Consider the parametric curve r = r(t) = t sin t i + 3 t j + t cos t k, for t >= 0. To find the tangent line at the point corresponding to t = π, the first step is to differentiate the curve. Mathematica can do that, but this formula is easy enough to differentiate by hand:

r'(t) = (sin t + t cos t) i + 3 j  + (cos t - t sin t) k,

r'(π) = - π i + 3 j - k.

When
t = π, note that x(t) = (0, 3 π, - π). Hence, the tangent line to the curve at the point P(0, 3π, - π) has vector equation

r = r(t) = (0, 3 π, - π) + t (- π, 3, - 1).

The following
Mathematica routine plots the curve in blue and the tangent line at the point (0, 3 π, - π) in red inside a coordinate box. As usual, to execute it, position the cursor at the end of the last blue line of code and press the Enter key, or press the Shift and Return keys together.

In[9]:=

${\mathrm{Curve}}{\text{}}{=}{\mathrm{ParametricPlot3D}}{\left[}{\text{}}{\left\{}{\text{}}\text{}\left\{\text{}t*\mathrm{Sin}\left[t\right],3*t,t*\mathrm{Cos}\left[t\right],\text{}\text{}\mathrm{RGBColor}\left[0,0,\text{}1\right]\right\}{,}\text{}\text{}\left\{\text{}-\mathrm{Pi}*t,\text{}3*t\text{}+\text{}3*\mathrm{Pi},\text{}-t\text{}-\text{}\mathrm{Pi},\text{}\text{}\mathrm{RGBColor}\left[1,\text{}0,\text{}0\right]\text{}\right\}\text{}{\right\}}{,}{\text{}}\left\{t,0,\text{}2*\mathrm{Pi}\right\}{,}\text{}\text{}\text{}\mathrm{AxesLabel}\text{}->\text{}\left\{x,\text{}y,\text{}z\right\}\text{}{\right]}{;}$

$\mathrm{Basepoint}\text{}=\mathrm{Graphics3D}\left[\left\{\text{}\mathrm{RGBColor}\left[1,0,1\right]\text{},\text{}\text{}\left\{\mathrm{PointSize}\left[0.025\right],\text{}\mathrm{Point}\left[\left\{0,\text{}3*\mathrm{Pi},\text{}-\mathrm{Pi}\right\}\right]\right\}\text{}\right\}\right]$

$\mathrm{Legend}\text{}=\mathrm{Graphics3D}\left[\text{}\left\{\mathrm{RGBColor}\left[1,0,0\right],\text{}\text{}\mathrm{Text}\left[\text{}"P",\text{}\left\{0,\text{}3*\mathrm{Pi},\text{}-\mathrm{Pi}\right\},\text{}\left\{-2,\text{}0\right\}\right]\right\}\right]$

$\mathrm{Show}\left[\mathrm{Curve},\mathrm{Basepoint},\text{}\mathrm{Legend}\right]$

Out[10]=

$⁃\mathrm{Graphics3D}⁃$

Out[11]=

$⁃\mathrm{Graphics3D}⁃$

Out[12]=

$⁃\mathrm{Graphics3D}⁃$

Here is a second example, for the curve in Exercise 24, Section 14.2, of Calculus, 4th Edition by James Stewart, Brooks/Cole ITP, 2001. The parametric equations given there say that

r = r(t) = (${t}^{2}-1$) i + (${t}^{2}+1$) j + ( t  + 1) k.

The question asks for the tangent line at the point P(-1, 1, 1), which clearly corresponds to t = 0. Again, although Mathematica can easily differentiate the parametric vector function r, it is so simple that hand differentiation is just as easy:

r'(t) = (2t) i + (2t) j + k  = k  at the point P.

The tangent plane to the graph of r at the point P thus has vector equation

r =
${r}_{0}$ + t k = < -1, 1, 1>  +   < 0, 0, t> = < -1, 1, 1 + t> .

The following routine graphs the curve and the tangent line at P.

${\mathrm{Curve}}{\text{}}{=}{\mathrm{ParametricPlot3D}}{\left[}{\text{}}{\left\{}{\text{}}\text{}\left\{\text{}t^2\text{}-1,t^2\text{}+\text{}1,t+1,\text{}\text{}\mathrm{RGBColor}\left[0,0,\text{}1\right]\right\}{,}\text{}\text{}\left\{\text{}-1,\text{}1,\text{}1+\text{}t\text{},\text{}\text{}\mathrm{RGBColor}\left[1,\text{}0,\text{}0\right]\text{}\right\}\text{}{\right\}}{,}{\text{}}\left\{t,-2,\text{}2\right\}{,}\text{}\text{}\text{}\mathrm{AxesLabel}\text{}->\text{}\left\{x,\text{}y,\text{}z\right\}\text{}{\right]}{;}$

$\mathrm{Basepoint}\text{}=\mathrm{Graphics3D}\left[\left\{\text{}\mathrm{RGBColor}\left[1,0,1\right]\text{},\text{}\text{}\left\{\mathrm{PointSize}\left[0.025\right],\text{}\mathrm{Point}\left[\left\{-1,\text{}1,1\right\}\right]\right\}\text{}\right\}\right]$

$\mathrm{Legend}\text{}=\mathrm{Graphics3D}\left[\text{}\left\{\mathrm{RGBColor}\left[1,0,0\right],\text{}\text{}\mathrm{Text}\left[\text{}"P",\text{}\left\{-1,\text{}1,1\right\},\text{}\left\{-2,\text{}0\right\}\right]\right\}\right]$

$\mathrm{Show}\left[\mathrm{Curve},\mathrm{Basepoint},\text{}\mathrm{Legend}\right]$

Converted by Mathematica  (June 5, 2003)