University Calculus: MCV4U Unit 2 Derivatives Assignment Solution

This document presents a detailed solution to a Unit 2 Derivatives assignment for a Calculus course (MCV4U). The assignment covers a wide range of derivative concepts, including finding derivatives of various functions using the power rule, product rule, quotient rule, and chain rule. It also includes implicit differentiation, finding higher-order derivatives, determining equations of tangent lines, and analyzing particle motion, including velocity, acceleration, and displacement calculations. Furthermore, the solution incorporates applications of derivatives, such as related rates problems involving the rate of change of volume and surface area, and optimization problems. The assignment also features questions on related rates and the interpretation of derivative values in context. The solution demonstrates step-by-step calculations and explanations for each problem, making it a valuable resource for students studying calculus.

Running head: UNIT 2: DERIVATIVES
UNIT 2: DERIVATIVES
Name of the Student
Name of the University
Author Note

1UNIT 2: DERIVATIVES
1.
a. f(x) = 13x^4 – 7x^3 + 5x^2 + 11x+ 75
=> f’(x) = 13*4x^3 + 7*3x^2 + 5*2x + 11
=> f’(x) = 52x^3 + 21x^2 + 10x + 11
b. f(x) = (x^3 + 2x^2 + 4)(x^4-3)
f’(x) = (x^3+2x^2 + 4)(3x^3) + (x^4-3)*(3x^2 + 4x) (by using product rule)
= 6x^6 + 6x^5 + 12x^3 + 3x^6 +4x^5 – 9x^2-12x
= 9x^6 + 10x^5 +12x^3 -9x^2 – 12x
c. f(x) = (3x^2 -6x+7)/(4x-1)
=> f’(x) = ( 4 x−1 )∗( 6 x−6 )− ( 3 x2−6 x+ 7 )∗4
( 4 x −1 )2 = 24 x2−30 x+6−12 x2 +24 x −28
( 4 x−1 )2 =
12 x2−6 x−22
( 4 x−1 ) 2
d. f(x) = (4x^3 -7x)^7
=> f’(x) = 7∗( 4 x3 −7 x ) 6
∗( 12 x2−7 )
e. 12 = y^2 + x^2
=> 0 = 2y*dy/dx + 2x
=> dy/dx = -x/y = −x
√12−x2
f. f(x) = 156
=> f’(x) = 0

2UNIT 2: DERIVATIVES
g. f(x) = (1-3x)^2 * (x^2 – 2)^3
f’(x) = 2 ( 1−3 x )∗( −3 ) ( x2−2 ) 3
+3∗2 x ( 1−3 x ) 2 ( x2−2 ) 2
= ( 18 x−6 ) ( x2−2 )3
+6 x ( 1−3 x )2 ( x2−2 )2
h. f(x) = ( 2 x2 +1
3 x3 +1 )2
=> f’(x) = 2*
2 x2+ 1
3 x3+ 1∗( 3 x3 +1 ) ∗4 x+ ( 2 x2 +1 )∗9 x2
( 3 x3+ 1 ) 2
= ( 4 x2+2)(12 x4 +4 x +18 x4 +9 x2)
( 3 x3 +1 )3 (By
quotient rule)
2.
f(x) = (3x^2-7x+4)^8
 f’(x) = 8∗( 3 x2−7 x+ 4 )
7
∗(6 x−7)
f’’(x) = 8∗( 3 x2−7 x+4 )7
∗6+8∗7 ( 3 x2−7 x +4 )6
( 6 x−7 )∗( 6 x −7)
= 48 ( 3 x2−7 x +4 )7
+56 ( 3 x2−7 x+ 4 )6
( 6 x−7 )2
3.
f(x) = 12x^10 – 3x^7 +4x^5 + 5x^2 + 6
 f’(x) = 120x^9 – 21x^6 + 20x^4 + 10x
 f’’(x) = 108x^8 – 126x^5 + 80x + 10
 f’’’(x) = 864x^7 – 630x^4 + 80
4.
y = x^2 -2x + 3

3UNIT 2: DERIVATIVES
 dy/dx = 2x +2
[ dy
dx ](2,3)
=2∗2+2=6 = m
Equation of the tangent line (y-3)/(x-2) = m
 (y-3)/(x-2) = 6
 y-3 = 6x-12
 y = 6x-9
5.
Equation of parabola y = -x^2 + 3x +4
 dy/dx = -2x + 3 = m
Hence, m =5 at
-2x + 3 = 5 => x = -1
Hence, y = -1-3+4 = 0
Thus the point where slope of the tangent is 5 is (-1,0)
6.
An example of decreasing rate of velocity of a particle is given by,
d2 x
d t2 =−k
The example of increasing rate of velocity of a particle is given by,
d2 x
d t2 =k
where, dx/dt = velocity of particle in m/sec and k>0 and t = time in seconds

4UNIT 2: DERIVATIVES
7.
When a vehicle is travelling at -35 km/h then it infers that the vehicle is moving 35 km per
hour at the opposite direction of the considered direction.
8.
dV/dt = rate of change of volume of the object with time.
9.
Given initial height of hammer is s(0) = 90 m
Height at time t is s(t) = 90-4.9t^2, t>0
a.
velocity s’(t) = -9.8t
velocity at t = 1 sec s’(1) = -9.8 m/sec
velocity at t =4 sec s’(4) = -9.8*4= -39.2 m/sec
b.
The hammer hits the ground at s = 0
90-4.9t^2 = 0
 t^2 = 90
 t = 9.4868 sec.
Hence, the hammer hits the ground at about 9.49 secs.
c.
The impact velocity when hitting ground is -9.8*9.49 = -93 m/sec.

