- Assignment Help
- Samples
- Mathematics
- Functions And Calculus Coursework Assignment Sample

Table of Contents

- Type Assignment
- Downloads500
- Pages9

Get free samples written by our Top-Notch subject experts for taking online Assignment Help services.

(a)

Radians to degrees

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)

Degrees to radians

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

= Pi /2 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

Gradient of y:

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)

(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.

References

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.

Conference Presentation on Practice in Business Assignment sample

Introduction Get free samples written by our Top-Notch subject experts for taking online Assignment...View and Download

Case Study Report: Farmbox Assignment Sample

Introduction Get free samples written by our Top-Notch subject experts for taking online Assignment...View and Download

Research in Practice Assignment Sample

Introduction Get free samples written by our Top-Notch subject experts for taking assignment help uk from Rapid...View and Download

Teaching In A Specialist Area Assignment Sample2

Introduction Get free samples written by our Top-Notch subject experts for taking online Assignment...View and Download

Changes On Air Canada During Covid19 Assignment Sample
Introduction Get free samples written by our Top-Notch subject experts for taking online Assignment...View and Download

Animal Experiment Research Is Ethical Or Not Assignment Sample
Introduction Get free samples written by our Top-Notch subject experts for taking online Assignment...View and Download

Copyright 2024 @ Rapid Assignment Help Services

×

**Hi!** We're here to answer your questions! Send us message, and we'll reply via WhatsApp

Back

Pleae enter your phone number and we'll contact you shortly via Whatsapp

We will contact with you as soon as possible on whatsapp.

- 6500+ Projects Delivered
- 503+ Experts 24*7 Online Help

offer valid for limited time only*