Math 21a
Tangent Lines and Planes
What do we know about the gradient ∇f ?
Tangent Lines to Curves in the Plane.
Fall, 2016
1. For each of the following curves, find the tangent line to the curve at the point (1, 1):
(a) The level curve f (x, y) = 2
for f (x, y) = x2 + y 2
(b) The level curve y − 2x2 − log x = −1
(c) The level curve f (x, y) = e
for f (x, y) = x2 ey
(d) The graph y = x2
(e) The level curve f (x, y) = −2
for f (x, y) = xy 2 − 3x3 y 5
(f) The parametric curve
~r(t) = ht2 + 1, et i
Tangent Planes to Surfaces in the Space.
2. For each of the following equations, find the tangent plane to the surface defined by the
equation at the given point.
(a) x2 + y 2 + z 2 = 3 at (1, 1, 1)
(b) xyz − x2 + yz = 11 at (1, 2, 3)
(c) x3 − 2y sin z = 8 at (2, 1, 0)
(d) z = x2 + 3xy + y 3 + 1 at (2, −1, −2)
3. (a) Find the tangent plane to the level surface F (x, y, z) = 2 at the point (1, 1, 2), where
F (x, y, z) = x2 + y 2 .
(b) Find the tangent plane to the graph z = f (x, y) at the point (1, 1, 2), where f (x, y) =
x2 + y 2 .
√
√
(c) Find the tangent plane to the parametric surface ~r(u, v) = h 2 cos u, 2 sin u, vi at the
point (u, v) = ( π4 , 2).
Tangent Lines and Planes – Answers and Solutions
1. (a) x + y = 2
i. As ∇f = h2x, 2yi, we have ∇f (1, 1) = h2, 2i. So the equation can be written in the
form 2x + 2y = d. As (1, 1) is on the line, we get d = 4. So 2x + 2y = 4 or x + y = 2
gives the tangent line.
ii. By implicit differentiation, we have
y 0 (x) = −
2x
x
fx
=−
=− .
fy
2y
y
So the slope at x = 1 is y 0 (1) = −1. From this we get y = −x + 2. This is the same
as x + y = 2.
√
√
iii. We parametrize
the √
curve by ~r(t) = h 2 cos t, 2 sin ti. Then ~r(π/4) = h1, 1i. As
√
~r 0 (t) = h1 − 2 sin t, 2 cos ti, we get ~r 0 (π/4) = h−1, 1i. So hx, yi = h1, 1i + th−1, 1i
gives a parametrization of the tangent line. As hx, yi = h1, 1i+th−1, 1i = h1−t, 1+ti,
we have
y−1
x−1
=t=
,
namely,
− (x − 1) = y − 1
or
x + y = 2.
−1
1
(b) Write y = 2x2 + log x − 1. As y 0 = 4x + x1 , y 0 (1) = 5. So y = 5x − 4 gives the tangent
line.
(c) As ∇f = h2xey , x2 ey i, we have ∇f (1, 1) = h2e, ei. So the equation can be written in the
form 2ex + ey = d. As (1, 1) is on the line, we get d = 3e. So 2ex + ey = 3e or 2x + y = 3
gives the tangent line.
(d) y = 2x − 1
(e) As ∇f = hy 2 − 9x2 y 5 , 2xy − 15x3 y 4 i, we have ∇f (1, 1) = h−8, −13i. So the equation
can be written in the form −8x − 13y = d. As (1, 1) is on the line, we get d = −21. So
−8x − 13y = −21 or 8x + 13y = 21 gives the tangent line.
(f) Note that ~r(0) = h1, 1i. A direction vector of the tangent line is given by the velocity
vector ~r 0 (0). As ~r 0 (t) = h2t, et i, so ~r 0 (0) = h0, 1i. Hence hx, yi = h1, 1i + th0, ei gives a
parametrization of the tangent line. This is the same as the line defined by x − 1 = 0.
2. (a) As ∇F = h2x, 2y, 2zi, we have ∇F (1, 1, 1) = h2, 2, 2i. So the equation can be written in
the form 2x + 2y + 2z = d. As (1, 1, 1) is on the surface, we get d = 6. So 2x + 2y + 2z = 6
or x + y + z = 3 gives the tangent plane.
(b) As ∇F = hyz − 2x, xz + z, xy + yi, we have ∇F (1, 2, 3) = h4, 6, 4i. So the equation can
be written in the form 4x + 6y + 4z = d. As (1, 2, 3) is on the surface, we get d = 28. So
4x + 6y + 4z = 28 or 2x + 3y + 2z = 14 gives the tangent plane.
