Differentiation to determine rate of change when voltage is given

This document explains how to use differentiation to determine the rate of change when voltage is given in calculus. It provides step-by-step solutions and examples for different scenarios. The document also includes tasks related to integration and finding maxima or minima points.

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.
8

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