Download Math 10c Practice Final Solutions: Integration, Series, and Vector Calculus and more Exams Calculus in PDF only on Docsity!
Math 10c - Practice Final Solutions
- Using the u-substitution u = − 0. 4 x,
p(x)dx =
- 4 e
− 0. 4 x dx = −
e
u du = −e
u
− 0. 4 x
We can now use the evaluation theorem to compute the integrals.
(a)
3
2
p(x)dx = (−e
− 0. 4 · 3 ) − (−e
− 0. 4 · 2 ) ≈. 15
(b)
2
0
p(x)dx = (−e
− 0. 4 · 2 ) − (−e
− 0. 4 · 0 ) ≈. 55
(c) P (t) =
t
0
p(x)dx = (−e
− 0. 4 ·t ) − (−e
− 0. 4 · 0 ) = 1 − e
− 0. 4 t
- (a)
2
- 5
p(x)dx =
1
4
2
- 5
x
3 dx =
1
4
2
4
4
- 5
4
4
(b)
2
0
x ·
x
3
4
dx = 1.6 hours.
(c) We solve for T :
1
2
T
0
x
3
4
dx =
1
16
T
4 − 0
4
⇒ 8 = T
4
So T =
4
8 ≈ 1 .68 hours.
- The series is geometric with a = 3 and r =
1
2
. Using the formula
∑ ∞
n=
ar
n
a
1 −r
for |r| < 1, we find that the sum of the series is 2.
- First compute four derivatives:
f
′ (x) = − sin x, f
′′ (x) = − cos x, f
(3) (x) = sin x, f
(4) (x) = cos x
Evaluating the function f at x = 0 and each of the derivatives as well we obtain
f (0) = 1, f
′ (0) = 0, f
′′ (0) = −1, f
(3) (0) = 0, f
(4) (0) = 1.
The corresponding approximation is P 4
(x) = 1 −
x
2
2
x
4
24
A cylinder of radius 3 that is centered about the z-axis.
(a) A plane parallel to the yz-plane that passes through the point (− 2 , 0 , 0).
(b) A cylinder of radius 1 centered about the y-axis.
- Let c = 2x
2
- y. Solving for y gives y = − 2 x
2
- c. The level curves will be
parabolas. (Be sure to sketch them and label them with the corresponding c
value).
- Substituting (0, 0 , 0) into the equation implies c = 0. Substituting (0, 1 , 3)
and solving for n gives n = 3. Finally substituting (− 1 , − 2 , 0) gives m = −6 so
that z = − 6 x + 3y.
i − 3
j +
k‖ =
- The vector
2 √
14
i −
3 √
14
j +
1 √
14
k
is a unit vector. Multiplying this vector by 3 gives the needed vector:
6 √
14
i −
9 √
14
j +
3 √
14
k
- Convert 200m/min to km/hr to obtain 200m/min = 12km/hr. Let ~v =
x 0
i + y 0
j + z 0
k be the velocity vector of the airplane. The plane is rising at a
rate of 12km/hr so that z 0
= 12. The airspeed of the plane is 200 km/hr so
‖~v‖ = 200. Computing ‖~v‖, we get the equation
x
2
0
2
0
2 = 200
The x-axis points east, the y-axis points north, and the plane is heading north-
west so that −x 0 = y 0 (drawing a picture may help you see this). Substituting
in for x 0 and solving for y 0 we get y 0 ≈ 141 .167. Therefore the velocity vector
of the plane is ~v = − 141. 167
i + 141. 167
j + 12
k.
- 2 ~v = 6
i − 14
j is parallel to ~v. If u = u 1
i + u 2
j, then ~u is perpendicular to ~v
if ~u · ~v = 0. This gives 3u 1 − 7 u 2 = 0. So 7
i + 3
j and − 7
i − 3
j are perpendicular
to ~v.
- Find the displacement vectors from P to Q and from P to R:
i +
j +
k
i + 2
j + 2
k
Then compute the cross product of these two vectors to get the vector 0
i+
j− 5
k.
The equation of the plane is 0 = 0(x − 1) + 5(y − 0) − 5(z − 1) which simplifies
to z = y + 1.
13) G
s
> 0: Our grade should improve if we increase the number of hours
studied. On the other hand, G t
< 0: Our grade will go down if we watch more
television.
- (a) f x
= 2 cos x cos 3y + y, f y
= −6 sin x sin 3y + x
(b) z x
= 14xy + 5y
- Let g(x, y) = 4x − 2 y. Then ∇f = 2x~i + 2y~j and ∇g = 4
i − 2
j. Solving
∇f = λ∇g and using the constraint, we obtain the equations:
2 x = 4λ 2 y = − 2 λ 4 x − 2 y = 15
Multiplying the second by -2 gives − 4 y = 4λ so combining this with the first
equation we obtain 2x = − 4 y which gives x = − 2 y. Substituting into the third
equation we find x = 3 and y = −
3
2
We compute f (3, −
3
2
) = 11.25 but we don’t know if this is a maximum or
a minimum so we compute f at another point that satisfies the constraint and
compare. The point (4,
1
2
) satisfies 4x − 2 y = 15 and f (4,
1
2
So there is a minimum at (3, −
3
2
). There is no maximum.
