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MA 242
Test #2 SOLUTIONS
Fall, 1999
L. K. Norris
1. (40 pts) Let f (x, y, z) = x2 y 3z 4 and P0 = (1, 1, 1).
(a) Find the directional derivative of f (x, y, z) at the point P0 in the direction from
P0 to P1 = (3, 1, 2).
SOLUTION: The gradient of f is ∇f =< 2xy 3 z 4 , 3x2 y 2 z 4 , 4x2 y 3 z 3 >, and at
P0 this gives the vector ∇f (P0 ) =< 2, 3, 4 >. The unit vector from P0 to P1 is
~u = √15 < 2, 0, 1 >. Hence the directional derivative at P0 in the direction from P0
to P1 is
√
1
D~u f (P0 ) = ~u · ∇f (P0 ) = √ < 2, 0, 1 > · < 2, 3, 4 > = 8/ 5
5
(b) What is the greatest rate of increase of f (x, y, z) at P0 ?
SOLUTION: The greatest rate of increase of f at P0 equals the magnitude of
the gradient vector at P0 . Hence
√
√
greatest rate of increase of f at P0 = |∇f (P0 )| = 4 + 9 + 16 = 29
(c) Find an equation for the tangent plane at the point P0 = (1, 1, 1) to the level
surface of f (x, y, z) that passes through P0 .
SOLUTION: We can take the gradient vector at P0 is the normal for the tangent
plane. Hence
equation of tangent plane is:
2(x − 1) + 3(y − 1) + 4(z − 1) = 0
(d) Find parametric equations for the normal line at the point P0 = (1, 1, 1) to the
level surface of f (x, y, z) that passes through P0 .
SOLUTION: We can take the gradient vector at P0 as the direction vector for
the line. Hence
equations of normal line are:
x = 1 + 2t, y = 1 + 3t, z = 1 + 4t
2. (10 pts) Let f (x, y) = x2 y 2z 2 , and x = u2 + v 2 , y = u2 v 2 , and z = u2 − v 2 .
(a) Compute ∂f
∂v , expressing your answers in terms of x, y, u and v.
SOLUTION:
∂f = ∂f ∂x + ∂f ∂y + ∂f ∂z = (2xy 2 z 2 )(2v) + (2yx2 z 2 )(2u2 v) + (2zx2 y 2 )(−2v)
∂v
∂x ∂v
∂y ∂v ∂z ∂v
(b) Compute ∂f
∂u , expressing your answers in terms of x, y, u and v.
SOLUTION:
∂f = ∂f ∂x + ∂f ∂y + ∂f ∂z = (2xy 2z 2 )(2u) + (2yx2z 2 )(2uv 2) + (2zx2 y 2 )(2u)
∂u
∂x ∂u ∂y ∂u ∂z ∂u
3. (30 pts) Let f (x, y) = −x3 + 4xy − 2y 2 + 1. Find all critical points of this function,
and then use the second derivative test to determine if each critical point corresponds to a local extreme value or a saddle point of the function. [ HINT: There are
exactly 2 critical points.]
SOLUTION: The two critical points are (0, 0) and (4/3, 4/3) . Computing the function D(x, y) = fxx fyy − [fxy ]2 we find D(x, y) = 24x − 16. Then
(a) For the critical point (0, 0) we find D(0, 0) = −16 < 0, so this point corresponds to
a saddle point .
(b) For the critical point (4/3, 4/3) we find D(4/3, 4/3) = 16 > 0 and fxx (4/3, 4/3) =
−8 < 0. Hence the critical point (4/3, 4/3) corresponds to a local maximum of f .
4. (10 pts) Let f (x, y) = sin(2x2 + 3xy). Find the largest set on which f (x, y) is continuous. You must justify your work here by quoting appropriate theorems.
SOLUTION: (In what follows the material that is in boldfaced type is the material
that must be stated in some form in order to get full credit.)
This function is the composition of the polynomial function (2x2 + 3xy) with the sine
function sin(x). Since polynomial functions are continuous on the entire xyplane and since the sine function is continuous on all of the real line, the
composition function f (x, y) = sin(2x2 + 3xy) is continuous at each point of
the xy-plane.
5. (10 pts) The plane y = 2 intersects the graph of the function f (x, y) = 9 − x2 − y 2 in
a curve (See the drawing on the board). Use the geometrical interpretation of partial
derivatives to find parametric equations of the tangent line to this curve at the point
P = (1, 2, 4).
SOLUTION: Since y is held constant, the slope of the line in the y = 2 plane is
given by the x-partial derivative fx (x, y) = −2x evaluated at the point (1, 2), so slope
= fx (1, 2) = −2. The equation for the line in the y = 2 plane is, using the point-slope
formula for a line, z − 4 = −2(x − 1), y = 2. Hence the direction vector for the line is
~v =< 1, 0, −2 >, so parametric equations are
x = 1 + t, y = 2, z = 4 − 2t
6. (10 pts) Write the exact Maple V command that will plot the graph of the equations
x2 − y 2 = 1 and x2 + y 2 = 4 on the same set of axes, with the x-range running from
x=-2 to x=5 and the y-range running from y=0 to y=9. Assume that the command
”with(plots):” has already been executed in your maple worksheet.
SOLUTION:
implicitplot({ x2 − y 2 = 1, x2 + y 2 = 4 },x=-2..5, y=0..9)
Note that the Maple V command plot cannot be used here.