CALCULUS Final SOLUTION

CALCULUS Final SOLUTION

Exam Set:

B

1. (i) To find the first partial derivative with respect to y,hold x constant to obtain

f_y(x, y) = ∂

∂y

px²+ y²= 2y 2p

x²+ y² = y px²+ y² The value of f_y(x, y) at the point (8, −6) is

f_y(6, 8) = 8

p(6)²+ (8)² = 8

√100 = 8 10 =4

5 = 0.8 (ii) Since limx→∞ln x

x =^∞_∞, you can apply L’Hopital’s Rule,as follows

x→∞lim lnx

x = lim

x→∞

d dx[lnx]

d dx[x]

= lim

x→∞

1 x

1 = lim

x→∞

1 x = 0

2. (i) Since f (x) = ax²(2 − x) is a probability density function on the interval [0, 2]. So, evaluate the following integral

1 = Z 2

0

ax²(2−x)dx = a Z 2

0

x²(2−x)dx = a 2 3x³−1

4x⁴

²

0

= a 16 3 − 4



= 4a 3 Thus, a = ³₄.

(ii)

P



0 ≤ x ≤1 2



= Z 1/2

0

3

4x²(2 − x)dx = 3 4

Z 1/2 0

x²(2 − x)dx = 3 4

 2 3x³−1

4x⁴

^1/2

0

= 3

4

 1 12− 1

64



= 13 256

3. (i) First you interchange the order of integration so that y is the outer variable ,then y will have constant bounds of integration given by 0 ≤ y ≤ 1.Solving for x in the equation y = √

x implies that the bounds for x are 0 ≤ x ≤ y².Thus

Z 1 0

Z 1

√x

sin y³+ 1 2



dydx = Z 1

0

Z y² 0

sin y³+ 1 2

 dxdy =

Z 1 0

sin y³+ 1 2

 y²dy

(u = y³+ 1

2 , y : 0 → 1 ⇒ du = 3

2y²dy, u : 1/2 → 1)

= Z 1

1/2

sin(u)2 3du = 2

3 Z 1

1/2

sin(u)du = −2

3 [cosu]¹_1/2

= −2

3 [cos(1) − cos(1/2)]

(ii) By integration by parts, we can written the integral as follows Z 3

0

xf⁰⁰(x)dx = [xf⁰(x)]³₀− Z 3

0

f⁰(x)dx = [xf⁰(x)]³₀− [f (x)]³₀= 3f⁰(3) − f (3) + f (0) By assumption f (0) = 5, f (3) = 5, and f⁰(3) = 4.Thus

Z 3 0

xf⁰⁰(x)dx = 3(4) − 5 + 5 = 12

4. (i) Let u = 1 + cos²t, du = −2sintcostdt,Then the integral Z sintcost

√1 + cos²tdt = −1 2

Z 1

√udu = −u^1/2+ C = −p

1 + cos²t + C

(ii)

Z ∞

−∞

e^x

(1 + e^x)³dx = lim

a→−∞

Z 0 a

e^x

(1 + e^x)³dx + lim

b→∞

Z b 0

e^x (1 + e^x)³dx (u = 1 + e^x, x : a → 0, x : 0 → b ⇒ du = e^xdx, u : 1 + e^a → 2, u : 2 → 1 + e^b)

= lim

a→−∞

Z 2 1+e^a

1

u³du + lim

b→∞

Z 1+e^b 2

1 u³du

= lim

a→−∞



− 1 2u²

2 1+e^a

+ lim

b→∞



− 1 2u²

1+e^b 2

= lim

a→−∞



−1

8 + 1

2(1 + e^a)²

 + lim

b→∞



− 1

2(1 + e^b)² +1 8



= −1

8 + lim

a→−∞

1

2(1 + e^a)²− lim

b→∞

1

2(1 + e^b)²+1 8

= −1 8 +1

2 +1 8 = 1

2 5. (i) y⁰ = y(1 − y) ⇒dy

dt = y(1 − y) ⇒ 1

y(1 − y)dy = dt Integrate both sides R dt = R _y(1−y)¹ dy.Then

t + C1= Z 1

ydy +

Z 1

1 − ydy = ln | y | −ln | 1 − y |= ln | y 1 − y |

⇒ lnC2e^t= ln | y

1 − y |⇒ Ce^t= y

1 − y ⇒ Ce^t− yCe^t= y

⇒ (1 + Ce^t)y = Ce^t⇒ y(t) = Ce^t 1 + Ce^t (ii) ¹₃ = y(0) = Ce⁽⁰⁾

1 + Ce⁽⁰⁾ = C

1 + C ⇒ C = ¹₂.Thus the solution for the equation with initial condition is y(t) = e^t

