7. -/0.5 points 0/5 submissions used\nat what point do the curves r₁(t) = (t, 2 - t, 35 + t²) and r₂(s) = (7…

Explanation:

Step1: Set the components equal

Set up the following system of equations by equating the corresponding components of $\mathbf{r}_1(t)$ and $\mathbf{r}_2(s)$: [ \begin{cases} t=7 - s\ 2 - t=s - 5\ 35 + t^{2}=s^{2} \end{cases} ] 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 $\mathbf{r}_1^\prime(t)=\langle1,-1,2t\rangle$, and when $t = 1$, $\mathbf{r}_1^\prime(1)=\langle1,-1,2\rangle$ The derivative of $\mathbf{r}_2(s)$ is $\mathbf{r}_2^\prime(s)=\langle - 1,1,2s\rangle$, and when $s = 6$, $\mathbf{r}_2^\prime(6)=\langle - 1,1,12\rangle$

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)