Homework 16 Solutions
Math 21a
Spring, 2014
Due Friday, March 14th (MWF) or Tuesday, March 25th (TTh)
Preparing for linear approximation.
1.
(a) (Stewart 11.4 #6 ) Following Example 1 in Section 11.4, find an equation of the tangent plane to the surface
z = ln(x − 2y) at the point (3, 1, 0).
Solution:
We compute the partial derivatives zx and zy at point (3, 1, 0). They are:
zx =
1
x − 2y
=⇒
zx (3, 1) = 1.
and
2
=⇒
zy (3, 1) = −2.
x − 2y
Thus, the equation of the tangent plane is z = x − 2y + C for some constant C. Since the plane passes through
(3, 1, 0), we find C = −1. So the equation for the tangent plane is z = x − 2y − 1.
p
(b) (Stewart 11.4 #16 ) Following Example 2 in Section 11.4, verify the linearization near (0, 0) of f (x, y) = y + cos2 x ≈
1 + 21 y, and use it to approximate f (0.1, −0.1).
zy = −
Solution:
Again, we need to compute the partial derivatives and evaluate them at (0, 0).
and
−2 cos x sin x
fx = p
2 y + cos2 x
=⇒
fx (0, 0) = 0.
1
fy = p
2 y + cos2 x
=⇒
fy (0, 0) =
1
.
2
Since f (0, 0) = 1, near (0, 0), f (x, y) ≈ 1 + 21 y as desired. In particular f (0.1, −0.1) ≈ 0.95.
Problems on Partial Differentiation, 11.3
2. Using Clairaut’s Theorem
(a) (Stewart 11.3 #58 ) By finding both partial derivatives uxy and uyx , verify that Clairaut’s Theorem holds for the
function u = tan(2x + 3y).
Solution:
If we first differentiate with respect to x and then with respect to y, we get:
ux =
2
cos2 (2x + 3y)
and
uxy = (ux )y =
12 sin(2x + 3y)
.
cos3 (2x + 3y)
If we do it the other way around, we get:
uy =
3
cos2 (2x + 3y)
and
uyx = (uy )x =
12 sin(2x + 3y)
.
cos3 (2x + 3y)
Remarkably, the two expressions coincide (as Clairaut’s Theorem assures us they must).
√
√
(b) (Stewart 11.3 #66 ) If g(x, y, z) = 1 + xz + 1 − xy, find gxyz .
Hint: Use a different order of differentiation for each term. (Please explain why you are allowed to do this.)
Solution: We find gxyz = 0. The first term is independent of y, so if we differentiate it first with respect to y,
we get 0. Further derivatives with respect to x and z are again 0. The second term is independent of z; so we
differentiate it first with respect to z and get 0. Further derivatives are also 0.
3. (Stewart 11.3 #68 ) In the textbook, the level curves are drawn for a function f near the point P . Determine whether
the following partial derivatives are positive or negative at the point P .
(a) fx
(b) fy
(d) fxy
(e) fyy
(c) fxx
Solution:
(a) fx is negative, as f is decreasing when we move to the right horizontally (in the x-direction) through P .
(b) fy is positive, as f is increasing when we move up vertically (in the y-direction) through P .
(c) fxx is positive, as fx is increasing as we move to the right through P . That is, f is decreasing slower and slower
as we move to the right (in the x direction) through P .
(d) fxy is negative. We know that fx is negative at P , but if we move up (vertically, in the positive y-direction) we
find that fx is even more negative (that is, f would decrease at a faster rate in the x-direction).
(e) fyy is positive, as fy is increasing as we move up from P . That is, f is increasing faster and faster as we move up
in the y direction from P . (We can see this clearly in the level curves – they are closer together as we move up.)
4. Based on (Stewart 11.3 #86 ) The paraboloid z = 6 − x − x2 − 2y 2 intersects the plane x = 1 in a parabola.
(a) Find parametric equations for the tangent line to this parabola at the point (1, 2, −4).
Solution: Plugging in x = 1 into the equation of the paraboloid, we get an equation of the parabola: z = 4 − 2y 2 .
The tangent line at the point (1, 2, −4) is given by the equations x = 1 and z = −8y + 12. We put the equation
of the tangent line in parametric form: r(t) = hx(t), y(t), z(t)i = h1, t, −8t + 12i.
(b) Use the Mathematica command
plane = ContourPlot3D[x==1,{x,-3,3},{y,-3,3},{z,-10,10}]
to plot the given plane, and the command
paraboloid = Plot3D[6-x-x^2-2y^2,{x,-3,3},{y,-3,3}]
to plot the paraboloid. Finally, replace x(t), y(t), z(t) by the parametric equations you found in (a) in the command
tangent = ParametricPlot3D[{x(t), y(t), z(t)},{t,-3,3}, PlotStyle->{Red,Thick}]
to plot the tangent line. And use
Show[plane,paraboloid,tangent]
to display all the pieces in a single plot.
Solution:
The plot is shown in Figure 1.
(c) Find a parameterization of the parabola that is the intersection of the plane and the paraboloid. Use the
ParametricPlot3D command to plot this curve, and the Show command to display all four pieces in one plot.
Solution: We put the equation of the parabola x = 1, z = 4−2y 2 in parametric form r(t) = hx, y, zi = h1, t, 4−2t2 i
and plot it in green in Figure 1.
(d) Explain in your own words (in complete English sentences) what is the geometric relationship between the tangent
line you found and the partial derivatives of f (x, y) = 6 − x − x2 − 2y 2 .
Solution: Since we restrict to the plane x = 1 where x is constant, the slope of the tangent line in this plane
(parallel to the yz-plane) is given by the partial derivative with respect to y.
Figure 1: Plots for Problem 4 parts (b) (on the left) and (c) (on the right)
5. (Stewart 11.3 #88 ) In a study of frost penetration it was found that the temperature T and time t (measured in days)
at a depth x (measured in feet) can be modeled by the function T (x, t) = T0 + T1 e−λx sin(ωt − λx), where ω = 2π/365
and λ is a positive constant. (T0 and T1 are also constants.)
(a) Find ∂T /∂x. What is its physical significance?
Solution:
The derivative is
∂T
= −λT1 e−λx sin(ωt − λx) − λT1 e−λx cos(ωt − λx)
∂x
= −λT1 e−λx sin(ωt − λx) + cos(ωt − λx) .
The physical significance is that
∂T
∂x
measures how temperature varies with depth.
(b) Find ∂T /∂t. What is its physical significance?
Solution:
The derivative is
∂T
= ωT1 e−λx cos(ωt − λx)
∂t
The physical significance is that ∂T
measures how temperature varies with time. In the graph below, we can see
∂t
the cyclical nature of temperature: it is warmer in the summer and cooler in the winter.
(c) Show that T (x, t) satisfies the heat equation Tt = kTxx for a certain constant k.
Solution:
If we differentiate with respect to x again, we get
∂2T
2
2
−λx
=
λ
T
sin(ωt
−
λx)
+
cos(ωt
−
λx)
+
λ
T
e
cos(ωt
−
λx)
−
sin(ωt
−
λx)
1
1
∂x2
= 2λ2 T1 e−λx cos(ωt − λx).
Comparing equations, we find that
∂T
T1 ω ∂ 2 T
ω ∂2T
=
·
=
·
2
2
∂t
2λ T1 ∂x
2λ2 ∂x2
as desired.
(d) If λ = 0.2, T0 = 0 and T1 = 10, use Mathematica to plot the graph of T (x, t). (Use the Plot3D command that you
used above and that was introduced in Homework 01.)
Solution:
A plot over two years: