Exam 1, Practice: MATH 251 Section 507
Jens Forsgård
February 2016
All answers need to be justified.
All use of electronic devices are prohibited.
Final answers should be boxed.
1. Find a parametric representation and the symmetric equations of the line passing through the points
P1 = P1 (1, −1, −7) and P2 = P2 (3, 0, 4).
2. Find a parametric representation and the symmetric equations of the line passing through the points
P1 = P1 (1, 1, 0) and P2 = P2 (0, 3, 4).
3. Find a parametric representation of the line which is the intersection of the hyperplanes x+5y +5z = 3
and 5x − 2y = 2.
4. Find a parametric representation of the line which is the intersection of the hyperplanes 2x+3y +z = 3
and x − z = 8.
5. Let C be the space curve with equations x(t) = 1, y(t) = et − 2 and z(t) = cos(sin(t)).
(a) Find the point where C intersects the xz-plane.
(b) Find a parametric representation of the tangent line to C at the point given by t = π.
6. Let C be the space curve with equations x(t) = cos(t), y(t) = t3 − 1 and z(t) = ln(t).
(a) Find the point where C intersects the xz-plane.
(b) Find a parametric representation of the tangent line to C at the point given by t = 1.
7. Find the intersection point and the angle of intersection between the two space curves
r1 (t) = ht, t2 + 1, t3 + 4t2 + 5ti and r2 (s) = hes − 1, 1, ln(s + 1)i.
8. Find the intersection point and the angle of intersection between the two space curves
r1 (t) = ht, t − 1, sin(πt)i and r2 (s) = hs2 , 1 + s, ln(s + 2)i.
9. Let f (x, y, z) = (x2 − 2y + z 2 )2 − 1.
1
(a) Find quadratic equations which describe the level curves of the function f (x, y, z).
(b) Identify the quadratic surfaces defined by the equations from (a).
(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.
10. Let f (x, y, z) = ln (2x2 + 2y 2 + z 2 )2 .
(a) Find quadratic equations which describe the level curves of the function f (x, y, z).
(b) Identify the quadratic surfaces defined by the equations from (a).
(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.
11. Let f (x, y, z) = ln(x2 − y 2 + z 2 ).
(a) Find quadratic equations which describe the level curves of the function f (x, y, z).
(b) Identify the quadratic surfaces defined by the equations from (a).
(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.
12. Find the domain of the function f (x, y) = ln(x2 − y 2 ), and sketch it.
13. Find the domain of the function f (x, y) = ln(x2 − 2y 2 + 1), and sketch it.
14. Let f (x, y) = ln(x2 ) + x40 y 2 .
(a) Compute the gradient ∇f for any (x, y).
(b) Find a parametric representation of the normal line to the surface with equation z = f (x, y) at the
point (1, 1, 1).
(c) Find an equation of the tangent plane to the surface with equation z = f (x, y) at the point (1, 1, 1).
15. Let f (x, y) = x14 + exy .
(a) Compute the gradient ∇f for any (x, y).
(b) Find a parametric representation of the normal line to the surface with equation z = f (x, y) at the
point (1, 1, 1 + e).
(b) Find the equation of the tangent plane to the surface S with equation z = f (x, y) at the point
(1, 1, 1 + e).
16. Let f (x, y) = x2 + xy 2 .
(a) Compute the gradient ∇f for any (x, y).
(b) Find a parametric representation of the normal line to the surface with equation z = f (x, y) at the
point (1, 1, 2).
2
(c) Find the equation of the tangent plane to the surface S with equation z = f (x, y) at the point (1, 1, 2).
17. Leia is situated on the surface z = f (x, y) = sin(x) ln(y) at the point (x, y) = (π, 2). In which
direction should she move to descend as fast as possible?
18. Let f (x, y) = x3 + xey + ln(x2 y) and let P = P (1, 1).
(a) Compute the directional derivative of f at P in the direction of v = h1, −1i.
(b) Find the maximal rate of change of f at P .
19. Let f (x, y) = x + xexy and let P = P (1, 1).
(a) Compute the directional derivative of f at P in the direction of v = h1, −1i.
(b) Find the maximal rate of change of f at P .
20. Let f (x, y) = x2 + xy 2 + ln(xy) and let P = P (1, 1).
(a) Compute the directional derivative of f at P in the direction of v = h1, −1i.
(b) Find the maximal rate of change of f at P .
21. Let K(t) denote a cube with side lengths x(t), y(t), and z(t) respectively. If x(t) increases by 1 cm/s
and y(t) decreases by 2 cm/s while z(t) is constant, what is the rate of change of the volume of K(t) at
t = 0 if x(0) = y(0) = z(0) = 2?
√
√
√
√
22. Give a rational approximation of the number ( 101− 120)2 which is better than ( 100− 121)2 =
(10 − 11)2 = 1.
23. Compute the length of the space curve with equations r(t) = ht3/2 , 2t, 2i for 0 ≤ t ≤ 1.
24. Compute the length of the space curve with equations r(t) = t3 , t2 ,
2t
3
for 0 ≤ t ≤ 1.
25. Compute the length of the space curve with equations r(t) = t, t2 , 32 t3 for 0 ≤ t ≤ 1.
26. Find and classify all critical points of the function f (x, y) = 2x3 + xy 2 + 5x2 + y 2 + 8.
27. Find the extremal values, and the points where those values are obtained, of the function
f (x, y) = x2 + 4xy + y 2
in the region D = {(x, y) | x2 + y 2 ≤ 1}.
3
28. Find the extremal values, and the points where those values are obtained, of the function
f (x, y) =
1
8 + 4x − 3y
in the region D = {(x, y) | x2 + y 2 ≤ 1}.
29. Find the extremal values, and the points where those values are obtained, of the function
f (x, y) =
1
8 + 4x2 − 3y 2
in the region D = {(x, y) | x2 + y 2 ≤ 1}.
30. Find the extremal values, and the points where those values are obtained, of the function f (x, y) =
ex (y 2 − 2x) under the restriction that y 2 + x2 = 2.
31. Find the extremal values, and the points where those values are obtained, of the function f (x, y) =
ex y 2 under the restriction that y 2 + 4x2 = 5.
Solutions to selected exercises:
2. To find a parametric representation L(t) = L0 + tv of a line L, we need (the position vector of) a
point L0 on the line, and a directional vector v parallel to the line. In this case, we choose L0 = h1, 1, 0i
and v = h0, 3, 4i − h1, 1, 0i = h−1, 2, 4i. To find the symmetric equations, we set hx, y, zi = L(t) and
z
solve componetwise for t. This yields t = 1 − x, t = y−1
2 , and t = 4 .
6. The directional vector of the line is given by the cross product of the normal vectors of the planes.
That is, v = h2, 3, 1i × h1, 0, −1i = h−3, 3, −3i. To find a point on the line, we choose z = 0. The
second equation then gives that x = 8. Plugging these two values into the first equation, we obtain that
3y = 3 − 2 · 8, that is, y = − 13
3 .
12. (a) The level curves of f is given by f (x, y, z) = k, where k is a constant. We obtain that ln(x2 − y 2 +
z 2 ) = k. To get rid of the irritation logarithm, we exponentiate both sides, obtaining x2 − y 2 + z 2 = ek .
(b) Since ek > 0 for all k, these are one-sheeted hyperboloids.
17. (a) The gradient is given by ∇f (x, y) = h2x + y 2 , 2xyi. (b) The normal vector is given by n =
h−fx0 (1, 1), −fy0 (1, 1), 1i = h−3, −2, 1i. This is the directional vector of the normal line. (c) Using the
normal vector from (b), we know that the plane has the equation −3x − 2y + z = k for some constant
k. Since (1, 1, 2) is contained in the plane, we find that k = −3 − 2 + 2 = −3.
√
21. (a) The vector vD is not a unit vector. Normalizing,
we obtain that u = h1, −1i/ 2. The gradient of
E
f is equal to ∇f = 2x + y 2 + x1 , 2xy + y1 . Thus, ∇f (1, 1) = h4, 3i. The directional derivative is given
√
√
by Du f (1, 1) = ∇f (1, 1) • u = h4, 3i • h1, −1i/ 2 = 1/ 2. (b) The maximal rate of change of f at P is
given by |∇f (1, 1)| = |h4, 3i| = 5.
