Question

Show that the vector-valued function \(\mathbf{r}(t)=e^{-t} \cos t \mathbf{i}+e^{-t} \sin t \mathbf{j}+e^{-t} \mathbf{k}\) lies on the cone \(z^{2}=x^{2}+y^{2}\). Sketch the curve.

Short answer

Yes, the vector valued function \(\mathbf{r}(t)=e^{-t} \cos t \mathbf{i}+e^{-t} \sin t \mathbf{j}+e^{-t} \mathbf{k}\) lies on the cone \(z^{2}=x^{2}+y^{2}\) as its coordinates fit the conical curve equation. The sketch of the curve would show the function \(\mathbf{r}(t)\) plotted within the conical structure.

Step-by-step solution

- Understanding the vector-valued function

The function is defined as \(\mathbf{r}(t)=e^{-t} \cos t \mathbf{i}+e^{-t} \sin t \mathbf{j}+e^{-t} \mathbf{k}\). Therefore, \(x= e^{-t} \cos t, y= e^{-t} \sin t\) and \(z = e^{-t}\).

- Substituting into the equation of the cone

Now, substitute the expressions for \(x, y, z\) into the cone's equation to verify if they hold true: \(z^{2}= x^{2} + y^{2}\) becomes \((e^{-t})^{2}= (e^{-t} \cos t)^{2} + (e^{-t} \sin t)^{2}\). This simplifies to \(e^{-2t} = e^{-2t} (\cos^2 t + \sin^2 t)\). Since \(\cos^2 t + \sin^2 t =1\), it means our function does fulfill the equation of the cone.

- Sketching the curve

To sketch the curve, start by sketching the cone \(z^{2}= x^{2}+y^{2}\) in 3D space. This cone is symmetric around the z-axis. Next, plot the vector function \(\mathbf{r}(t)\) inside this cone.