OpenStax 3권 · 3장 복습 · 10번대학 수준우리말 번역 다음 곡선을 나타내는 벡터함수를 구하세요. 원뿔 z=x2+y2z=\sqrt{{x}^{2}+{y}^{2}}z=x2+y2 과 평면 z=y−4z=y-4z=y−4 의 교선 영어 원문 보기 Find a vector function that describes the following curves. Intersection of the cone z=x2+y2z=\sqrt{{x}^{2}+{y}^{2}}z=x2+y2 and plane z=y−4z=y-4z=y−4 풀이 공책 펴기한 줄씩 식 확인정답·풀이 보기r(t)=⟨t,2−t28,−2−t28⟩\text{r}(t)=\langle t,2-\frac{{t}^{2}}{8},-2-\frac{{t}^{2}}{8}\rangler(t)=⟨t,2−8t2,−2−8t2⟩ 원본 교재 풀이(영어)r(t)=⟨t,2−t28,−2−t28⟩\text{r}(t)=\langle t,2-\frac{{t}^{2}}{8},-2-\frac{{t}^{2}}{8}\rangler(t)=⟨t,2−8t2,−2−8t2⟩ 한 번 더 — 비슷한 문제OpenStax 3권 · 3장 복습 · 12번우리말 번역 u(t),\text{u}(t),u(t), u′(t),{\text{u}}^{'}(t),u′(t), u′(t) × u(t),{\text{u}}^{'}(t)\,\times \,\text{u}(t),u′(t)×u(t), u(t) × u′(t),\text{u}(t)\,\times \,{\text{u}}^{'}(t),u(t)×u′(t), u(t)⋅u′(t)\text{u}(t)\cdot {\text{u}}^{'}(t)u(t)⋅u′(t) 의 도함수를 구하세요. 단위접벡터를 구하세요. u(t)=⟨t2,2t+6,4t5−12⟩\text{u}(t)=\langle {t}^{2},2t+6,4{t}^{5}-12\rangleu(t)=⟨t2,2t+6,4t5−12⟩ 풀이 공책 펴기한 줄씩 식 확인다 풀었으면 정답 확인u′(t)=⟨2t,2,20t4⟩,{\text{u}}^{'}(t)=\langle 2t,2,20{t}^{4}\rangle ,u′(t)=⟨2t,2,20t4⟩, u′′(t)=⟨2,0,80t3⟩,\text{u}''(t)=\langle 2,0,80{t}^{3}\rangle ,u′′(t)=⟨2,0,80t3⟩, ddt[u′(t) × u(t)]=⟨−480t3−160t4,24+75t2,12+4t⟩,\frac{d}{dt}[{\text{u}}^{'}(t)\,\times \,\text{u}(t)]=\langle -480{t}^{3}-160{t}^{4},24+75{t}^{2},12+4t\rangle ,dtd[u′(t)×u(t)]=⟨−480t3−160t4,24+75t2,12+4t⟩, ddt[u(t) × u′(t)]=⟨480t3+160t4,−24−75t2,−12−4t⟩,\frac{d}{dt}[\text{u}(t)\,\times \,{\text{u}}^{'}(t)]=\langle 480{t}^{3}+160{t}^{4},-24-75{t}^{2},-12-4t\rangle ,dtd[u(t)×u′(t)]=⟨480t3+160t4,−24−75t2,−12−4t⟩, ddt[u(t)⋅u′(t)]=720t8−9600t3+6t2+4\frac{d}{dt}[\text{u}(t)\cdot {\text{u}}^{'}(t)]=720{t}^{8}-9600{t}^{3}+6{t}^{2}+4dtd[u(t)⋅u′(t)]=720t8−9600t3+6t2+4, 단위접벡터: T(t)=2t400t8+4t2+4i+2400t8+4t2+4j+20t4400t8+4t2+4k\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}T(t)=400t8+4t2+42ti+400t8+4t2+42j+400t8+4t2+420t4k 이 문제로 이어서 풀기 # 벡터함수의 미분과 적분계산자동 분류