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문제 은행 · 미적분 · 대학 미적분 · 벡터와 공간

벡터함수의 미분과 적분

먼저 혼자 풀어 보고, 풀이로 확인한 다음, 아래 「한 번 더」 문제로 이어서 연습하세요.

CLP-4 · 1.1절 · 23번대학 수준우리말 번역

CC 를 포물기둥 x=y2x = y^2 과 쌍곡포물면 3z=2xy3z = 2xy 의 교선 중 y≥0y\ge 0 인 부분이라 하자.

(a) xx 를 매개변수로 하여 CC 의 벡터 매개변수 방정식을 쓰세요. (b) 원점과 점 (9,3,18)(9, 3, 18) 사이에 있는 CC 부분의 길이를 구하세요. (c) 입자가 xx 가 증가하는 방향으로 CC 를 따라 움직인다. 이 입자가 일정한 속력 9로 움직일 때, 점 (1,1,23)(1, 1, \frac{2}{3}) 에 있을 때의 속도 벡터를 구하세요. (d) (c)의 입자가 점 (1,1,23)(1, 1, \frac{2}{3}) 에 있을 때의 가속도 벡터를 구하세요.

영어 원문 보기

Let CC be the part of the curve of intersection of the parabolic cylinder x=y2x = y^2 and the hyperbolic paraboloid 3z=2xy3z = 2xy with y≥0y\ge 0.

(a) Write a vector parametric equation for CC using xx as the parameter. (b) Find the length of the part of CC between the origin and the point (9,3,18)(9, 3, 18). (c) A particle moves along CC in the direction for which xx is increasing. If the particle moves with constant speed 9, find its velocity vector when it is at the point (1,1,23)(1, 1, \frac{2}{3}). (d) Find the acceleration vector of the particle of part (c) when it is at the point (1,1,23)(1, 1, \frac{2}{3}).

정답·풀이 보기

(a) y=xy=\sqrt{x} 이고 z=23xy=23x3/2z=\frac{2}{3}xy=\frac{2}{3}x^{3/2} 이므로,

r(x)=x ı^+x ȷ^+23x3/2 k^\begin{aligned} \mathbf{r}(x)&=x\,\hat{\boldsymbol{\imath}}+\sqrt{x}\,\hat{\boldsymbol{\jmath}}+\frac{2}{3}x^{3/2}\,\hat{\mathbf{k}} \end{aligned}

이 문제의 나머지 부분을 위해 다음도 필요하다.

r′(x)=ı^+12x ȷ^+x k^r′′(x)=−14x3/2 ȷ^+12x k^dsdx=∣r′(x)∣=1+14x+x=(12x+x)2=12x+xdsdx(1)=32\begin{aligned} \mathbf{r}'(x)&=\hat{\boldsymbol{\imath}}+\frac{1}{2\sqrt{x}}\,\hat{\boldsymbol{\jmath}}+\sqrt{x}\,\hat{\mathbf{k}}\\ \mathbf{r}''(x)&=-\frac{1}{4x^{3/2}}\,\hat{\boldsymbol{\jmath}}+\frac{1}{2\sqrt{x}}\,\hat{\mathbf{k}}\\ \frac{\mathrm{d}s}{\mathrm{d}x}=|\mathbf{r}'(x)| &=\sqrt{1+\frac{1}{4x}+x} =\sqrt{\left(\frac{1}{2\sqrt{x}}+\sqrt{x}\right)^2} =\frac{1}{2\sqrt{x}}+\sqrt{x}\\ \frac{\mathrm{d}s}{\mathrm{d}x}(1)&=\frac{3}{2} \end{aligned}

(b)

∫C ds=∫09dsdx dx=∫09(12x+x) dx=[x+23x3/2]09=3+18=21\int_C \ ds =\int_0^9 \frac{\mathrm{d}s}{\mathrm{d}x}\ \mathrm{d}x =\int_0^9 \left(\frac{1}{2\sqrt{x}}+\sqrt{x}\right)\ \mathrm{d}x =\left[\sqrt{x}+\frac{2}{3}x^{3/2}\right]_0^9 =3+18 =21

(c) 다음과 같이 놓는다.

  • r(x)\mathbf{r}(x): 입자의 첫째 좌표가 xx 일 때의 위치,
  • R(t)\mathbf{R}(t): 시각 tt 에서의 입자의 위치,
  • x(t)x(t): 시각 tt 에서의 입자의 xx–좌표,
  • s(x)s(x): 원점에서 r(x)\mathbf{r}(x) 까지의 곡선의 길이.