2 + e^t

6. (i) When the ball hits the ground the first time, it has traveled a distance of

D₁= 18

.Between the first and second times it hits the ground, it has traveled an additional distance of

D2= 18 7 10



+ 18 7 10



= 36 7 10



Between the second and third times the ball hits the ground, it has traveled an additional distance of

D3= 18 7 10

  7 10



+ 18 7 10

  7 10



= 36 7 10

²

Continuing this process,you obtain a total vertical distance traveled of

D = 18 + 36 7 10



+ 36 7 10

2

+ · · · = −18 + 36 + 36 7 10



+ 36 7 10

2

+ · · ·

= −18 +

∞

X

n=0

36 7 10

n

= −18 + 36 1 − 7

10

 = −18 + 120 = 102

(ii) The infinite series

∞

X

k=1

1

√3

k² = 1 1^2/3 + 1

2^2/3+ 1 3^2/3 + · · ·

is a p-series with p = ²₃.Because p < 1,you can conclude that the series divergence.

7. (i) The first derivative of f is f⁰(x) = 3x². Thus ,the iterative formula for Newton’s Method is

x_n+1= x_n− f (xn)

f⁰(xn) = x_n−x³− 66 3x²

The calculations for two iterations are shown in the table.

n x_n f (x_n) f⁰(x_n) _f^{f (x}0(xⁿn⁾) x_n−_f^{f (x}0(xⁿn⁾)

1 4 −2 48 0.04167 4.04167

2 4.04167 0.02107 49.00529 0.00043 4.0412 3 4.0412

Thus,the approximation is √³

67 = 4.0412

(ii) Begin by finding several derivatives of f and evaluation each at c = 64 g(x) = x^1/3 g(64) = 4

g⁰(x) = 1

3x^−2/3 g⁰(64) = 1

3(0.0625) = 0.02083 g⁰⁰(x) = −2

9 x^−5/3 g⁰⁰(64) = −2

9 (0.00098) = −0.00011 Thus, the two-degree Taylor polynomial is

g(x) = g(64) + g⁰(64)(x − 64) +g⁰⁰(64)(x − 64)²

2! = 4 + 0.02083(x − 64) − 0.00011(x − 64)² To evaluate the series when x = 66.

g(66) = 4+0.02083(66−64)−0.00011(66−64)²= 4+0.02083(2)−0.00011(4) = 4.04122

8. (i)

E[X] = 0P (X = 0) + 1P (X = 1) + 2P (X = 2) + 3P (X = 3) + · · ·

= e⁻²2¹

1! + 2e⁻²2²

2! + 3e⁻²2³

3! + · · · = 2e⁻²(1 + e⁻²2¹

1! + e⁻²2² 2! + · · ·)

= 2e⁻²

∞

X

n=0

2ⁿ

n! = 2e⁻²e²= 2 (ii)

