Unit-2: Derivatives | Assignment

Please not that one is calculus and one is advanced functions and it is clearly indicated in the title.

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