모든 tt 에 대해 ∣R′(t)∣=9|\mathbf{R}'(t)|=9 임이 주어져 있다. 따라서

R(t)=r(x(t))  ⟹  R′(t)=r′(x(t))dxdt(t)  ⟹  9=∣R′(t)∣=dsdx(x(t))dxdt(t)=(12x(t)+x(t))dxdt(t)\begin{aligned} \mathbf{R}(t)=\mathbf{r}\big(x(t)\big)&\implies \mathbf{R}'(t)=\mathbf{r}'\big(x(t)\big)\frac{\mathrm{d}x}{\mathrm{d}t}(t) \\ &\implies 9=|\mathbf{R}'(t)|=\frac{\mathrm{d}s}{\mathrm{d}x}\big(x(t)\big) \frac{\mathrm{d}x}{\mathrm{d}t}(t) =\left(\frac{1}{2\sqrt{x(t)}}+\sqrt{x(t)}\right)\frac{\mathrm{d}x}{\mathrm{d}t}(t) \end{aligned}

특히 입자가 시각 00 에 (1,1,23)(1,1,\frac{2}{3}) 에 있다면 x(0)=1x(0)=1 이고

9=(121+1)dxdt(0)  ⟹  dxdt(0)=69=\left(\frac{1}{2\sqrt{1}}+\sqrt{1}\right)\frac{\mathrm{d}x}{\mathrm{d}t}(0) \implies \frac{\mathrm{d}x}{\mathrm{d}t}(0)=6

이므로

R′(0)=r′(1)dxdt(0)=(ı^+12 ȷ^+k^)6= 6 ı^+3 ȷ^+6 k^\mathbf{R}'(0)=\mathbf{r}'(1)\frac{\mathrm{d}x}{\mathrm{d}t}(0) =\left(\hat{\boldsymbol{\imath}}+\frac{1}{2}\,\hat{\boldsymbol{\jmath}}+\hat{\mathbf{k}}\right)6 =\ 6\,\hat{\boldsymbol{\imath}}+3\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}}

이다.

(d) 곱의 미분법과 연쇄 법칙에 의해,

R′(t)=r′(x(t))dxdt(t)  ⟹  R′′(t)=r′′(x(t))(dxdt(t))2+r′(x(t))d2xdt2(t)\mathbf{R}'(t)=\mathbf{r}'\big(x(t)\big)\frac{\mathrm{d}x}{\mathrm{d}t}(t)\implies \mathbf{R}''(t)=\mathbf{r}''\big(x(t)\big)\left(\frac{\mathrm{d}x}{\mathrm{d}t}(t)\right)^2 +\mathbf{r}'\big(x(t)\big)\frac{\mathrm{d^2}x}{\mathrm{d}{t}^2}(t)

(c)에서 9=∣R′(t)∣=(12x(t)+x(t))dxdt(t)9=|\mathbf{R}'(t)|=\Big(\frac{1}{2\sqrt{x(t)}}+\sqrt{x(t)}\Big)\frac{\mathrm{d}x}{\mathrm{d}t}(t) 임을 보았으므로

dxdt(t)=9(12x(t)+x(t))−1\frac{\mathrm{d}x}{\mathrm{d}t}(t)=9\left(\frac{1}{2\sqrt{x(t)}}+\sqrt{x(t)}\right)^{-1}

이다. 이를 미분하면

d2xdt2(t)=−9(12x(t)+x(t))−2(−14x(t)3/2+12x(t))dxdt(t)\frac{d^2x}{dt^2}(t)=-9\left(\frac{1}{2\sqrt{x(t)}}+\sqrt{x(t)}\right)^{-2} \left(-\frac{1}{4x(t)^{3/2}}+\frac{1}{2\sqrt{x(t)}}\right)\frac{\mathrm{d}x}{\mathrm{d}t}(t)

이다. 특히 t=0t=0 일 때 x(0)=1x(0)=1 이고 dxdt(0)=6\frac{\mathrm{d}x}{\mathrm{d}t}(0)=6 이므로

d2xdt2(0)=−9(32)−2(14)6=−6\frac{\mathrm{d^2}x}{\mathrm{d}{t}^2}(0)=-9\left(\frac{3}{2}\right)^{-2} \left(\frac{1}{4}\right)6=-6

