Math 1C
Mathematica Handout #2
(Tangent lines to a surface.)
The purpose of the following example is to show you how to use calculus, together with Mathematica to
draw a tangent line to a surface in space. Read the example carefully and use it as a guide for solving
Problem #94 from section 14.3 in the text.
Example: The plane x = −2 intersects the
hyperbolic paraboloid f (x, y) = x 2 − y 2 in a
parabola. Find parametric equations for the tangent
line to this parabola at the point (–2, 1, 3) and graph
the surface, the parabola, and the tangent line on the
same coordinate system.
Solution: We begin by finding the vector form of the parametric equations for the parabola. Since
x = −2 , we have z = f (x, y) = 4 − y 2 . Letting y = t (the parameter), we have r(t) = −2, t, 4 − t 2 . Note
that t = 1 gives the point (–2, 1, 3). Next, we’ll use Plot3D and ParametricPlot3D to graph the surface and
the curve.
hyperbolicparaboloid = Plot3D[x^2 – y^2, {x, –5, 5}, {y, –5, 5}]
intersection = ParametricPlot3D[{–2, t, 4 – t^2}, {t, -5, 5}, PlotStyle –> {Blue, Thick}]
Show[hyperbolicparaboloid, intersection]
To find the parametric equations of the tangent line, note that r′(t) = 0, 1, − 2t . So r′(1) = 0, 1, − 2 .
Thus, x = −2, y = 1 + t, and z = 3 − 2t are the parametric equations of the line.
Putting it all together in Mathematica, with the following statements, produces the graph shown above.
tangent = ParametricPlot3D[{–2, 1 + t, 3 – 2 t}, {t, –5, 5}, PlotStyle –> {Red, Thick}]
Show[hyperbolicparaboloid, intersection, tangent]
The equation of the tangent line can also be found by using partial differentiation. Note that since x is
∂z
. This will give the slope of the tangent line in the plane
∂y
∂ 2
∂z
[x − y 2 ] = −2y and at (–2, 1, 3),
= −2 . This corresponds to the ratio of the z and y
x = −2 . Now,
∂y
∂y
constant in the plane x = −2 , we will find
components of r ′(1) .