Question

1. -/0.5 points 0/5 submissions used

consider the curve r(t) = ⟨8 cos(t), 8 sin(t), e^t⟩, 0 ≤ t ≤ π.
find r(t).
r(t) = ⟨-8 sin(t),8 cos(t),e^t⟩
consider the plane √3x + y = 1. determine the normal vector n of the plane.
n = ⟨√3,1,0⟩
find the point on the curve r(t) = ⟨8 cos(t), 8 sin(t), e^t⟩, 0 ≤ t ≤ π, where the tangent line is parallel to the plane √3x + y = 1.
(x,y,z) = ( )

Explanation:

Step1: Recall the condition for parallelism

If the tangent line to the curve $\mathbf{r}(t)$ is parallel to the plane $\sqrt{3}x + y=1$, then the tangent vector $\mathbf{r}'(t)$ is orthogonal to the normal vector $\mathbf{n}$ of the plane. The dot - product of $\mathbf{r}'(t)$ and $\mathbf{n}$ is zero. We know $\mathbf{r}'(t)=\langle - 8\sin(t),8\cos(t),e^{t}
angle$ and $\mathbf{n}=\langle\sqrt{3},1,0
angle$. The dot - product formula is $\mathbf{a}\cdot\mathbf{b}=a_1b_1 + a_2b_2+a_3b_3$. So, $\mathbf{r}'(t)\cdot\mathbf{n}=-8\sqrt{3}\sin(t)+8\cos(t)+0\times e^{t}=0$.

Step2: Solve the trigonometric equation

We have $-8\sqrt{3}\sin(t)+8\cos(t) = 0$. Rearrange it to get $8\sqrt{3}\sin(t)=8\cos(t)$, then $\tan(t)=\frac{1}{\sqrt{3}}$. Since $0\leq t\leq\pi$, we find that $t = \frac{\pi}{6}$.

Step3: Find the point on the curve

Substitute $t=\frac{\pi}{6}$ into $\mathbf{r}(t)=\langle8\cos(t),8\sin(t),e^{t}
angle$. We have $x = 8\cos(\frac{\pi}{6})=8\times\frac{\sqrt{3}}{2}=4\sqrt{3}$, $y = 8\sin(\frac{\pi}{6})=8\times\frac{1}{2}=4$, and $z = e^{\frac{\pi}{6}}$.

Answer:

$(4\sqrt{3},4,e^{\frac{\pi}{6}})$