이고, 따라서

R′′(0)=r′′(1) (6)2+r′(1)(−6)=36(−14 ȷ^+12 k^)−6(ı^+12 ȷ^+k^)=−6 ı^−12 ȷ^+12 k^\begin{aligned} \mathbf{R}''(0)&=\mathbf{r}''(1)\ \big(6\big)^2+\mathbf{r}'(1)\big(-6\big) =36\left(-\frac{1}{4}\,\hat{\boldsymbol{\jmath}}+\frac{1}{2}\,\hat{\mathbf{k}}\right) -6\left(\hat{\boldsymbol{\imath}}+\frac{1}{2}\,\hat{\boldsymbol{\jmath}}+\hat{\mathbf{k}}\right) \\ &=-6\,\hat{\boldsymbol{\imath}}-12\,\hat{\boldsymbol{\jmath}}+12\,\hat{\mathbf{k}} \end{aligned}

이다.

원본 교재 풀이(영어)

(a) Since y=xy=\sqrt{x} and z=23xy=23x3/2z=\frac{2}{3}xy=\frac{2}{3}x^{3/2},

r(x)=x ı^+x ȷ^+23x3/2 k^\begin{aligned} \mathbf{r}(x)&=x\,\hat{\boldsymbol{\imath}}+\sqrt{x}\,\hat{\boldsymbol{\jmath}}+\frac{2}{3}x^{3/2}\,\hat{\mathbf{k}} \end{aligned}

For the remaining parts of this problem we will also need

r′(x)=ı^+12x ȷ^+x k^r′′(x)=−14x3/2 ȷ^+12x k^dsdx=∣r′(x)∣=1+14x+x=(12x+x)2=12x+xdsdx(1)=32\begin{aligned} \mathbf{r}'(x)&=\hat{\boldsymbol{\imath}}+\frac{1}{2\sqrt{x}}\,\hat{\boldsymbol{\jmath}}+\sqrt{x}\,\hat{\mathbf{k}}\\ \mathbf{r}''(x)&=-\frac{1}{4x^{3/2}}\,\hat{\boldsymbol{\jmath}}+\frac{1}{2\sqrt{x}}\,\hat{\mathbf{k}}\\ \frac{\mathrm{d}s}{\mathrm{d}x}=|\mathbf{r}'(x)| &=\sqrt{1+\frac{1}{4x}+x} =\sqrt{\left(\frac{1}{2\sqrt{x}}+\sqrt{x}\right)^2} =\frac{1}{2\sqrt{x}}+\sqrt{x}\\ \frac{\mathrm{d}s}{\mathrm{d}x}(1)&=\frac{3}{2} \end{aligned}

(b)

∫C ds=∫09dsdx dx=∫09(12x+x) dx=[x+23x3/2]09=3+18=21\int_C \ ds =\int_0^9 \frac{\mathrm{d}s}{\mathrm{d}x}\ \mathrm{d}x =\int_0^9 \left(\frac{1}{2\sqrt{x}}+\sqrt{x}\right)\ \mathrm{d}x =\left[\sqrt{x}+\frac{2}{3}x^{3/2}\right]_0^9 =3+18 =21

(c) Denote by

  • r(x)\mathbf{r}(x) the position of the particle when its first coordinate is xx,
  • R(t)\mathbf{R}(t) the position of the particle at time tt,
  • x(t)x(t) the xx–coordinate of the particle at time tt, and
  • s(x)s(x) the arc length of the curve from the origin to r(x)\mathbf{r}(x).

We are told that ∣R′(t)∣=9|\mathbf{R}'(t)|=9 for all tt. So

R(t)=r(x(t))  ⟹  R′(t)=r′(x(t))dxdt(t)  ⟹  9=∣R′(t)∣=dsdx(x(t))dxdt(t)=(12x(t)+x(t))dxdt(t)\begin{aligned} \mathbf{R}(t)=\mathbf{r}\big(x(t)\big)&\implies \mathbf{R}'(t)=\mathbf{r}'\big(x(t)\big)\frac{\mathrm{d}x}{\mathrm{d}t}(t) \\ &\implies 9=|\mathbf{R}'(t)|=\frac{\mathrm{d}s}{\mathrm{d}x}\big(x(t)\big) \frac{\mathrm{d}x}{\mathrm{d}t}(t) =\left(\frac{1}{2\sqrt{x(t)}}+\sqrt{x(t)}\right)\frac{\mathrm{d}x}{\mathrm{d}t}(t) \end{aligned}