µ =

Z ∞ 0

x 5e^−5x
dx = lim

a→∞

Z a 0

x 5e^−5x
dx

= lim

a→∞



−xe^−5x 

a 0+

Z a 0

e^−5xdx



= lim

a→∞



−ae^−5a−1 5e^−5x

a

0



= lim

a→∞



−ae^−5a− 1 5e^5a +1

5



= lim

a→∞

−a e^−5a − lim

a→∞

1 5e^5a +1

5

= − lim

a→∞

1 5e^5a +1

5 = 1 5 Z ∞

0

x² 5e^−5x
dx = lim

a→∞

Z a 0

x² 5e^−5x
dx

= lim

a→∞



−x²e^−5x 

a 0−2

5 xe^−5x 

a 0+2

5 Z a

0

e^−5xdx



= lim

a→∞

 −a² e^−5a − 2a

5e^5a −2 5 e^−5x

a 0



= lim

a→∞

−a² e^5a − lim

a→∞

2a

5e^5a − lim

a→∞

2 25e^5a + 2

25

= − lim

a→∞

2a

5e^4a − lim

a→∞

2 25e^5a + 2

25 = − lim

a→∞

2 25e^5a + 2

25

= 2

25 Thus

σ = r2

25− (1 5)²=

r1 25= 1

5 9. Finding several derivatives of f and evaluation each at c = 0

f (x) = ln

1+x 1−x

= ln(1 + x) − ln(1 − x) f (0) = 0 f⁰(x) = 1

1 + x− −11 − x = 2

1 − x² f⁰(0) = 2 f⁰⁰(x) = 0 − (−2x)

1 − x^2
2 = 2x

1 − x^2
2 f⁰⁰(0) = 0 f⁽³⁾(x) = 2 1 − x²
2

− 2x(2(1 − x²)(−2x))

1 − x²
4 = 2 + 6x²

1 − x²
3 f⁽³⁾(0) = 2 Continuing this process,we can see that f⁽²ⁿ⁻²⁾(0) = 0, f⁽²ⁿ⁻¹⁾(0) = 2, Thus the Taylor series

f (x) = f (0) + f⁰(0)x +f⁰(0)x²

2! +f⁰(0)x³

3! +f⁰(0)x⁴ 4! + · · ·

= 2x + 2 3!x³+ 2

5!x⁵+ · · · =

∞

X

n=0

2x²ⁿ⁺¹ (2n + 1)!

For this power series, an= 2 (2n + 1)!

n→∞lim     

an+1x^2(n+1)+1 a_nx²ⁿ⁺¹

= lim

n→∞

2

(2n + 3)!x²ⁿ⁺³ 2

(2n + 1)!x²ⁿ⁺¹        

= lim

n→∞

x² (2n + 3)(2n + 2)

= 0

So,by the Ration Test, this series converges for all x and the radius of convergence is ∞

10. Let T (x, y, z) = 8x²yz and g(x, y, z) = x²+ y²+ z²− 1.Then, define a new function F (x, y, x, λ) by

F (x, y, x, λ) = T (x, y, z) − λ(x, y, z) = 8x²yz − λ x²+ y²+ z²− 1
To find the critical numbers of F , set the partial derivatives of F with respect to x,y,z,and λ equal to zero and obtain

F_x(x, y, x, λ) = 16xyz − 2λx = 0, F_y(x, y, x, λ) = 8x²z − 2λy = 0, Fz(x, y, x, λ) = 8x²y − 2λz = 0 , Fλ(x, y, x, λ) = −x²− y²− z²+ 1 = 0

Then

Fx

F_y : 2y x =x

y,Fy

F_z : z y =y

z,Fx

F_z : 2z x =x

z

⇒ x²= 2y², z²= y²

Substitute this into the equation Fλ(x, y, x, λ) = −x²− y²− z²+ 1 = 0 and solve y

0 = Fλ(x, y, x, λ) = −2y²− y²− y²+ 1 ⇒ y²= 1 4 = ±1

2 Using this y-value,you can conclude that the critical values are

y = 1

2 ⇒ x = ± 1

√2, z = ±1 2 x = −1

2 ⇒ x = ± 1

√2, z = ±1 2 which implies that the temperature value is

T



± 1

√ 2,1

2,1 2



= 8(1/2)(1/2)(1/2) = 1, T



± 1

√ 2, −1

2,1 2



= 8(1/2)(−1/2)(1/2) = −1,

T



± 1

√2,1 2, −1

2



= 8(1/2)(1/2)(−1/2) = −1, T



± 1

√2, −1 2, −1

2



= 8(1/2)(1/2)(1/2) = 1, Thus,the point(s) on the sphere at which the temperature is greatest is



± 1

√2,1 2,1

2

 ,



± 1

√2, −1 2, −1

2



the point(s) on the sphere at which the temperature is least is



± 1

√ 2, −1

2,1 2

 ,



± 1

√ 2,1

2, −1 2