Question

Question 6 of 10

Answer 1

Answer:

You need to get a tutor!!!!!

Step-by-step explanation:

√

7. [c

2 = a

2 + b

2 − 2ab cos C.] (5 marks)

2. x

−4/3 +

y

16 = 1. [Verify that the point is on the curve. Find slope

dy

dx = 12 (at that point)

and the tangent y + 8 = 12(x + 2). (5 marks)

Rearrange the equation to get it in intercept form, or solve y = 0 for x-intercept and x = 0

for y-intercept.] (5 marks)

3. One. [Show that g

0

(t) < 0 ∀t ∈ R, hence g is strictly decreasing. Show existence. Show

uniqueness.] (5 marks)

Comment: Graphical solution is also acceptable, but arguments must be rigorous.

4. 4π(9 − 2

√

6) or 16.4π or 51.54 cm3

. [V =

R 2π

0

R Rh

0

2

√

R2 − r

2 rdrdθ, R = 3, Rh =

√

3.

Sketch domain. (5 marks)

V = 4π

R Rh

0

√

R2 − r

2 rdr, substitute R2 − r

2 = t

2

. (5 marks)

Evaluate integral V = −4π

R

√

R2−R2

h

R t

2dt.] (5 marks)

5. (a) (2,1,8). [Consider p~ − q~ = ~s − ~r.] (5 marks)

(b) p

3/5. [cosQ =

QP~ ·QR~

kQP~ k kQR~ k

.] (5 marks)

(c) 9

5

(j + 2k). [Vector projection = (QP~ · QRˆ )QRˆ .] (5 marks)

(d) 6

√

6 square units. [Vector area = QP~ × QR~ .] (5 marks)

(e) 7x + 2y − z = 8.

[Take normal n in the direction of vector area and n · (x − q).] (5 marks)

Comment: Parametric equation q + α(p − q) + β(r − q) is also acceptable.

(f) 14, 4, 2 on yz, xz and xy planes. [Take components of vector area.] (5 marks)

6. 1.625%. [A = πab ⇒ dA

A = da

a + db

b

. (5 marks)

Put a = 10, b = 16 and da = db = 0.1.] (5 marks)

7. 9/2. [Sketch domain: region inside the ellipse x

2 + 4y

2 = 9 and above the x-axis. (5 marks)

By change of order, I =

R 3

−3

R 1

2

√

9−x2

0

ydydx