R1
26. The length is given by L = 0 |r0 (t)|dt. We have that r0 (t) = h1, 2t, 2t2 i. Thus, |r0 (t)| =
p
√
1 + 4t2 + 4t4 = (1 + 2t2 )2 = |1 + 2t2 | = 1 + 2t2 , since the last expression is aways positive. Therefor
1
R1
L = 0 1 + 2t2 dt = t + 32 t3 0 = 1 + 32 = 53 .
Answers:
1. L(s) = h1, −1, −7i + sh2, 1, 11i and
answers.
x−1
2
2. L(t) = h1, 1, 0i+th−1, 2, 4i and 1−x =
=y+1=
y−1
2
z+7
11 .
Note that this exercise has several correct
= z4 .Note that this exercise has several correct answers.
3. L(s) = 0, −1, 85 + sh10, 25, −27i. Note that this exercise has several correct answers.
4. L(t) = 8, − 13
3 , 0 + th−3, 3, −3i. Note that this exercise has several correct answers.
5. (a) h1, 0, cos(sin(ln(2)))i. (b) L(s) = h1, eπ − 2, 1i + sh0, eπ , 0i. Note that (b) has several correct
answers.
6. (a) hcos(1), 0, 0i. (b) L(s) = hcos(1), 0, 0i + sh− sin(1), 3, 1i. Note that this exercise has several
correct answers.
7. The intersection point is P (0, 1, 0). The angle of intersection is θ = arccos √313 .
8. The intersection point is P (1, 0, 0). The angle of intersection is θ = arccos
9. (a) x2 − 2y + z 2 =
√
√−π−1
12+6π 2
.
k + 1, where k + 1 ≥ 0. (b) These are elliptic paraboloids. (c) –
k
10. (a) 2x2 + 2y 2 + z 2 = e 2 . (b) These are ellipsoids. (c) –
11. (a) x2 − y 2 + z 2 = ek . (b) These are one-sheeted hyperboloids. (c) –
12. The domain is x2 − y 2 > 0. Notice that x2 − y 2 = (x − y)(x + y). Thus, the boundary of the domain
is two lines through the origin. The domain consist of the interior of the two cones containing the
x-axis.
13. The domain is x2 − 2y 2 + 1 > 0. Its boundary is a hyperbola, and the domain is the region
containing the origin.
14. (a) ∇f (x, y) = x2 + 40x39 y 2 , 2x40 y . (b) L(s) = h1, 1, 1i + sh−42, −2, 1i. (c) −42x − 2y + z = −43.
Note that (b) and (c) has several correct answers.
15. (a) ∇f = h14x13 + yexy , xexy i. (b) L(s) = h1, 1, 1 + ei + sh−14 − e, −e, 1i. (c) −(14 + e)x − ey + z =
−13 − e. Note that (b) and (c) has several correct answers.
16. (a) ∇f = h2x + y 2 , 2xyi. (b) L(t) = h1, 1, 2i + th−3, −2, 1i. (c) −3x − 2y + z = −3. Note that (b)
and (c) has several correct answers.
17. u = h1, 0i (which is the unit vector with the same direction as, e.g., hln(2), 0i).
√
√
18. (a) 2 2 (b) 2e2 + 12e + 26.
√
√ . (b)
1 + 4e + 5e2 .
19. (a) 1+e
2
√
20. (a) 1/ 2 (b) 5.
21. −4 cm3 /s.
22. 1 −
23.
61
27 .
24.
5
3.
25.
5
3.
1
10
−
1
11 .
26. (0, 0) is a local minimum, (−5/3, 0) is a local maximum, (−1, −2) and (−1, 2) are saddle points.
27. Maxima is 3, which is obtained at
√1 , − √1
√1 , √1 .
and
−
2
2
2
2
√1 , √1
2
2
and −
√1 , √1
2
2
28. Maxima is 13 , which is obtained at − 54 , 35 . Minima is
1
13 ,
. Minima is −1, which is obtained at
which is obtained at
4
3
5, −5
.
1
, which is obtained at (1, 0) and
29. Maxima is 15 , which is obtained at (0, 1) and (0, −1). Minima is 12
(−1, 0).
√
√ √2
√
30. Maxima
√ is 2, which is obtained at (0, 2) and (0, − 2). Minima is −2 2e , which is obtained
at ( 2, 0).
√
1
1
31. Maxima
is
4
,
2
and
,
−2
. Minima is 0, which is obtained at
e,
which
is
obtained
at
2
2
√ 5
± 2 ,0 .
7