CLP-4 · 1.4절 · 6번대학 수준우리말 번역CCC 를 다음과 같은 공간곡선이라 하세요. r(t)=(et−e−t) ı^+(et+e−t) ȷ^+2t k^\begin{aligned} \mathbf{r}(t) = \big(e^t - e^{-t}\big)\,\hat{\boldsymbol{\imath}} + \big(e^t + e^{-t}\big)\,\hat{\boldsymbol{\jmath}} +2t\,\hat{\mathbf{k}} \end{aligned}r(t)=(et−e−t)^+(et+e−t)^+2tk^ (a) r′\mathbf{r}'r′, r′′\mathbf{r}''r′′ 와 CCC 의 곡률을 구하세요. (b) r(0)\mathbf{r}(0)r(0) 과 r(1)\mathbf{r}(1)r(1) 사이의 곡선의 길이를 구하세요. 영어 원문 보기Let CCC be the space curve r(t)=(et−e−t) ı^+(et+e−t) ȷ^+2t k^\begin{aligned} \mathbf{r}(t) = \big(e^t - e^{-t}\big)\,\hat{\boldsymbol{\imath}} + \big(e^t + e^{-t}\big)\,\hat{\boldsymbol{\jmath}} +2t\,\hat{\mathbf{k}} \end{aligned}r(t)=(et−e−t)^+(et+e−t)^+2tk^ (a) Find r′\mathbf{r}'r′, r′′\mathbf{r}''r′′ and the curvature of CCC. (b) Find the length of the curve between r(0)\mathbf{r}(0)r(0) and r(1)\mathbf{r}(1)r(1). 풀이 공책 펴기한 줄씩 식 확인정답·풀이 보기(a) 다음과 같으므로 r′(t)=(et+e−t) ı^+(et−e−t) ȷ^+2 k^dsdt(t)=∣r′(t)∣=4+2e2t+2e−2t=2(et+e−t)r′′(t)=(et−e−t) ı^+(et+e−t) ȷ^r′(t)×r′′(t)=−2(et+e−t) ı^+2(et−e−t) ȷ^+4 k^\begin{aligned} \mathbf{r}'(t)&=\big(e^t + e^{-t}\big)\,\hat{\boldsymbol{\imath}} + \big(e^t - e^{-t}\big)\,\hat{\boldsymbol{\jmath}} +2\,\hat{\mathbf{k}} & \frac{\mathrm{d}s}{\mathrm{d}t}(t) &=|\mathbf{r}'(t)|= \sqrt{4+2e^{2t}+2e^{-2t}} =\sqrt{2}\big(e^t+e^{-t}\big)\\ \mathbf{r}''(t)&=\big(e^t - e^{-t}\big)\,\hat{\boldsymbol{\imath}} + \big(e^t + e^{-t}\big)\,\hat{\boldsymbol{\jmath}} & \mathbf{r}'(t)\times\mathbf{r}''(t)&=-2\big(e^t + e^{-t}\big)\,\hat{\boldsymbol{\imath}} + 2\big(e^t - e^{-t}\big)\,\hat{\boldsymbol{\jmath}} +4\,\hat{\mathbf{k}} \end{aligned}r′(t)r′′(t)=(et+e−t)^+(et−e−t)^+2k^=(et−e−t)^+(et+e−t)^dtds(t)r′(t)×r′′(t)=∣r′(t)∣=4+2e2t+2e−2t=2(et+e−t)=−2(et+e−t)^+2(et−e−t)^+4k^ 곡률은 다음과 같다. κ(t)=∣v(t)×a(t)∣(dsdt)3=24+2e2t+2e−2t[4+2e2t+2e−2t]3/2=12+e2t+e−2t\begin{aligned} \kappa (t) &= \frac{|\mathbf{v}(t)\times\mathbf{a}(t)|}{\big(\frac{\mathrm{d}s}{\mathrm{d}t}\big)^3} =\frac{2\sqrt{4+2e^{2t}+2e^{-2t}}}{[4+2e^{2t}+2e^{-2t}]^{3/2}} =\frac{1}{2+e^{2t}+e^{-2t}} \end{aligned}κ(t)=(dtds)3∣v(t)×a(t)∣=[4+2e2t+2e−2t]3/224+2e2t+2e−2t=2+e2t+e−2t1 (b) r(0)\mathbf{r}(0)r(0) 과 r(1)\mathbf{r}(1)r(1) 사이의 CCC 의 길이는 다음과 같다. ∫01dsdt(t) dt=2∫01(et+e−t) dt=2[et−e−t]01=2[e−1e]\begin{aligned} \int_0^1 \frac{\mathrm{d}s}{\mathrm{d}t}(t) \ \mathrm{d}t &= \sqrt{2} \int_0^1 (e^t+e^{-t}) \ \mathrm{d}t =\sqrt{2}\Big[e^t-e^{-t}\Big]_0^1 =\sqrt{2}\Big[e-\frac{1}{e}\Big] \end{aligned}∫01dtds(t) dt=2∫01(et+e−t) dt=2[et−e−t]01=2[e−e1] 원본 교재 풀이(영어)(a) As r′(t)=(et+e−t) ı^+(et−e−t) ȷ^+2 k^dsdt(t)=∣r′(t)∣=4+2e2t+2e−2t=2(et+e−t)r′′(t)=(et−e−t) ı^+(et+e−t) ȷ^r′(t)×r′′(t)=−2(et+e−t) ı^+2(et−e−t) ȷ^+4 k^\begin{aligned} \mathbf{r}'(t)&=\big(e^t + e^{-t}\big)\,\hat{\boldsymbol{\imath}} + \big(e^t - e^{-t}\big)\,\hat{\boldsymbol{\jmath}} +2\,\hat{\mathbf{k}} & \frac{\mathrm{d}s}{\mathrm{d}t}(t) &=|\mathbf{r}'(t)|= \sqrt{4+2e^{2t}+2e^{-2t}} =\sqrt{2}\big(e^t+e^{-t}\big)\\ \mathbf{r}''(t)&=\big(e^t - e^{-t}\big)\,\hat{\boldsymbol{\imath}} + \big(e^t + e^{-t}\big)\,\hat{\boldsymbol{\jmath}} & \mathbf{r}'(t)\times\mathbf{r}''(t)&=-2\big(e^t + e^{-t}\big)\,\hat{\boldsymbol{\imath}} + 2\big(e^t - e^{-t}\big)\,\hat{\boldsymbol{\jmath}} +4\,\hat{\mathbf{k}} \end{aligned}r′(t)r′′(t)=(et+e−t)^+(et−e−t)^+2k^=(et−e−t)^+(et+e−t)^dtds(t)r′(t)×r′′(t)=∣r′(t)∣=4+2e2t+2e−2t=2(et+e−t)=−2(et+e−t)^+2(et−e−t)^+4k^ the curvature κ(t)=∣v(t)×a(t)∣(dsdt)3=24+2e2t+2e−2t[4+2e2t+2e−2t]3/2=12+e2t+e−2t\begin{aligned} \kappa (t) &= \frac{|\mathbf{v}(t)\times\mathbf{a}(t)|}{\big(\frac{\mathrm{d}s}{\mathrm{d}t}\big)^3} =\frac{2\sqrt{4+2e^{2t}+2e^{-2t}}}{[4+2e^{2t}+2e^{-2t}]^{3/2}} =\frac{1}{2+e^{2t}+e^{-2t}} \end{aligned}κ(t)=(dtds)3∣v(t)×a(t)∣=[4+2e2t+2e−2t]3/224+2e2t+2e−2t=2+e2t+e−2t1 (b) The length of CCC between r(0)\mathbf{r}(0)r(0) and r(1)\mathbf{r}(1)r(1) is ∫01dsdt(t) dt=2∫01(et+e−t) dt=2[et−e−t]01=2[e−1e]\begin{aligned} \int_0^1 \frac{\mathrm{d}s}{\mathrm{d}t}(t) \ \mathrm{d}t &= \sqrt{2} \int_0^1 (e^t+e^{-t}) \ \mathrm{d}t =\sqrt{2}\Big[e^t-e^{-t}\Big]_0^1 =\sqrt{2}\Big[e-\frac{1}{e}\Big] \end{aligned}∫01dtds(t) dt=2∫01(et+e−t) dt=2[et−e−t]01=2[e−e1] 한 번 더 — 비슷한 문제CLP-4 · 1.4절 · 7번우리말 번역점 (2,4,8)(2,4,8)(2,4,8) 에서 r(t)=(t,t2,t3)\mathbf{r}(t)=(t,t^2,t^3)r(t)=(t,t2,t3) 의 열률을 구하세요. 풀이 공책 펴기한 줄씩 식 확인답 쓰고 채점채점다 풀었으면 정답 확인점 (2,4,8)(2,4,8)(2,4,8) 은 t=2t=2t=2 일 때이다. v(t)=(1,2t,3t2)v(2)=(1,4,12)a(t)=(0,2,6t)a(2)=(0,2,12)dadt(t)=(0,0,6)dadt(2)=(0,0,6)v(2)×a(2)=(24,−12,2)∣v(2)×a(2)∣=2181\begin{aligned} \mathbf{v}(t)&=(1,2t,3t^2) & \mathbf{v}(2)&=(1,4,12)\\ \mathbf{a}(t)&=(0,2,6t) & \mathbf{a}(2)&=(0,2,12)\\ \frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(t)&=(0,0,6) & \frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(2)&=(0,0,6)\\ && \mathbf{v}(2) \times \mathbf{a}(2)&=(24,-12,2)\\ && |\mathbf{v}(2) \times \mathbf{a}(2)|&=2\sqrt{181} \end{aligned}v(t)a(t)dtda(t)=(1,2t,3t2)=(0,2,6t)=(0,0,6)v(2)a(2)dtda(2)v(2)×a(2)∣v(2)×a(2)∣=(1,4,12)=(0,2,12)=(0,0,6)=(24,−12,2)=2181 이제 열률 공식을 사용한다: τ(t)=(v(t)×a(t))⋅dadt(t)∣v(t)×a(t)∣2τ(2)=(24,−12,2)⋅(0,0,6)(2181)2=3181\begin{aligned}\tau(t)&=\dfrac{(\mathbf{v}(t) \times \mathbf{a}(t))\cdot\frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(t)}{|\mathbf{v}(t) \times \mathbf{a}(t)|^2}\\ \tau(2)&=\frac{(24,-12,2)\cdot(0,0,6)}{(2\sqrt{181})^2}=\frac{3}{{181}} \end{aligned}τ(t)τ(2)=∣v(t)×a(t)∣2(v(t)×a(t))⋅dtda(t)=(2181)2(24,−12,2)⋅(0,0,6)=1813 이 문제로 이어서 풀기 # 단위접선·법선벡터계산●●●보통자동 분류