Calculus Assignment: Rate of Change, Integration, and Maxima/Minima

This calculus assignment covers various applications of differentiation and integration. Task 1 focuses on determining the rate of change of voltage using differentiation, with examples involving different voltage functions. Task 2 explores integration, calculating the integral of a current function over time and finding the current value. Task 3 involves definite integration to calculate areas under curves. Finally, Task 4 focuses on finding maxima and minima of given functions and plotting their graphs. The assignment demonstrates the application of calculus concepts to solve practical problems. The solutions are step-by-step and include calculations for different scenarios involving voltage, current, and areas under curves. The assignment provides a comprehensive overview of calculus concepts and their applications.

Calculus

Table of Contents
TASK 1............................................................................................................................................1
Differentiation to determine rate of change when voltage is given.............................................1
TASK 2............................................................................................................................................4
PART 1........................................................................................................................................4
PART 2........................................................................................................................................5
TASK 3............................................................................................................................................6
PART 1........................................................................................................................................6
PART 2........................................................................................................................................6
TASK 4............................................................................................................................................7
PART 1........................................................................................................................................7
PART 2........................................................................................................................................8

TASK 1
Differentiation to determine rate of change when voltage is given
(a) v = (t2 + 6)2
r = dv/dt
= d (t2 + 6)2
dt
= 2 (t2 + 6). 2t
= 4t (t2 + 6)
at t = 5 sec, rate of change will be
r = 4 x 5 (52 + 6)
= 20 (31) = 620 units
(b) v = (3t3 – 4t + 6)3
r = dv/dt
= d (3t3 – 4t + 6)3
dt
= 3 (3t3 – 4t + 6)2. (9t2 – 4)
= 3(9t2 – 4) (3t3 – 4t + 6)2
at t = 5 sec, rate of change will be
r = 3(9.52 – 4) (3.53 – 4.5+ 6)2
= 3 (221) (361)2
≈ 8.6 x 107 units
1

(c) v = loge (2t)
r = dv/dt
= d [loge (2t)]
dt
= 1 . 2 = 1
2t t
at t = 5 sec, rate of change will be
r = 1/5 = 0.2 units
(d) v = 4 e-0.5t
r = dv/dt
= d (4 e-0.5t)
dt
= 4 (-0.5) e-0.5t
= -2 e-0.5t
at t = 5 sec, rate of change will be
r = -2 e-0.5x5
= -2 e-2.5
≈ -0.16 units
(e) v = sin (2t3 + 4t – 2)
r = dv/dt
= d [sin (2t3 + 4t – 2)]
dt
= cos (2t3 + 4t – 2). (6t2 + 4)
at t = 5 sec, rate of change will be
r = cos (2.53 + 4.5 – 2). (6.52 + 4)
= cos (268) . 154
≈ -5.37 units
2

(f) v = cos (3t4 – 5t + 4)
r = dv/dt
= d [cos (3t4 – 5t + 4)]
dt
= -sin (3t4 – 5t + 4). (12t3 – 5)
at t = 5 sec, rate of change will be
r = -sin (3.54 – 5.5 + 4). (12.53 – 5)
= -sin (604) . (1495)
≈ 1343.7 units
3

TASK 2
PART 1
i = E e-t/RC
R
∫ i.dt = ∫ E e-t/RC
R
Let -t/RC = x
dt = - RC. dx
so,
I = -RC . E ∫ ex dx
R
= -EC .ex
= -EC e-t/RC
at, E = 10V C = 1.0μF and R = 1.0MΏ and t = 0 sec
then,
I = -10 x 1.0 e-t/1.0 x 1.0
= -10 units
4

PART 2
Given,
iL.dt = 1 ∫ cos(100t).dt
L
= 1 [sin(100t)]
100 L
at L = 10mH and t = 0.9sec
iL = 1/10 sin (100 x 0.9)
= 0.1 sin (90)
= 0.1 unit
5

TASK 3
PART 1
Given,
4
y = ∫2 (1 – e-t). dt
= [ t + e-t]24
= [ 4 – 2 + e-4 – e-2]
= 2 + 0.018 – 0.135
≈ 1.9 sq. unit
PART 2
Given,
4
y = ∫2 (– e-t). dt
= [ e-t]24
= [ e-4 – e-2]
= 0.018 – 0.135
≈ -0.12 sq. unit
6

TASK 4
PART 1
Given
y = 3x2 – 5x
Plotting the graph –
differentiate with respect to x
dy = 6x – 5
dx
Maxima or minima point can find out at dy/dx = 0
6x – 5 = 0
x = 5/6
so, at x = 5/6 = 0.8
y(5/6) = 3(0.8)2 – 5(0.8)
= -2 (approx.)
so, maxima or minima coordinates of given equation is (0,0) and (0, -2)
7

PART 2
Given,
y = x3 – 4x + 6
differentiate with respect to x
dy = 3x2 – 4x
dx
Maxima or minima point can find out at dy/dx = 0
3x2 – 4x = 0
x (3x – 4) = 0
x = 0, 4/3
so, at x = 0
y(0) = (0)3 – 4(0) + 6
= 6
at x = 4/3 = 1.3
y(1.3) = (1.3)3 – 4(1.3) + 6
= 3 (approx.)
so, at x = 0 the given function will get its minimum value.
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