벡터함수의 미분과 적분 연습문제 — OpenStax 3권 · 3장 복습 · 12번

OpenStax 3권 · 3장 복습 · 12번대학 수준우리말 번역

$\text{u}(t),$ ${\text{u}}^{'}(t),$ ${\text{u}}^{'}(t)\,\times \,\text{u}(t),$ $\text{u}(t)\,\times \,{\text{u}}^{'}(t),$ $\text{u}(t)\cdot {\text{u}}^{'}(t)$ 의 도함수를 구하세요. 단위접벡터를 구하세요.

$\text{u}(t)=\langle {t}^{2},2t+6,4{t}^{5}-12\rangle$

영어 원문 보기

Find the derivatives of $\text{u}(t),$ ${\text{u}}^{'}(t),$ ${\text{u}}^{'}(t)\,\times \,\text{u}(t),$ $\text{u}(t)\,\times \,{\text{u}}^{'}(t),$ and $\text{u}(t)\cdot {\text{u}}^{'}(t).$ Find the unit tangent vector.

$\text{u}(t)=\langle {t}^{2},2t+6,4{t}^{5}-12\rangle$

정답·풀이 보기

${\text{u}}^{'}(t)=\langle 2t,2,20{t}^{4}\rangle ,$ $\text{u}''(t)=\langle 2,0,80{t}^{3}\rangle ,$ $\frac{d}{dt}[{\text{u}}^{'}(t)\,\times \,\text{u}(t)]=\langle -480{t}^{3}-160{t}^{4},24+75{t}^{2},12+4t\rangle ,$ $\frac{d}{dt}[\text{u}(t)\,\times \,{\text{u}}^{'}(t)]=\langle 480{t}^{3}+160{t}^{4},-24-75{t}^{2},-12-4t\rangle ,$ $\frac{d}{dt}[\text{u}(t)\cdot {\text{u}}^{'}(t)]=720{t}^{8}-9600{t}^{3}+6{t}^{2}+4$ , 단위접벡터: $\text{T}(t)=\frac{2t}{\sqrt{400{t}^{8}+4{t}^{2}+4}}\text{i}+\frac{2}{\sqrt{400{t}^{8}+4{t}^{2}+4}}\text{j}+\frac{20{t}^{4}}{\sqrt{400{t}^{8}+4{t}^{2}+4}}\text{k}$

원본 교재 풀이(영어)

${\text{u}}^{'}(t)=\langle 2t,2,20{t}^{4}\rangle ,$ $\text{u}''(t)=\langle 2,0,80{t}^{3}\rangle ,$ $\frac{d}{dt}[{\text{u}}^{'}(t)\,\times \,\text{u}(t)]=\langle -480{t}^{3}-160{t}^{4},24+75{t}^{2},12+4t\rangle ,$ $\frac{d}{dt}[\text{u}(t)\,\times \,{\text{u}}^{'}(t)]=\langle 480{t}^{3}+160{t}^{4},-24-75{t}^{2},-12-4t\rangle ,$ $\frac{d}{dt}[\text{u}(t)\cdot {\text{u}}^{'}(t)]=720{t}^{8}-9600{t}^{3}+6{t}^{2}+4,$ unit tangent vector: $\text{T}(t)=\frac{2t}{\sqrt{400{t}^{8}+4{t}^{2}+4}}\text{i}+\frac{2}{\sqrt{400{t}^{8}+4{t}^{2}+4}}\text{j}+\frac{20{t}^{4}}{\sqrt{400{t}^{8}+4{t}^{2}+4}}\text{k}$

한 번 더 — 비슷한 문제OpenStax 3권 · 3장 복습 · 14번우리말 번역

다음 적분의 값을 구하세요.

$\int_{1}^{4}\text{u}(t)dt$ , 단, $\text{u}(t)=\langle \frac{\ln (t)}{t},\frac{1}{\sqrt{t}},\sin \left(\frac{t\pi }{4}\right)\rangle$

다 풀었으면 정답 확인

$\frac{\ln {(4)}^{2}}{2}\text{i}+2\text{j}+\frac{2\left(2+\sqrt{2}\right)}{\pi }\text{k}$

이 문제로 이어서 풀기

# 벡터함수의 미분과 적분계산자동 분류