Question

Find the points (if they exist) at which the following planes and curves intersect. $$y=2 x+1 ; \quad \mathbf{r}(t)=\langle 10 \cos t, 2 \sin t, 1\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Short answer

Answer: No, there are no points of intersection between the plane and curve in the given domain.

Step-by-step solution

Express the plane equation in terms of the position vector function

Since the position vector function \(\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle = \langle 10 \cos t, 2 \sin t, 1 \rangle\), the x and y values can be represented as \(x(t) = 10 \cos t\) and \(y(t) = 2 \sin t\). Substitute these into the plane equation: $$2 \cdot 10 \cos t + 1 = 2 \sin t$$

Solve for t

To solve for the variable \(t\), first isolate the sine and cosine terms: $$20 \cos t - 2 \sin t = -1$$ Now use the trigonometric identity \(\cos^2 t + \sin^2 t = 1\): $$\sin t = \sqrt{1 - \cos^2 t}$$ Substitute this expression into the previous equation: $$20 \cos t - 2 \sqrt{1 - \cos^2 t} = -1$$ Now square both sides to remove the square root: $$\left(20\cos t - 2\sqrt{1 - \cos^2 t}\right)^2 = 1$$ Expand and simplify the equation: $$400\cos^2 t - 80\cos t \cdot \sqrt{1 - \cos^2 t} + 4(1 - \cos^2 t) = 1$$ Now, let \(u = \cos t\). So the equation becomes: $$400u^2 - 80u\sqrt{1 - u^2} + 4(1 - u^2) = 1$$ Notice that this equation has no elementary solutions for \(u\). Therefore, there are no points of intersection for the given plane and curve in the domain \(0 \leq t \leq 2 \pi\).