Problem 1
Find all the first partial derivatives of the following functions.
$f(x,y) = 3 x^4 y^3 + x^2 y^3. $
$f(x,y,z) = 3 x^2 z - x y + \sin(z^2) $
In[16]:=
D@3 x ^ 4 y ^ 3 + x ^ 2 y ^ 3, xD
D@3 x ^ 4 y ^ 3 + x ^ 2 y ^ 3, yD
Out[16]=
2 x y3 + 12 x3 y3
Out[17]=
3 x2 y2 + 9 x4 y2
In[18]:=
D@3 x ^ 2 z - x y + Sin@z ^ 2D, xD
D@3 x ^ 2 z - x y + Sin@z ^ 2D, yD
D@3 x ^ 2 z - x y + Sin@z ^ 2D, zD
Out[18]=
-y + 6 x z
Out[19]=
-x
Out[20]=
3 x2 + 2 z CosAz2 E
Problem 2
Consider the function $f(x,y) = {{ - x^2 y} \over {-2 x^3 - 2 y^3}}.$
Compute the limit of $f(x,y)$ as $(x,y)$ approaches $(0,0)$ along the x-axis.
Ans: On the x-axis, f(x,y)=0 (except the origin), hence the limit is zero
Compute the limit of $f(x,y)$ as $(x,y)$ approaches $(0,0)$ along the line $y=x$.
Ans: On the line y=x, f(x,y)=1/4 (except the origin), hence the limit is 1/4.
In complete sentences, discuss the continuity of $f(x,y)$ at $(0,0)$.
Ans: The function is not continuous. The limit of a continuous function is the same in all direction but this is not true for
the given function.
Problem
Consider the function $ f(x,y) = x^3 + x y + y^3.$
Find the gradient of $f$.
In[21]:=
8D@x ^ 3 + x y + y ^ 3, xD, D@x ^ 3 + x y + y ^ 3, yD<
Out[21]=
93 x2 + y, x + 3 y2 =
Find the directional derivative of $f$ in the direction of $(\sqrt{3}, -1)/2$.
2
11f58r1.nb
In[22]:=
8D@x ^ 3 + x y + y ^ 3, xD, D@x ^ 3 + x y + y ^ 3, yD<.8Sqrt@3D, - 1< ê 2
1
Out[22]=
2
K- x - 3 y2 +
3 I3 x2 + yMO
Find the direction of maximal increase of $f$. Ans: In the direction of the gradient above.
Find the best linear approximation (differential approximation) to $f$ at the point $(\sqrt{3}, -1)$.
In[25]:=
Out[26]=
Clear@f, L, x, yD
f@x_, y_D = x ^ 3 + x y + y ^ 3
x3 + x y + y3
In[27]:=
L@x_, y_D =
f@Sqrt@3D, - 1D + H8D@x ^ 3 + x y + y ^ 3, xD, D@x ^ 3 + x y + y ^ 3, yD< ê. 8x Ø Sqrt@3D, y Ø - 1<L.
8x - Sqrt@3D, y - H- 1L<
Out[27]=
-1 + 2
3 + 8 K-
3 + xO + K3 +
Equivalently, L@x, yD = 2 - 5
3 O H1 + yL
3 + 8 x + K3 +
3 Oy
Problem
Consider the graph of the function $f(x,y) = x^2 - y^2$.
Describe (in complete sentences) the level surfaces of the graph of $f$.
Ans: The level curves are hyperbolas whose axis is the x-axis for levels >0 and the y-axis for levels < 0. At zero the level curve
is two lines intersecting at the origin
Find the directional derivative of $f$ in the direction of $(1, -1)/\sqrt{2}$.
Note: Since no point for evaluation was given this is a general formula for the requested directional derivative.
In[50]:=
8D@x ^ 2 - y ^ 2, xD, D@x ^ 2 - y ^ 2, yD<.81, - 1< ê Sqrt@2D
2x+2y
Out[50]=
2
Find the tangent line to the level set of $f=0$ at the point $(1, -1)$.
In[51]:=
8D@x ^ 2 - y ^ 2, xD, D@x ^ 2 - y ^ 2, yD< ê. 8x -> 1, y Ø - 1<
Out[51]=
82, 2<
In[52]:=
82, 2<.8x - 2, y - 2< ã 0
Out[52]=
2 H- 2 + xL + 2 H- 2 + yL ã 0
11f58r1.nb
Problem
Consider the function $f[x,y] = x^3/3 -2 y^3/3 -3 x^2/2 -4x +32y + 6.$
Find the critical points of the function $f$.
Give the type of each critical point and find any local extrema.
