Question

In Exercises 24–29 a vector function and a point on the graph of the function are given. Find an equation for the line tangent to the curve at the specified point, and then find an equation for the plane orthogonal to the tangent line containing the given point.

$\mathbf{29}\mathbf{.}\mathbf{}\mathbf{r}\left(t\right)=\cos 3t\mathbf{i}+\sin 4t\mathbf{j}+t\mathbf{k},(0,0,\frac{\pi }{2})$

Short answer

The equation for the plane orthogonal to the tangent line containing the given point is$3x+4y+z=\frac{\pi }{2}$

Step-by-step solution

Step 1. Given data 

We have a point on the graph of the function are given $\mathbf{r}\left(t\right)=\cos 3t\mathbf{i}+\sin 4t\mathbf{j}+t\mathbf{k},(0,0,\frac{\pi }{2})$

We have to find an equation for the line tangent to the curve and find an equation for the plane orthogonal to the tangent line at the given point.

Step 2. The line tangent to the curve 

We have,

$z=z\left(t\right)=t=\frac{\pi }{2}$

$r^{\mathrm{\prime }}\left(t\right)=⟨-3\sin 3t,4\cos 4t,1⟩$

At$t=\frac{\mathrm{\pi }}{2}$

$$\begin{gathered} \begin{array}{rl}r\left(t\right)& =r\left(\frac{\pi }{2}\right)\\ & =⟨0,0,\frac{\pi }{2}⟩\end{array} \\ \begin{array}{rl}{r}^{\mathrm{\prime }}\left(t\right)& =r^{\mathrm{\prime }}\left(\frac{\pi }{2}\right)\\ & =⟨3,4,1⟩\end{array} \end{gathered}$$

Here, the tangent line of the vector curve defined by $r\left(t\right)$at $r\left(t_0\right)$is,

Therefore, the line tangent to the curve at the given point is

$x=3t,y=4t,z=\frac{\pi }{2}+t$

Step 3. The equation for the plane orthogonal to the tangent line 

The equation for the plane orthogonal to the tangent line containing the given point is

$\begin{array}{rl}\frac{d}{dt}\left(3t\right)x+\frac{d}{dt}\left(4t\right)y+\frac{d}{dt}(\frac{\pi }{2}+t)z& =c_1\\ 3\frac{d}{dt}\left(t\right)x+4\frac{d}{dt}\left(t\right)y+1\left(z\right)& =c_1\\ 3x+4y+z& =c_1\end{array}$

The plane passes through $(0,0,\frac{\pi }{2})$and substituting this in the equation we get,

$$\begin{gathered} 3\left(0\right)+4\left(0\right)+\frac{\pi }{2}=c_1 \\ {c}_1=\frac{\pi }{2} \end{gathered}$$

That is,$3x+4y+z=\frac{\pi }{2}$

Therefore, the equation for the plane orthogonal to the tangent line containing the given point is$3x+4y+z=\frac{\pi }{2}$