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7. -/0.5 points 0/5 submissions used at what point do the curves r₁(t) …

Question

  1. -/0.5 points 0/5 submissions used

at what point do the curves r₁(t) = (t, 2 - t, 35 + t²) and r₂(s) = (7 - s, s - 5, s²) intersect?
(x, y, z) = (
find their angle of intersection, θ, correct to the nearest degree.
θ =

Explanation:

Step1: Set the components equal

Set up the following system of equations by equating the corresponding components of $\mathbf{r}_1(t)$ and $\mathbf{r}_2(s)$:

$$ LATEXBLOCK0 $$

From the first - equation $t = 7 - s$, substitute $t$ into the second equation:

$$2-(7 - s)=s - 5$$
$$2 - 7 + s=s - 5$$
$$-5 + s=s - 5$$

Substitute $t = 7 - s$ into the third equation:

$$35+(7 - s)^{2}=s^{2}$$

Step2: Expand and solve for s

Expand $(7 - s)^{2}$:

$$35 + 49-14s+s^{2}=s^{2}$$
$$84-14s = 0$$
$$14s=84$$
$$s = 6$$

Step3: Find the value of t

Since $t=7 - s$, when $s = 6$, then $t=7 - 6=1$

Step4: Find the intersection point

Substitute $t = 1$ into $\mathbf{r}_1(t)$ (or $s = 6$ into $\mathbf{r}_2(s)$).
For $\mathbf{r}_1(t)$ with $t = 1$, we have $\mathbf{r}_1(1)=(1,2 - 1,35+1)=(1,1,36)$

Step5: Find the tangent vectors

The derivative of $\mathbf{r}_1(t)$ is $\mathbf{r}_1^\prime(t)=\langle1,-1,2t
angle$, and when $t = 1$, $\mathbf{r}_1^\prime(1)=\langle1,-1,2
angle$
The derivative of $\mathbf{r}_2(s)$ is $\mathbf{r}_2^\prime(s)=\langle - 1,1,2s
angle$, and when $s = 6$, $\mathbf{r}_2^\prime(6)=\langle - 1,1,12
angle$

Step6: Use the dot - product formula to find the angle

The dot - product formula is $\mathbf{a}\cdot\mathbf{b}=\vert\mathbf{a}\vert\vert\mathbf{b}\vert\cos\theta$
\(\mathbf{r}_1^\prime(1)\cdot\mathbf{r}_2^\prime(6)=1\times(-1)+(-1)\times1 + 2\times12=-1-1 + 24=22\)
\(\vert\mathbf{r}_1^\prime(1)\vert=\sqrt{1^{2}+(-1)^{2}+2^{2}}=\sqrt{1 + 1+4}=\sqrt{6}\)
\(\vert\mathbf{r}_2^\prime(6)\vert=\sqrt{(-1)^{2}+1^{2}+12^{2}}=\sqrt{1 + 1+144}=\sqrt{146}\)
\(\cos\theta=\frac{\mathbf{r}_1^\prime(1)\cdot\mathbf{r}_2^\prime(6)}{\vert\mathbf{r}_1^\prime(1)\vert\vert\mathbf{r}_2^\prime(6)\vert}=\frac{22}{\sqrt{6}\times\sqrt{146}}=\frac{22}{\sqrt{876}}\approx\frac{22}{29.6}\approx0.74\)
\(\theta=\cos^{-1}(0.74)\approx42^{\circ}\)

Answer:

\((x,y,z)=(1,1,36)\)
\(\theta = 42\)