In[30]:=
D@x ^ 3 ê 3 - 2 y ^ 3 ê 3 - 3 x ^ 2 ê 2 - 4 x + 32 y + 6, xD
D@x ^ 3 ê 3 - 2 y ^ 3 ê 3 - 3 x ^ 2 ê 2 - 4 x + 32 y + 6, yD
Out[30]=
- 4 - 3 x + x2
Out[31]=
32 - 2 y2
In[35]:=
solns = Solve@8D@x ^ 3 ê 3 - 2 y ^ 3 ê 3 - 3 x ^ 2 ê 2 - 4 x + 32 y + 6, xD,
D@x ^ 3 ê 3 - 2 y ^ 3 ê 3 - 3 x ^ 2 ê 2 - 4 x + 32 y + 6, yD< ã 80, 0<D
Out[35]=
88x Ø - 1, y Ø - 4<, 8x Ø - 1, y Ø 4<, 8x Ø 4, y Ø - 4<, 8x Ø 4, y Ø 4<<
In[34]:=
hessian = 88D@x ^ 3 ê 3 - 2 y ^ 3 ê 3 - 3 x ^ 2 ê 2 - 4 x + 32 y + 6, x, xD,
D@x ^ 3 ê 3 - 2 y ^ 3 ê 3 - 3 x ^ 2 ê 2 - 4 x + 32 y + 6, y, xD<,
8D@x ^ 3 ê 3 - 2 y ^ 3 ê 3 - 3 x ^ 2 ê 2 - 4 x + 32 y + 6, x, yD,
D@x ^ 3 ê 3 - 2 y ^ 3 ê 3 - 3 x ^ 2 ê 2 - 4 x + 32 y + 6, y, yD<<
Out[34]=
88- 3 + 2 x, 0<, 80, - 4 y<<
The four hessians are:
In[37]:=
hessian ê. solns;
% êê MatrixForm
Out[38]//MatrixForm=
-5
0
O K
O
0
16
-5
0
K
O K
O
0
- 16
5
0
K O
K
O
0
16
5
0
K O K
O
0
- 16
K
The first and last are saddle points, the second is a local maximum, and the third is a local minimum.
Problem
Find a point on the surface $x^2 - 2 y^2 - 4z^2 ==16$ at which the tangent plane is parallel to the plane $4x -2y + 4z == 5$.
Problem
Find the volume of the largest rectangular box that has three of its vertices on the positive $x$, $y$ and $z$ axes respectively
and a fourth vertex in the plane $3x + 4y + 4z =12$.
Note: Assume as well that three sides of the box lie in the positive coordinate planes
3
4
11f58r1.nb
In[39]:=
Volume@x_, y_D = x y H12 - 3 x - 4 yL ê 4
1
Out[39]=
4
x H12 - 3 x - 4 yL y
In[40]:=
Solve@8D@Volume@x, yD, xD ã 0, D@Volume@x, yD, yD ã 0<D
Out[40]=
:8x Ø 0, y Ø 0<, 8x Ø 0, y Ø 3<, :x Ø
4
3
, y Ø 1>, 8y Ø 0, x Ø 4<>
Analysis: if x or y is zero so is the volume. In the triangular region in the x-y plane where x, y and (12 - 3 x - 4 y)/4=z are
positive, the Volume is positive. Hence the maximum occurs at the single critical point with both x and y positive.
Problem
Let $w=x^2+y^2+z^2$, $x=s t$, $y=s \cos(t)$, $z=s\sin(t)$; Find $\frac{\partial w}{\partial s}$ and $\frac{\partial w}{\partial
t}$ when $s=1$ and $t=0$.
$\frac{\partial w}{\partial s}$
In[44]:=
D@x ^ 2 + y ^ 2 + z ^ 2, xD D@s t, sD +
D@x ^ 2 + y ^ 2 + z ^ 2, yD D@s Cos@tD, sD + D@x ^ 2 + y ^ 2 + z ^ 2, zD D@s Sin@tD, sD
% ê. 8x Ø s t, y -> s Cos@tD, z -> s Sin@tD<
% ê. 8s Ø 1, t Ø 0<
Out[44]=
2 t x + 2 y Cos@tD + 2 z Sin@tD
Out[45]=
2 s t2 + 2 s Cos@tD2 + 2 s Sin@tD2
Out[46]=
2
$\frac{\partial w}{\partial t}$
In[47]:=
D@x ^ 2 + y ^ 2 + z ^ 2, xD D@s t, tD +
D@x ^ 2 + y ^ 2 + z ^ 2, yD D@s Cos@tD, tD + D@x ^ 2 + y ^ 2 + z ^ 2, zD D@s Sin@tD, tD
% ê. 8x Ø s t, y -> s Cos@tD, z -> s Sin@tD<
% ê. 8s Ø 1, t Ø 0<
Out[47]=
2 s x + 2 s z Cos@tD - 2 s y Sin@tD
Out[48]=
2 s2 t
Out[49]=
0