(c) As ∇F = h3x2 , −2 sin z, −2y cos zi, we have ∇F (2, 1, 0) = h12, 0, −2i. So the equation
can be written in the form 12x − 2z = d. As (2, 1, 0) is on the surface, we get d = 24.
So 12x − 2z = 24 or 6x − z = 12 gives the tangent plane.
(d)
i. First let’s rewrite the equation in the form F (x, y, z) = c. Put F (x, y, z) = x2 +3xy+
y 3 +1−z (the right hand side minus the left hand side). Then the surface is defined by
F (x, y, z) = 0. As ∇F = h2x+3y, 3x+3y 2 , −1i, we have ∇F (2, −1, −2) = h1, 9, −1i.
So the equation can be written in the form x + 9y − z = d. As (2, −1, −2) is on the
surface, we get d = −5. So x + 9y − z = −5 gives the tangent plane.
ii. The equation is of graph type: if you put f (x, y) = x2 + 3xy + y 3 + 1, the equation
is z = f (x, y). We want to approximate this equation by a linear equation. So we
use the linear approximation L(x, y) of f (x, y) at (x, y) = (2, −1):
L(x, y) = f (2, −1) + fx (2, −1) x − 2 + fy (2, −1) y − (−1)
= −2 + 1(x − 2) + 9(y + 1) = x + 9y + 5.
Then the tangent plane to z = f (x, y) is given by the equation z = L(x, y), i.e.,
z = x + 9y + 5. This is the same as x + 9y − z = −5.
3. Parts (a) and (b) are intended to clarify our method.
(a) As ∇F = h2x, 2y, 0i, we have ∇F (1, 1, 2) = h2, 2, 0i. So the equation can be written in
the form 2x + 2y = d. As (1, 1, 2) is on the surface, we get d = 4. So 2x + 2y = 4 or
x + y = 2 gives the tangent plane. Note that F is a function of three variables x, y
and z. So ∇F has three components, and it is a normal vector to the surface.
(b) Put F (x, y, z) = x2 + y 2 − z. Then the surface is defined by F (x, y, z) = 0 and our
method works. As ∇F = h2x, 2y, −1i, we have ∇F (1, 1, 2) = h2, 2, −1i. So the equation
can be written in the form 2x + 2y − z = d. As (1, 1, 2) is on the surface, we get d = 2.
So 2x + 2y − z = 2 gives the tangent plane.
A typical confusion arises when you just compute ∇f . This has only two components!
We need three numbers a, b and c for ax + by + cz = d! Of course, one of the right
ways is to consider F (x, y, z) = x2 + y 2 − z.. Or as in Problem 2 (d), we can use
lineariztion of f (x, y). You will find L(x, y) = 2 + 2(x − 1) + 2(y − 1) = 2x + 2y − 2. So
z = L(x, y) = 2x + 2y − 2 is an equation for the tangent plane.
(c) Note that ~r(π/4, 2) = h1, 1, 2i. (As ~r(u, v) is a parametrization of the surface x2 +y 2 = 2,
we are trying to find the same tangent plane as in (a).) First we consider grid curves
~r(u, 2) and ~r(π/4, v) to find
(their direction vectors) parallel to the tangent
√ two vectors
√
plane. As ~ru (u, v) = h− 2 sin u, 2 cos u, 0i, we have ~ru (π/2, 2) = h−1, 1, 0i. This is a
direction vector of the tangent line to the grid curve ~r(u, 2) at u = π/2. As ~rv (u, v) =
h0, 0, 1i, we have ~rv (π/2, 2) = h0, 0, 1i. This is a direction vector of the tangent line to
the grid curve ~r(π/4, v) at v = 2.
So hx, y, zi = h1, 1, 2i + uh−1, 1, 0i + vh0, 0, 1i gives a parametrization of the tangent
plane.
We can also find an equation of the form ax + by + cz = d. For this let’s recall ha, b, ci is
a normal vector to the plane. We know such a vector, namely, h−1, 1, 0i × h0, 0, 1i! As
h−1, 1, 0i × h0, 0, 1i = h1, 1, 0i, the equation can be written in the form x + y = d. As
(1, 1, 2) is on the tangent plane, we get d = 2. So x + y = 2 gives the tangent plane (and
is the same as the answer for (a)).