Memorandum MTHS211 Semester test 1 2021
p. 877. This table is very important - you should know the name of each surface,
the equation and you should be able to draw the surfaces.
1. Identify the surfaces.
(a) z = 3 − x2 − y 2 Elliptic paraboloid 3
p
(b) z = 2x2 + 2y 2 Cone (top half) 3
(c) z = 3 − x Plane 3
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1
p. 875
2. (a) Identify the surface 9x2 + 4y 2 + z 2 = 1
2
2
x2
+ y1 2 + z12 = 1 Ellipsoid 3
12
3
2
(b) Find and identify the traces with the xy−plane, z = 1 and the horizontal trace where k = 4.
(i) With xy plane z = 0
2
x2
+ y1 2 = 13 Ellipse 3
12
3
2
(ii) If z = 1,
x2
12
3
+
y2
12
2
x2
12
3
+
y2
12
2
+1=1
=03
→ (x, y) = (0, 0). This is a point (0, 0, 1) (the red dot in the figure) 3
(iii) If z = 4,
=⇒
x2
12
3
+
y2
12
2
x2
12
3
+
y2
1 2 42
+ 2
2
1
=1
= −153
No intersection with the surface =⇒ No trace.3
2
(c) Sketch the surface.
3 Show intercepts!
[8]
p. 890
3. Find the vector function equation that represent the curve of intersection, C of the two surfaces
3
z = x2 − y 2 and x2 + y 2 = 1.
The projection of the curve C of the intersection on the xy-plane =⇒ x2 + y 2 = 1, is a circle.
Use hx, yi = hcos t, sin ti, t ∈ [0, 2π]3.
But C also lies on the surface z = x2 − y 2 .
=⇒ the vector function hx, y, zi = hcos t, sin t, cos2 t − sin2 ti333OR
hcos t, sin t, cos(2t)i, t ∈ [0, 2π]
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4. Suppose that z = f (x, y) is a differentiable function of x and y, where x = g(t) and y = h(t) are
both differentiable functions of t. that z is a differentiable function of t and
dz
∂f dx ∂f dy
=
+
dt
∂x dt
∂y dt
Chain rule (Case I), pg 978
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4
p. 971
5. (a) Determine the first order Taylor polynomial of f (x, y) = ex sin y at the point (0, 0).
L(x, y) = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b)
f (0, 0) = e0 sin 0 = 0
fx = ex sin y =⇒ fx (0, 0) = e0 sin 0 = 0
fx yex cos y =⇒ fy (0, 0) = e0 cos 0 = 1
=⇒ T1 (0, 0) = 0 + 0(x − 0) + 1(y − 0) = y
(b) Determine the linear approximation of f (x, y) = f ( 12 , π6 ).
L( 21 , π6 ) = T1 ( 12 , π6 ) = π6 3≈ 0.523
6. State Fubini’s Theorem.
If f is continuous on the rectangle R = {(x, y)|a ≤ x ≤ b, c ≤ y ≤ d} then
RR
RbRd
RdRb
R f (x, y)dA = a c f (x, y)dydx = c a f (x, y)dxdy33
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[2]
p. 990 When calculating the directional derivative, the direction vector should be a unit vector .
5
p. 993
7. (a) Find the rate of change of the voltage at point (3, 4, 5) in the direction of the vector
h1, 1, −1i.
∇V (x, y, z) = h10x − 3y + yz, −3x + xz, xyi
∇V (3, 4, 5) = h38, 6, 12i
√
u = h1,1,−1i
3
√
Du (x, y, z) = h38, 6, 12i · h1,1,−1i
= √13 (38 + 6 − 12)
3
√
38
38
− √63 OR √
− 2 3 (= 18.4752V )
=√
3
3
(b) In which direction does the voltage change most rapidly at point (3, 4, 5)?
h38, 6, 12i 3or any vector in the same direction e.g √38h38,6,12i
2 +62 +122
(c) What is the maximum
rate of change of the voltage at point (3, 4, 5)?