In particular, if the particle is at (1,1,23)(1,1,\frac{2}{3}) at time 00, then x(0)=1x(0)=1 and

9=(121+1)dxdt(0)  ⟹  dxdt(0)=69=\left(\frac{1}{2\sqrt{1}}+\sqrt{1}\right)\frac{\mathrm{d}x}{\mathrm{d}t}(0) \implies \frac{\mathrm{d}x}{\mathrm{d}t}(0)=6

so that

R′(0)=r′(1)dxdt(0)=(ı^+12 ȷ^+k^)6= 6 ı^+3 ȷ^+6 k^\mathbf{R}'(0)=\mathbf{r}'(1)\frac{\mathrm{d}x}{\mathrm{d}t}(0) =\left(\hat{\boldsymbol{\imath}}+\frac{1}{2}\,\hat{\boldsymbol{\jmath}}+\hat{\mathbf{k}}\right)6 =\ 6\,\hat{\boldsymbol{\imath}}+3\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}}

(d) By the product and chain rules,

R′(t)=r′(x(t))dxdt(t)  ⟹  R′′(t)=r′′(x(t))(dxdt(t))2+r′(x(t))d2xdt2(t)\mathbf{R}'(t)=\mathbf{r}'\big(x(t)\big)\frac{\mathrm{d}x}{\mathrm{d}t}(t)\implies \mathbf{R}''(t)=\mathbf{r}''\big(x(t)\big)\left(\frac{\mathrm{d}x}{\mathrm{d}t}(t)\right)^2 +\mathbf{r}'\big(x(t)\big)\frac{\mathrm{d^2}x}{\mathrm{d}{t}^2}(t)

We saw in part (c) that 9=∣R′(t)∣=(12x(t)+x(t))dxdt(t)9=|\mathbf{R}'(t)|=\Big(\frac{1}{2\sqrt{x(t)}}+\sqrt{x(t)}\Big)\frac{\mathrm{d}x}{\mathrm{d}t}(t) so that

dxdt(t)=9(12x(t)+x(t))−1\frac{\mathrm{d}x}{\mathrm{d}t}(t)=9\left(\frac{1}{2\sqrt{x(t)}}+\sqrt{x(t)}\right)^{-1}

Differentiating that gives

d2xdt2(t)=−9(12x(t)+x(t))−2(−14x(t)3/2+12x(t))dxdt(t)\frac{d^2x}{dt^2}(t)=-9\left(\frac{1}{2\sqrt{x(t)}}+\sqrt{x(t)}\right)^{-2} \left(-\frac{1}{4x(t)^{3/2}}+\frac{1}{2\sqrt{x(t)}}\right)\frac{\mathrm{d}x}{\mathrm{d}t}(t)

In particular, when t=0t=0, x(0)=1x(0)=1 and dxdt(0)=6\frac{\mathrm{d}x}{\mathrm{d}t}(0)=6

d2xdt2(0)=−9(32)−2(14)6=−6\frac{\mathrm{d^2}x}{\mathrm{d}{t}^2}(0)=-9\left(\frac{3}{2}\right)^{-2} \left(\frac{1}{4}\right)6=-6

so

R′′(0)=r′′(1) (6)2+r′(1)(−6)=36(−14 ȷ^+12 k^)−6(ı^+12 ȷ^+k^)=−6 ı^−12 ȷ^+12 k^\begin{aligned} \mathbf{R}''(0)&=\mathbf{r}''(1)\ \big(6\big)^2+\mathbf{r}'(1)\big(-6\big) =36\left(-\frac{1}{4}\,\hat{\boldsymbol{\jmath}}+\frac{1}{2}\,\hat{\mathbf{k}}\right) -6\left(\hat{\boldsymbol{\imath}}+\frac{1}{2}\,\hat{\boldsymbol{\jmath}}+\hat{\mathbf{k}}\right) \\ &=-6\,\hat{\boldsymbol{\imath}}-12\,\hat{\boldsymbol{\jmath}}+12\,\hat{\mathbf{k}} \end{aligned}

한 번 더 — 비슷한 문제APEX · 11.2절 · 2번우리말 번역

변화율 함수의 정적분은 _____을(를) 나타낸다.

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