QUESTION
WAIKATO PATHWAYS COLLEGE
Mathematics with Calculus (CAFS004-11B)
Assignment 4
DATE DUE: Thursday, 13 October All working must be shown.
Question 1
a) i) Find the equation of the tangent to the curve when x = 1.
ii) Find the gradient of at the point where x = 3.
b) i) Find the coordinates of the turning points of the curve , and determine the nature of each one.
ii) Sketch the graph of .
c) The height of a seedling at the end of its first month is given by: for where H is the height, in cm, of the seedling at the end of the month, and w is the amount of water the seedling receives (litres per m2 of soil surface). Find the amount of water required for the seedling to grow to the maximum height in its first month. [You may assume that ]
Question 2
a) An air pump is blowing up a spherical balloon. The volume of the balloon
increases at a constant rate of 0.9 m3/s. Calculate the rate at which the surface area of the balloon is increasing when the radius is 2 m.
b) An island has a population of possums which are being trapped. The number of possums on the island is given by where P is the number of possums, and t is the time in months after the trapping begins. Find the rate of change in the population of possums after 3 months of trapping.
Question 3
a) Integrate with respect to :
i) ; ii) .
b) Find
i) ii) .
iii) .
c) A curve has gradient and passes through the point . Find the
equation of the curve.
Question 4 a) Evaluate: (i) ; (ii) . b) Use the substitution to evaluate.
c) Find the area enclosed by
i) the -axis and the curve in the range
.
ii) Find the area enclosed between the curves y = x2 – 2 and y = x.
iii) The area in c(i) above is completely rotated about the -axis. Find the volume
generated, giving your answer in terms of .
d) The area enclosed between y = x2– 9 is rotated about the x-axis. Calculate the
volume of the solid formed.
Question 5
a) Find the general solution for
i) ; ii) .
b) Find particular solutions for the following differential equations
i) , given that x = 2 when y = 5;
ii) , given that y = when x = 0.
Mathematics with calculus (CAFS004-11B)
Question 1 :
(a) (i) Tangent to the curve is given by dy/dx
Now,When x = 1 , y = 3 /((1)^2 -2) = 3/(1-2) = -3
Hence tangent passes through point (1,-3)
Now , when y = 3/(x^2 -2)
d(y)/ dx = d{3/(x^2-2)} = 3 ( 1/(x^2-2)^2) *d(x^2)/dx
= 3(2x)/(x^4 -4x+4)
= 6x /(x^4-4x+4)
When x = 1 , dy/dx = 6(1)/(1 -4 +4) = 6 /1 = 6
We got slope of tangent (at x = 1 ) = 6
Then,Equation of tangent through (1,-3) with slope 6
y – (-3) = 6(x-1) => y +3 = 6x -6 => y = 6x -9 Answer
(ii) Gradiant of line = slope of line = dy/dx
Now,Given that y = In( x^2-3)
So, dy/dx = d (In(x^2-3)/dx = 1/ (x^2-3)*d(x^2)/dx
= 2x/(x^2 -3)
So,Gradiant at x = 1 , dy/dx at x = 1 = 2(1)/(1^2 -3) = 2/(1-3) = 2/-2
Gradiant = -1 Answer
b.(i) y = 8x^2- x^4
dy/dx = 8(2x) – (4x^3) => dy/dx = 16x – 4x^3
Now,When dy/dx = 0 ,
16x-4x^3 = 0 , 4x(4-x^2) = 0 => either x = 0 or x = +2,-2
Now,When x = 0 , y = 8(0)^2 – 0^4 = 0 so, Point = (x,y) = (0,0)
Again ,when x = 2 , y = 8(2)^2 – 2^4 = 16
Point = (2,16)
Similarly ,when x = -2 , y = 8(-2)^2 –(-2)^4 = 8(4) -16 = 16
Point = (-2,16)
Hence turning points = (-2,16),(0,0) and (2,16) Answer
(i) Graph of y = 8x^2 –x^4
Step:1 Making of table.
|
X |
y |
| -3 | -9 |
| -2 | 16 |
| -1 | 7 |
| 0 | 0 |
| 1 | 7 |
| 2 | 16 |
| 3 | -9 |
Step: 2 , when we connect all these point ,we ill get the graph of given equation.
(c) H(w) = 12w – 1.2 e^0.2w
d(H(w))/dw = d(12w- 1.2e^0.2w)/dw
= 12 – 1.2 (0.2)e^0.2w
= 12 – 0.24 e^0.2w
When d(H(w))/dw = 0 , 12- 0.24e^0.2w = 0
e^0.2w = 12/0.24 = 50
0.2w = In(50) = 3.91 = > w = 3.91/0.2 = 19.56
Now,d^2(H(w))/dw^2 = – 1.2(0.2)(0.2)e^0.2w = 0.048e^0.2w
For w= 19.56 , 0.048e^0.2w > 0
Hence for w = 19.56 , H(w) is maximum,
And
Maximum H(w) = 12 (19.56) – 1.2 e^ 3.91
= 234.72 – 1.2(2.71)^3.91
= 234.72 – 59.168 = 175.55
And amount of water required = 19.56 liters /m^2 Answer
Question (2)
(a) dv/dt = 0.9 m^3/s
Surface area of balloon S = 4 (pi) r^2 , where r = radius (given that r =2 m)
And from formula of volume of spherical balloon
V = 4/3(pi) r^3
dv/dt = 4/3 (pi) (3r^2)dr/dt
0.9 = 4pi r^2 dr/dt
dr/dt = 0.9 /4(pi) (2^2) = 0.9 /(8*3.14) = 0.0358
Now s = 4 pi r^2
So, ds/dt = 4 pi (2r) dr/dt = 8 pi dr/dt
ds/dt = 8*3.14* 0.0358 = 1.8 m^2/s Answer
(b) p(t) = 5000 – 4t^2 – 1500In(2t+1)
Rate of change = dp(t)/dt
d p(t)/dt = d{5000- 4t^2 -1500In(2t+1)}/dt
= -8t – 1500/(2t+1)*2
= -8t -3000/(2t+1)
Now,When t = 3
Rate of change = dp(t)/dt = -8(3) – 3000(2*3+1) = -24 – 428.57 = – 552.57 Answer
Here ,negative sign mean that ,possums are decresing with time.