√
2
|∇V (3, 4, 5)| = 38 + 62 + 122 3
6
[4]
p. 979
8. The temperature at a point (x, y) is T (x, y), √
measured in degrees Celsius. A bug crawls so that
its position after t seconds is given by x = 1 + t, y = 2 + 13 t, where x and y is in cm. The
temperature function satisfies Tx (2, 3) = 4 and Ty (2, 3) = 3. Determine how fast the temperature
is rising on the √
bug’s path after 3s?
t = 3 =⇒ x = 1 + 3 = 2cm, y = 2 + 13 3 = 3cm,
dx
dx
1 dy
1
√1
dt = 2 1+t =⇒ dt |t=3 = 4 , dt = 3
∂T dx
Chainrule: dT
dt = ∂x dt +
1
1
∂T
∂x = = 4 4 + 3 3 3= 23
∂T dy
∂y dt
[2]
p. 1046
9. Consider the integral
R1R2
0
1
1
dxdy.
4
2y 3 x +1
Evaluate the integral by reversing the order of integra7
tion.
You have to plot the region!
(3for plotting the region)
R2R
x3
8
0
0
=
R2
=
=
=
=
1
dydx(33for
1+x4
correct limits)
x3
8
y
0 [ 1+x4 ]0 dx (3for integration)
R
1 2 x3
8 0 1+x4 dx3
1 1
4 2
8 ( 4 [Ln|1 + x |]0 ) (3for integration)
1
32 (Ln(17) − Ln(1))3
Ln(17)
32
[7] (p. 1055 Paragraph 15.3 nr. 17)
10. Convert the equation x2 + (y − 1)2 = 1 to polar coordinates.
(a) r2 (cos2 θ + sin2 θ) − 2r sin θ + 1 = 1
r2 = 2r sin θ
r = 2 sin θ3
You have to plot the area!
(3for plotting the area)
Determine θ1 and θ2 :
1 = 2 sin θ
8
θ1 = π6 and θ2 = π − π6 = 5π
6 because the sine function is positive in the first and second
quadrants OR (easier) use θ2 = π2 and double calculated area. If you did not calculate
the angles correctly you would have only lost 1 mark!
RR
RR
1
dA
=
rdrdθ just as to cal(b) To calculate Area with a double integral A
=
RRR
culate Volume with a triple integral: V =
1 dV
R θ= π R r=2 sin θ
rdrdθ( 33for correct limits)
A = 2 θ= π2 r=1
R θ=6π2 1 2 r=2 sin θ
A = 2 θ= π [ 2 r ]r=1
dθ( 3for integration)
R θ=6π2 1
A = 2 θ= π ( 2 (4 sin2 θ) − 12 )dθ
R θ= π2 6
A = θ= π (4 sin2 θ − 1)dθ
R θ=6π
A = θ= π2 (2 − 2 cos(2θ) − 1)dθ (3for using sin2 θ = 1−cos(2θ)
)
2
R θ=6π2
A = θ= π (1 − 2 cos(2θ))dθ(simplify)
6
θ= π
A = [θ − sin 2θ]θ= π2 ( for integration)
6
A = ( π2 − sin(π) − ( π6 − sin( π3 ))
√
A=
3
2
+
π
3
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OR
A=
R r=2 sin θ
R θ= 5π
6
A=
R θ= 5π
6
A=
A=
A=
θ= π6
r=1
rdrdθ
1 2 r=2 sin θ
dθ
θ= π6 [ 2 r ]r=1
R θ= 5π
1
1
2
6
θ= π6 ( 2 (4 sin θ) − 2 )dθ
R θ= 5π
1
2
6
θ= π6 (2 sin θ − 2 )dθ
R θ= 5π
1
t
6
θ= π6 (1 − cos 2θ − 2 )dθ = [ 2
θ=5π/6
− 21 sin 2t]θ= π
2
√
=
3
4
+
5π
12
√
− − 43 +
π
12
√
=
3
2
+
π
3
(p. 1048 Paragraph 15.2 nr. 25)
11. Set up the integral to determine the volume of the solid under the surface z = xy and above the
triangle with vertices (1, 1), (4, 1) and (1, 2). You do not have to evaluate the integral.
9
You have to plot the region!
(3for plotting the region)
− 31 x
R x=4 R y=− 13 x+ 37
7
3 3 x=1 y=1
Type I: y =
+
xydydx (33for correct limits)
OR
R x=2 R x=7−3y
Type II: x = 4 − 3y then y=1 x=1
xydydx
10
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