\
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&\implies \left [ \langle 4\sqrt{t+1},-3t^3,\frac{-3}{1+t^2} \right ]_0^3=\langle 8-4,-81, \tfrac{-3}{10}+3 \rangle = \langle 4,-81,\tfrac{27}{10} \rangle
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\end{align*}
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This corresponds with answer choice C.
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}
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\qs{}{
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\begin{align*}
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u \coloneqq t^2+1\;\mathrm{du}=2t\mathrm{d}t \quad \implies \left \langle \left [8t^3-3t \right ]_0^1,\int_0^1{\frac{4}{u} \mathrm{d}u}, -\int_0^1{\frac{1}{2\sqrt{u}}\mathrm{d}u} \right \rangle \\
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\left [ \left \langle 8t^3-3t, 4\ln{(t^2+1)},-\sqrt{t^2+1} \right \rangle \right ]_0^1 = \langle 8-3, 4\ln{2}-0,-\sqrt{2}+1\rangle
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\end{align*}
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This corresponds with answer choice A.
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}
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\chapter{Lab 8}
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\section{Work}
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\qs{}{
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\begin{align*}
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} =\vec{v}=\langle -14t,\tfrac{1}{7}t^2 \rangle
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\end{align*}
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This corresponds with answer choice B.
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}
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\qs{}{
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\begin{align*}
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} =\vec{v} = \langle -3\sin(3t),5\cos{t} \rangle \\
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\frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} =\vec{a} = \langle -9\cos(3t),-5\sin{t} \rangle
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\end{align*}
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This corresponds with answer choice D.
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}
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\qs{}{
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\begin{align*}
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\vec{v}(t)=\langle 18t+4, -15t^2,-2t \rangle \\
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\vec{v}(3)=\langle 58,-135,-6\rangle
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\end{align*}
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This correspon ds with answer choice B.
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}
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\qs{}{
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\begin{align*}
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} =\vec{v} = \langle -2\sin(2t),\frac{7}{t-3},-\tfrac{1}{3}t^2\rangle \\
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\vec{v}(0)=\langle 0,\tfrac{-7}{3},0 \rangle
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\end{align*}
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This corresponds with answer choice D.
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}
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\qs{}{
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\begin{align*}
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\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} &= \vec{v}
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= \langle 8\cos(2t), \; 10\sin(2t), \; -6\csc(2t)\cot(2t) \rangle \\[6pt]
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\frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} &= \vec{a}
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= \langle -16\sin(2t), \; 20\cos(2t), \; 12(\csc(2t)\cot^2(2t) + \csc^3(2t)) \rangle \\[6pt]
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\vec{a}\!\left(\tfrac{\pi}{4}\right) &= \langle -16\sin\!\big(\tfrac{\pi}{2}\big), \; 20\cos\!\big(\tfrac{\pi}{2}\big), \; 12(1\cdot 0^2 + 1^3) \rangle \\[6pt]
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&= \langle -16,\; 0,\; 12 \rangle
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\end{align*}
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This corresponds with answer choice B.
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}
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\qs{}{
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\begin{align*}
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\vec{v}(0) &= \langle \sqrt{2}, 0, \sqrt{2} \rangle, &
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\vec{a}(0) &= \langle 0, 0, \tfrac{\pi}{2} \rangle, \\[2mm]
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\vec{a}\cdot\vec{v} &= 0\cdot \sqrt{2} + 0\cdot 0 + \tfrac{\pi}{2}\cdot \sqrt{2} = \tfrac{\pi\sqrt{2}}{2}, \\[1mm]
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\|\vec{a}\| &= \tfrac{\pi}{2}, &
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\|\vec{v}\| &= \sqrt{(\sqrt{2})^2 + 0^2 + (\sqrt{2})^2} = \sqrt{4} = 2, \\[1mm]
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\cos\theta &= \frac{\vec{a}\cdot\vec{v}}{\|\vec{a}\|\|\vec{v}\|}
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= \frac{\tfrac{\pi\sqrt{2}}{2}}{\tfrac{\pi}{2}\cdot 2} = \frac{\sqrt{2}}{2}, \\[1mm]
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\theta &= \arccos\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}.
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\end{align*}
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}
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\end{document}
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