Question 3:
(a) (i) f(x) = 8e^4x +6sin(3x)
Integration of f(x) = Int{8e^4x +6sin(3x)}dx
8e^4x/4 +6(-cos(3x)/3 + c , Where c = Integretion constant
= 2e^4x -2cos(3x) + c Answer
(ii) f(x) = (x^4 +4)^2/4x^6
= (x^8 +8x^4 +16)/4x^6 => ¼{ x^8/x^6 +8x^4/x^6 + 16/x^6}
= ¼ {x^2 +8x^-2 +16x^-6}
Now, Integretion of f(x) = Integretion of ¼ {x^2 +8x^-2 +16x^-6}
= ¼ { x^3/3 -8x^-1 -16/5 x^-5} +c
( x^3)/12 – 2/x – 4/(5 x^5) + c Answer
b) (i) = 8x^4/4 – 6 /(-3x^3) +c
= 2x^4 +2/x^3 +c Answer
(ii) . = Int of { cos 4theta + cos theta } d (theta)
= sin(4 theta)/4 + sin(theta) +c Answer
(iii) . , Let (x^3 +7 ) = z , Then dz = 3x^2 ,dx , so ,x^2dx= dz/3,
Now Int of 1 /sqrtz dz = 2/3sqrtz +c
= 2/3 sqrt (x^3+7) + c Answer
(c) dy/dx = 8cos (2x)
Integrating both sides with respect to x , we get
y = 8 sin (2x)/2 +c = > y = 4sin(2x) +c
Now,Given that curve passes through point (pi/12 ,9)
So, 9 = 4 sin(2*pi/12) = 4 sin(pi/6) +c
9 = 4(1/2) +c
9 = 2+c = > c = 9-2 = 7
Hence Curve equation y = 4sin(2x) + 7 Answer
Question 4 :
(a) (i) ; = Int {4x^2 +20x +25} dx from x = -2 to x = 1
=[ 4/3x^3 +10x^2 +25x ] from x = -2 to x = 1
= 4/3 +10 +25 -(-32/3 +40 -50) = 57 Answer
(ii) = [9(-cot(3x)/3 ] from x = pi/18 to x = pi/9
-3 cot(pi/3) + 3cot (pi/6)
– 3/sqrt3 +3 sqrt3 = 2 sqrt3 Answer
(b) . Let Inx = z , dz = dx/x
Int z dz = z^2/2 = In(x)/2] from x = 1 to x = 2
= In2/2 – 0 = 0.346 Answer
(c) (1) Area A = Int of sin (x) dx from x = pi to x = 2 pi
= -cos(x) rom x = pi to x = 2pi
= – cos(2pi) +cos(pi) = -1 -1 = -2 = 2 units Answer
(ii) When y = x , x = x^2 -2 = > x^2 –x -2 = 0
(x-2)(x+1) = 0 , so , x = -1 and 2
Now , Area between 2 curves = x^3/3 -2x – x^2/2 from x = -1 to x = 2
Area = 8/3 -4 -2 + 1/3 -2 +1/2 = 3/2 Answer
(iii) Vollume = Pi (area)^2
= pi (2^2) = 4pi Answer
(d) Vollume v = pi (area)^2
V = pi (x^3/3 -9x +c)^2 Answer
Here c = Integretion constation.
Question 5:
(a)(i) ; dy/dx = 1+y
dy/(1+y) = dx
Now ,Integreting both sides ,We get
In(1+y) = x +c or, 1+y = e^x +c
Y = e^x +c-1 Y = e^x + c Answer
Here C = Integretion Constant .
(ii) . dy = – (x/x+2) dx
y = Int of –{ 1 -2/x+2} dx = – x +2 In(x+2) +c
Y = 2In(x+2) –x +c Answer
(b) (i) or 4ydy = xdx = > 2y^2 = x^2 /2 +c
Given That x = 2 y = 5 ,
2(5)^2 = (2)^2/2 +c , c = 48
So, 2y^2 = x^2/2 +48 or , y = ½ sqrt (x^2+96) Answer
(ii) => sin(y) = x^2/2 +c
Given that x = 0 , y = pi/2 sin(pi/2) = x^2/2 +c
C = 1 ,
Sin(y) = x^2/2 +1 Answer
GI51
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