# Functions And Calculus Coursework Assignment Sample

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## Functions And Calculus Coursework Assignment Sample

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## Introduction

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### Question 1

(a)
To convert from radians to the degrees, an angle in terms of radians has to be converted in the degrees by multiplying the angle by (180°/ Pi ).
Pi / 7 radians = (180^0/ Pi ) * Pi /7
= [25.714]^0
-1.245 radians = [(180]^0/ Pi ) * (-1.245)
= -71.333 °
(b)
To convert from the degrees to radians, an angle in terms of degree has to be converted in the radian by multiplying the angle by ( Pi /180^0).
50^0= [50^0 * Pi / 180^0] radians
200^0= [200^0 * Pi / 180^0] radians
= 10 Pi /9

### Question 2

y= - x^2 + x+2 … (i)
Roots: The roots of the above quadratic equation can be found be factorizing the avove equation.
y= - x^2 + x+2
= - x^2 + (2-1) x+2
= - x^2 +2 x - x +2
= -x(x-2) - 1(x-2)
= (x-2)(-x-1)
Either, x-2 = 0
Therefore, x=2
Or, -x-1 = 0
Therefore, x=-1
The roots of the quadratic equation are 2 and -1.
Sketching of the curve:

Graph 1: graph of y= - x^2 + x+2
(Source: self created on geogebra.org)
The graph of a quadratic equation is a parabola as shown in the above graph. By equating the equation (i) with the equation of quadratic faction ax^2 +bx+c = 0, get, a = -1, b= 1, and c = 2. By putting all the values the above graph is determine.
X intercepts- x-intercepts are +2 and -1
The gradient of y is the slope of the equation which can be found by differentiating the equation.
By differentiating equation (i) with respect to x, get
dy/dx = -2x +1 (Since, d/dx (x^n) = nx^(n-1) and d/dx (k)= 0, where k is a constant)
Therefore, the gradient of y (dy/dx) at x= 2
= -2*2 +1
= -4 + 1
= -3
The gradient of y (dy/dx) at x= -1
= -2*(-1) + 1
= 2+1
= 3

### Question 3

(a)
y= 2x^3 - 4x+ 3
By differentiating equation with respect to x, get
dy/dx = 6 x^2 - 4 (Since, d/dx (x^n) = nx^(n-1) and d/dx (k)= 0, where k is a constant)
= 6 x^2 - 4
(b)
y = sinx - 2 e^x
By differentiating equation with respect to x, get
dy/dx = cosx - 2 e^x (Since, d/dx (sinx) = cosx and d/dx (e^x) = e^x)
= cosx - 2 e^x
(c)
y = - 3/x^2 + √x - lnx
= -3x^(-2) + x^(1/2) - lnx
By differentiating equation with respect to x, get
= -3*(-2) x^(-2-1) + ½ x^(1/2-1) - 1/x (since, d/dx (x^n) = nx^(n-1) and d/dx (lnx) = 1/x)
= 6 x^(-3) + ½ x^(-1/2) - 1/x
= 6/x^3 + 1/2 √x - 1/x

### Question 4

(a)
f(t) = (t^2-2) cost
= t^2 cost - 2 cost
By differentiating equation with respect to t, and apply the product rule of differentiation, i.e. d/dx(f(x) g(x)) = g(x)d/dx (f(x) + f(x) d/dx g(x), get
d/dt (f(t)) = 2t cost + (-sint) t^2 - 2 (-sint) (since, d/dx (x^n) = nx^(n-1) and d/dx (cosx = - sinx)
d/dt (f(t)) = 2t cost - t^2 sint + 2 sint
= (- t^2 + 2) sint + 2t cost
(b)
g(t) = (tant)e^t
By differentiating equation with respect to t, and apply the product rule of differentiation, i.e. d/dx(f(x) g(x)) = g(x)d/dx (f(x) + f(x) d/dx g(x), get
d/dt (g(t)) = e^t [sec]^(2 )t + (tant)e^t (since, d/dx(tanx) = [sec]^(2 )x and d/dx (e^x) = e^x )
= e^t([sec]^(2 )t +tant)

### Question 5

(a)
y = lnx/ (7x - 1)
By differentiating equation with respect to t, and apply the quotient rule of differentiation, i.e.
d/dx(f(x)/g(x)) = [g(x)d/dx (f(x) - f(x) d/dx g(x)]/ (g(x))^2, get
d/dx (y)
= [(7x - 1) d/dx (lnx) - lnx d/dx(7x - 1)] / [(7x - 1) ]^2
= [(7x - 1) 1/x - lnx (7)]/ [(7x - 1) ]^2 (since, d/dx (x^n) = nx^(n-1) and d/dx (lnx) = 1/x, and d/dx (k) = 0, where k is a constant)
= [(7x - 1)/x - 7lnx]/ [(7x - 1) ]^2
(b)
y = 3√x/ cosx
By differentiating equation with respect to t, and apply the quotient rule of differentiation, get
dy/dx = d/dx (3√x/ cosx) = 3 [cosx d/dx (x^(1/2)) - √x d/dx (cosx)] [[(cos][x)]]^2
= 3[cosx ½ x^(1/2-1) - √x ([-sin]x)]/ [[(cos][x)]]^2 (since, d/dx (x^n) = nx^(n-1) and d/dx (cos[x)] = -sinx)
= [[cos]x/2√x + √x sinx]/ [[(cos][x)]]^2
= [[cos]x + 2xsinx]/ 2√x [[(cos][x)]]^2\

