Question

Find the points (if they exist) at which the following planes and curves intersect. $$y+x=0 ; \mathbf{r}(t)=\langle\cos t, \sin t, t\rangle, \text { for } 0 \leq t \leq 4 \pi$$

Short answer

Answer: The intersection points are ⟨(√2)/2, -(√2)/2, 𝜋/4⟩, ⟨-(√2)/2, (√2)/2, (3𝜋)/4⟩, ⟨(√2)/2, -(√2)/2, (5𝜋)/4⟩, and ⟨-(√2)/2, (√2)/2, (7𝜋)/4⟩.

Step-by-step solution

Write down the given equations

The plane equation is: $$y+x=0$$ And the curve equation is given by the parametric representation: $$\mathbf{r}(t)=\langle\cos t, \sin t, t\rangle$$ where \(0 \leq t \leq 4\pi\).

Plug the curve components into the plane equation

Substitute the components of the parametric representation of the curve into the plane equation: $$(\sin t) + (\cos t) = 0$$

Solve for t

We need to solve the equation \((\sin t) + (\cos t) = 0\) with the restriction \(0\leq t\leq 4\pi\). This equation can be rewritten as: $$\sin t = -\cos t$$ $$\tan t = -1$$ The solution for this equation is \(t = (2n+1)\frac{\pi}{4}\), where \(n\) is any integer, and within the given range. Now we have to find the solution for t within the interval \(0\leq t\leq 4\pi\).

Restrict the solutions for t within the given range

Taking into consideration the restriction on the range of t, namely \(0\leq t\leq 4\pi\), we can determine the possible values of n: For n=0, t=\(\frac{\pi}{4}\) For n=1, t=\(\frac{3\pi}{4}\) For n=2, t=\(\frac{5\pi}{4}\) For n=3, t=\(\frac{7\pi}{4}\) All four of these values of t satisfy the condition, and therefore, are valid values for the points of intersection.

Find the intersection points

Now, we can find the intersection points by plugging the value of t back into the parametric equation: For \(t=\frac{\pi}{4}\): \(\mathbf{r}\left(\frac{\pi}{4}\right) = \langle\cos\left(\frac{\pi}{4}\right), \sin\left(\frac{\pi}{4}\right), \frac{\pi}{4}\rangle = \langle\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, \frac{\pi}{4}\rangle\) For \(t=\frac{3\pi}{4}\): \(\mathbf{r}\left(\frac{3\pi}{4}\right) = \langle\cos\left(\frac{3\pi}{4}\right), \sin\left(\frac{3\pi^4}\right), \frac{3\pi}{4}\rangle = \langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, \frac{3\pi}{4}\rangle\) For \(t=\frac{5\pi}{4}\): \(\mathbf{r}\left(\frac{5\pi}{4}\right) = \langle\cos\left(\frac{5\pi}{4}\right), \sin\left(\frac{5\pi}{4}\right), \frac{5\pi}{4}\rangle = \langle\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, \frac{5\pi}{4}\rangle\) For \(t=\frac{7\pi}{4}\): \(\mathbf{r}\left(\frac{7\pi}{4}\right) = \langle\cos\left(\frac{7\pi}{4}\right), \sin\left(\frac{7\pi}{4}\right), \frac{7\pi}{4}\rangle = \langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, \frac{7\pi}{4}\rangle\) Thus, the intersection points between the curve and the plane are: $$\langle\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, \frac{\pi}{4}\rangle, \langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, \frac{3\pi}{4}\rangle, \langle\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, \frac{5\pi}{4}\rangle, \text{and} \langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, \frac{7\pi}{4}\rangle$$