5UNIT 2: DERIVATIVES
10.
Particle motion s(t) = 2t^3 – 15t^2 +33t + 17
a) Particle is at rest when its velocity is zero.
s’(t) = 0
 6t^2 – 30t + 33 = 0
 t = 1.634 and 3.366 secs.
b) Now, velocity is 6t^2 – 30t + 33 is zero at t = 1.634 and 3.366 sec.
(t – 1.634)(t-3.366) > 0
 t > 1.634 and t >3.366
or t < 1.634 and t < 3.366
Hence, velocity is positive when t > 3.366 and t < 1.634.
c) The motion of the particle in first 10 secs is given below.

6UNIT 2: DERIVATIVES
d) The distance travelled in forst 10 seconds can be obtained by integrating the velocity in
first 10 seconds.
Hence, distance = ∫
0
10
(6 t2 – 30 t +33)dt = [ 2 t3 – 15 t2+33 t ]0
10
= 2∗103 – 15∗102 +33∗10 = 830 meters.
11.
Given surface area decrement rate = 0.5 cm^2/min
 d(SA)/dt = 0.5
 d(4πr^2)/dt = 0.5
 8πr*(dr/dt) = 0.5
 (dr/dt) = 0.5/8πr
 (dr/dt) | (r = 4) = 0.5/32π = 0.005
When the radius is 4 cm then radius decreases at 0.005 cm/min.
12.
Distance between origin and Brent’s car at time t hour
D1 = 0.4 – 60*t
Distance between origin and Aaron’s car at time t hour
D2 = 0.3 – 70*t
Thus distance between both cars at time t is
D = √ ( 0.4 – 60t ) 2 + ( 0.3 – 70 t ) 2

7UNIT 2: DERIVATIVES
Rate of change of distance =
(½)( ( 0.4 – 60 t )¿ ¿2+ ( 0.3 – 70 t )2 )0.5∗(−120 ( 0.4−60 t ) −140(0.3−70 t)) ¿
Thus rate of change of distance when Brent’s car is 0.4 km and Aaron’s is 0.3 km from the
intersection or at t = 0
Rate(t=0) = (½)( ( 0.4 ) ¿¿ 2+ ( 0.3 ) 2 )0.5∗(−120 ( 0. 4 ) −140(0. 3))¿ = -22.5 km/hr.
13.
Cone height h = 5.2 m
Cone radius r= 2.5 m
At any moment from the similarity principle of the cone triangle
r/2.5 = h/5.2
 r = (2.5/5.2)h = 0.48h
Now, when there is 8π m^3 water then height to which water is filled
(1/3)π*r^2*h = 8π
 h^3 = 8/((1/3)*(0.48)^2) = 104.17 m => h = 4.705 m
Now, the water filling rate = volume increasing rate = 1.2 m^3/min.
Hence, (d/dh)(1/3*π*0.48^2*h^3) = 1.2
 0.24127*3h^2(dh/dt) = 1.2
 (dh/dt) = 1.2/(0.24127*3*4.705^2) = 0.075.
14.
f(x) = ax^3 + bx^2 -5x + 9

8UNIT 2: DERIVATIVES
f(-1) = 12
 -a + b -5 + 9 = 0
 b-a = -4 => b = a-4
f’(x) = 3ax^2 + 2bx – 5
f’(-1) = 3
 3a + 2b -5 = 3
 3a +2b = 8
 3a + 2*(a-4) = 8
 5a = 16
 a = 16/5
15.
y = x^2 + 1
 dy/dx = 2x = m
Let P(a,b) is any point on parabola.
Hence, gradient at P is
m = 2a
b = a^2 + 1 (as P lies on parabola)
Now, using point-slope form
y- b = m*(x-a)
 y-(a^2+1) = (2x)*(x-a)
Now, if this line passes through point (-1,-3) then

9UNIT 2: DERIVATIVES
 -3-(a^2+1) = (-2)*(-1-a)
 -3 – a^2 -1 = 2 + 2a
 a^2+2a+6 = 0
As a become imaginary hence there exist no tangent that passes through point (-1,3) in the
parabola.
16.
s1 = 4t – t^2
 s1’(t) = 4 – 2t
 s1’’(t) = -2
s2 = 5t^2 – t^3
 s2’(t) = 10t – 3t^2
 s2’’(t) = 10 – 6t
Hence, s2’’(t) = s1’’(t)
 10 – 6t = -2
 6t = 12 =>t = 2
Now, at t = 2 velocity of s1 is s1’(2) = 0.
velocity of s2 is
s2’(t) = 10*2 – 3*2^2 = 20-12 = 8.

10UNIT 2: DERIVATIVES