### Question 6

(a)
y = sin[(t^3- 2 t^2+ 1)]
dy/dt = d/dt (sin[(t^3- 2 t^2+ 1)])
= cos[(t^3- 2 t^2+ 1)] * d/dt (t^3- 2 t^2+ 1) (since, d/dx (sinx) = cosx)
= cos[(t^3- 2 t^2+ 1)] * [3 t^2 - 2*2 t] (Since, d/dx (x^n) = nx^(n-1) and d/dx (k)= 0, where k is a constant)
= cos[(t^3- 2 t^2+ 1)](3t^2 - 4t)
= t cos[(t^3- 2 t^2+ 1)] (3t - 4)
(b)
y = e^(-√t)
dy/dt = d/dt (e^(-√t))
= e^(-√t) * d/dt (-√t) (since, d/dx (e^x) =e^x)
= e^(-√t) [- ½t^(1/2-1)] (Since, d/dx (x^n) = nx^(n-1))
= e^(-√t) [- ½t^(-1/2)]
= - e^(-√t)/ 2√t

### Question 7

(a)
y = ln[(4x+1)]/ x^5
By differentiating equation with respect to x, and apply the quotient rule of differentiation, get
dy/dx = d/dx (ln[(4x+1)]/ x^5)
= [x^5 d/dx (ln[(4x+1)]) - ln[(4x+1)] d/dx (x^5)]/ (x^5)) ^2
= [x^5 ln[(4x+1)] d/dx (4x +1) - ln[(4x+1)] 5 x^(5-1)] / x^10 (since, d/dx (x^n) = nx^(n-1) and d/dx (lnx) = 1/x, and d/dx (k) = 0, where k is a constant)
= [x^5 ln[(4x+1)] 4 - 5x^4 ln[(4x+1)]]/ x^10
= x^4 [4x ln[(4x+1)] - 5 ln[(4x+1)]]/ x^10
= [4x ln[(4x+1)] - 5 ln[(4x+1)]]/ x^6
(b)
y = 3xe^cosx
By differentiating equation with respect to x, and apply the product rule of differentiation, i.e. d/dx(f(x) g(x)) = g(x)d/dx (f(x) + f(x) d/dx g(x), get
dy/dx = d/dx (3xe^cosx )
= 3[x d/dx (e^cosx ) + e^cosx d/dx (x)]
= 3[x e^cosx d/dx (cosx) + e^cosx * 1] (since, d/dx (e^x) =e^x and dx/dx = 1)
= 3[xe^cosx * (-sinx) +e^cosx ] (since, d/dx (cos[x)] = -sinx)
= 3e^cosx (1- xsinx)

### Question 8

The displacement, x = 4.9t^2 + 2/t^2
(a)
The displacement of the item from its starting point at t = 2.5 seconds
x_(t=2.5) = 4.9 * [2.5]^2 + 2/ [2.5]^2
= 0.784 + 0.32
= 1.104 meters
(b)
The velocity, v = dx/dt = the rate of change of displacement with respect to time.
= d/dt (4.9t^2 + 2/t^2)
= 2*4.9 t + 2 (-2) * t^(-2-1) (since, d/dx (x^n) = nx^(n-1))
= 9.8 t - 4t^(-3)
= 9.8 t - 4/t^3
Therefore, the velocity at 2.5 seconds
v_(t=2.5) = 9.8 * 2.5 - 4/ [2.5]^3
= 24.5 - 0.256
= 24.244 meter/seconds
(c)
The acceleration, a = dv/dt = the rate of change of velocity with respect to time.
a = dv/dt
= d/dt (9.8 t - 4/t^3)
= 9.8 1 - 4 (-3)* t^(-3-1) (since, d/dx (x^n) = nx^(n-1))
= 9.8 + 12t^(-4)
= 9.8 + 12/t^4
Therefore, the acceleration at 2.5 seconds
a_(t=2.5) = 9.8 + 12/[2.5]^4
= 9.8 + 0.307
= 10.107 meter/ seconds^2

