MATH 251, Exam 1
Jens Forsgård
February 19, 2016
1. Let L be the line passing through the points P1 = P1 (0, 1, 2) and P2 = P2 (3, 3, 3).
(a) Find a parametric representation of L.
2 pts.
A directional vector of the line is given by v = h3, 3, 3i − h0, 1, 2i = h3, 2, 1i. A point on the line is given
by, e.g., P1 = (0, 1, 2). Hence, one correct answer is
L(t) = h0, 1, 2i + th3, 2, 1i.
(b) Use your answer from (a) to find the symmetric equations of the line L.
2 pts.
We have that x(t) = 3t, that y(t) = 1 + 2t and that z(t) = 2 + t. Solving for t we obtain
t=
y−1
x
=
= z − 2.
3
2
2. Find a directional vector of the line which is the intersection of the hyperplanes x + y + z = 7 and
3x − 2y = 0.
2 pts.
The directional vector is given by the cross product of the normal vectors of the planes. We have that
n1 = h1, 1, 1i and n2 = h3, −2, 0i, so that
v = n1 × n2 = h2, 3, −5i.
3. Let C be the curve with equations x(t) = 1 − t, y(t) = et − 2, and z(t) = cos(t).
(a) Find the point where C intersects the xz-plane.
1 pts.
t
The point is given by y(t) = 0, which gives e − 2 = 0, or t = log(2). Hence, the point is
(x(log(2)), y(log(2)), z(log(2)) = (1 − log(2), 0, cos(log(2))).
(b) Find a tangent vector of the curve C at the point given by t = π.
0
0
0
0
2 pts.
0
0
The tangent vector at is given by r (t) = hx (t), y (t), z (t)i. We have that x (t) = −1, that y (t) = et ,
and that z 0 (t) = − sin(t). Hence
r0 (π) = h−1, eπ , 0i.
(c) Find a parametric representation of the tangent line to C at the point given by t = π.
2 pts.
The directional vector is given by the tangent vector from (b). A point is given by plugging in t = π.
Hence
L(s) = h1 − π, eπ − 2, −1i + sh−1, eπ , 0i.
1
4. Let f (x, y, z) = (x2 − 4y 2 + z 2 )2 .
(a) Find the quadratic equations which describe the level surfaces of the function f (x, y, z).
2 pts.
A level curve is given by f (x, y, z) = k for some constant k. By taking square roots (suuming k ≥ 0) we
obtain
√
x2 − 4y 2 + z 2 = k.
(b) Identify the quadratic surfaces defined by the equations from (a).
2 pts.
For k > 0 these are one-sheeted hyperboloids. For k = 0 it is a cone.
(c) Sketch one of the level curves of f (x, y, z) of your choice. For maximal score coordinate axes should
be added in a correct manner.
3 pts.
–
5. Let f (x, y, z) = x4 + x5 + yx and let P = P (1, 1)
(a) Compute the gradient of f .
2 pts.
we have that ∇f = h4x3 + 5x4 + y, xi.
(b) Compute the directional derivative of f in the direction of v = h−1, −1i.
2 pts.
√
The vector v is not a unit vector. Let u = v/|v| = h−1, −1i/ 2. From (a) we note that ∇f (1, 1) =
h10, 1i. The directional derivative of f in the direction of v is then given by
√
Du f (1, 1) = ∇f (1, 1) · u = −11/ 2.
6. Find and classify critical points of the function f (x, y) = xy 3 + 24y 2 − 8x + 2.
3
5 pts.
2
We have that ∇f = hy − 8, 3xy + 48yi. Thus, ∇f = h0, 0i gives y = 2 and x = −8. Hence, the only
critical point of f is P = (−8, 2).
00
00
00
00
= 6xy + 48. The second
= 3y 2 , and fyy
= fyx
= 0, fxy
We have the second order derivatives fxx
derivative test gives
0
3y 2
4
3y 2 6xy + 48 = −9y .
For x = −8 and y = 2 we obtain −9 · 16 which is negative. Hence, P is a saddle point.
7. Compute the length of the curve with equations r(t) = hcos(t), sin(t), 3ti for 0 ≤ t ≤ 4π.
0
5 pts.
The length of the curve is given by the arc length integral. We have that r (t) = h− sin(t), cos(t), 3i, and
hence
q
√
|r0 (t)| = sin2 (t) + cos2 (t) + 9 = 10.
Hence,
Z
4π
0
Z
|r (t)| dt =
L=
0
0
2
4π
√
√
10 dt = 2π 10.
8. Find the extremal values, and the points where those values are obtained, of the function f (x, y) =
ln(x2 − y + 4) in the closed region D = {(x, y) | x2 + y 2 ≤ 1}.
8 pts.
We have that
∇f =
2x
−1
, 2
2
x −y+4 x −y+4
.
Let us first find interior points. These are critical points of the function f . For such points it holds that
∇f = h0, 0i. After multiplication with x2 − y + 4 we obtain
2x = 0
and − 1 = 0.
Since the latter equation is a contradiction, there are no critical points of the function f .
Let us now find the maximum and minimum on the boundary of D. The boundary is given by x2 +y 2 = 1.
Set g(x, y) = x2 + y 2 − 1 so that the boundary is given by g(x, y) = 0. Using Lagrange multipliers we
should solve the system ∇f = λ∇g and g = 0. This gives

2x
 x2 −y+4 = λ2x
−1
= λ2y
2
 x2−y+42
x +y =1
If x = 0 then the first equation says 0 = 0. Plugging x = 0 into the third equation we obtain y = ±1.
If x 6= 0, then we can divide the first equation by 2x. This gives that λ = 1/(x2 − y + 4). Plugging this
into the second equation we obtain
2y
= −1
x2 + y 2 = 1
√
Hence, y = −1/2 and x = ± 3/2.
All in all we have obtained four points to consider. We write the following table.
x
y
0
1
0
−1
√
√3/2 −1/2
− 3/2 −1/2
f (x, y)
ln(3)
ln(5)
ln(21/4)
ln(21/4)
As 21/4
√ > 20/4 = 5, we find that the maximum is ln(21/4), which is obtained at the two points
x = ± 3/2 and y = −1/2, while the minimum is ln(3) which is obtained at the point (0, 1).
3