### Question 9

y = 3x^5 - 5 x^3
Coordinates of stationary points
To determine the stationary points of y needs to differentiate y with respect to x.
dy/dx = 3 5 x^4 - 5 3 x^2 (since, d/dx (x^n) = nx^(n-1))
= 15x^4 - 15 x^2
= 15x^2 (x^2- 1)
At stationary points dy/dx = 0
Therefore, 15x^2 (x^2- 1) = 0
Or, 15x^2 (x+1) (x-1) = 0
Therefore, the values of x will be, x=0, x=1, and x=-1
Now, the corresponding y values are,
For, x = 0
y = 3(0) ^5 - 5 (0) ^3
= 0
Therefore, (0, 0) will be a stationary point.
For, x = 1
y = 3(1) ^5 - 5 (1) ^3
= 3-5
= -2
Therefore, (1, -2) will be a stationary point.
For, x = -1
y = 3(-1) ^5 - 5 (-1) ^3
= -3 +5
= 2
Therefore, (-1, 2) will be a stationary point.
The coordinates of the stationary points are (0, 0), (1, -2), and (-1, 2).
Classification of stationary points
The stationary points can be classify by finding the d^2y/ dx^2
Therefore, d^2y/ dx^2 = d/dx (dy/dx)
= d/dx (15x^4 - 15x^2)
= 15 4 x^3 - 15 2 x (since, d/dx (x^n) = nx^(n-1))
= 60 x^3 - 30 x
Now, for, x = 0
d^2y/ dx^2 = 0
As, d^2y/ dx^2= 0, the stationary point (0, 0) is the point of inflection.
For x = 1
d^2y/ dx^2 = 60 -30
= 30
As, 30 > 0, this point is a maximum.
For x = -1
d^2y/ dx^2 = -60 + 30
= - 30
As, -30 < 0, this point is a minimum.
Therefore,
Stationary point (0, 0), a inflection
Stationary point (1, -2), a maximum
Stationary point (-1, 2), a minimum.
Question 10
The volume of the cylindrical tank (V) = 50 Pi m^3
Cost of material for base and top = £ 0.80 per m^2
Cost of material for sides of the barrels = £ 0.50 per[ m]^2
(a)
Height of the tank = h (in m)
Radius of the tank = r (in m)
The volume of a cylindrical tank (V) = Pi r^2h
Therefore, 50 Pi = Pi r^2h
Or, h = 50/ r^2 (hence proved)
(b)
Let C is the cost function of the materials.
The area of the bottom and top of the tank is 2 Pi r^2 and the area of the sides of the tank is 2 Pi rh
Therefore, the cost function (in £) of h and r is
C (r, h) = 0.80 (2 Pi r^2) + 0.50 (2 Pi rh)
= 1.6 Pi r^2 + Pi r * 50/r^2 (putting the value of h)
= 1.6 Pi r^2 + 50 Pi r^(-1)
Therefore, C = 1.6 Pi r^2 + 50 Pi r^(-1) (hence proved)
(c)
C = 1.6 Pi r^2 + 50 Pi r^(-1)
The differentiation of cost function C with respect to r,
dC/dr = d/dr (1.6 Pi r^2 + 50 Pi r^(-1))
= 1.6 Pi 2 r + (-1) 50 Pi r^(-1-1) (since, d/dx (x^n) = nx^(n-1))
= 3.2 Pi r - 50 Pi r^(-2)
The critical number can be find when dC/dr = 0
Therefore, 0 = 3.2 Pi r - 50 Pi r^(-2)
and since, r > 0
0 = 3.2 Pi r^3 - 50 Pi
Or, 3.2 Pi r^3 = 50 Pi
Or, r^3 = 50/3.2
Or, r^3 = 15.625
Or, r = 2.5
Now, d^2C/ dr^2 = 3.2 Pi r - (-2) * 50 Pi r^(-2-1)
= 3.2 Pi r + 100 Pi r^(-3)
And Since r > 0, then by STD
The critical number r = 2.5 m will give a minimum cost.
Therefore, height (h) = 50/ [2.5]^2
= 8 m
Therefore radius 2.5 m and height 8 m will minimise the manufacturing cost.
(d)
The minimum manufacturing cost is
C(r=2.5) = 1.6 Pi [2.5]^2 + 50 Pi [2.5]^(-1)
= Pi (1.6/6.25 + 50/2.5)
= Pi (2.56 + 20)
= Pi * 22.56
= 70.903
Therefore the minimum manufacturing cost is £70.903.

Reference List
Journals
Fernandez, A., Baleanu, D. and Srivastava, H.M., 2019. Series representations for fractional-calculus operators involving generalised Mittag-Leffler functions. Communications in Nonlinear Science and Numerical Simulation, 67, pp.517-527.
Ali, M.A., Budak, H., Zhang, Z. and Yildirim, H., 2021. Some new Simpson's type inequalities for coordinated convex functions in quantum calculus. Mathematical Methods in the Applied Sciences, 44(6), pp.4515-4540.
Srivastava, H.M., 2021. An introductory overview of fractional-calculus operators based upon the Fox-Wright and related higher transcendental functions. Journal of Advanced Engineering and Computation, 5(3), pp.135-166.
Ciripoi, D., Löhne, A. and Weißing, B., 2018. Calculus of convex polyhedra and polyhedral convex functions by utilizing a multiple objective linear programming solver. Optimization.
Yang, Z., Garg, H. and Li, X., 2021. Differential Calculus of Fermatean Fuzzy Functions: Continuities, Derivatives, and Differentials. Int. J. Comput. Intell. Syst., 14(1), pp.282-294.

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