Math Assignment.

Math Assignment 1
Math Assignment
Student’s Name
Course
Professor’s Name
University
City (State)
Date

Math Assignment 2
Question 1 (a)
dy
dx = d
dx ( 6 x3−5 x )
1
2
Let 6 x3−5 x=u
( 6 x3−5 x )
1
2 =u
1
2
d
du = 1
2 u
1
2 −1
= 1
2 u
−1
2 = 1
2 ( 6 x3−5 x )
−1
2
du
dx =6 ( 3 ) x2−5=18 x2−5
dy
dx = d
du × du
dx = 1
2 ( 6 x3−5 x )
−1
2 ( 18 x2−5 )= 1
2 ( 18 x2−5 ) ( 6 x3−5 x )
−1
2
dy
dx = 18 x2−5
2 ( 6 x3−5 x )
1
2
Question 1 (b)
Rate of change (x=5.75) , dy
dx = 18 x2−5
2 ( 6 x3 −5 x )
1
2
= 18(5.75)2−5
2 ( 6(5.75)3−5(5.75) )
1
2
Rate of change ( x=5.75)= 590.125
66.6905 =8.8487
Question 2 (a)
v=5 e(sin2 πt)
dv
dt = d
dt ( 5 e(sin 2 πt ) )=5 d
dt e
(sin2 πt)

Math Assignment 3
let u=sin 2 πt so that e(sin 2 πt)=eu
d
du ( eu ) =eu=e(sin 2 πt)
du
dt = d
dt ( sin 2 πt )=2 πcos 2 πt
dv
dt = d
du × du
dt =e(sin2 πt )× 2 πcos 2 πt=2 π e(sin2 πt) cos 2 πt
dv
dt =2 π e(sin 2 πt) cos 2 πt
Question 2 (b)
Rate of change (t=0.25)= dv
dt =2 π e(sin2 πt )cos 2 πt
dv
dt (t=0.25)=2 π e(sin 2 π (0.25)) cos 2 π (0.25)
¿ 2 π e(sin π
2 )
cos π
2 =2 π × 1× 0=0
Question 3 (a)
v=2 ¿
We apply the product rule: ( f ( t ) . g ( t ) ) '=g ( t ) . f ( t ) '+ f ( t ) . g ( t ) '
dv
dt =g ( t ) . f ( t )' + f (t ) . g ( t )'
Let g ( t )=2 sin(2 πt)∧f ( t )=cos (t ) so that:
g ( t )'=2× 2 π cos ( 2 πt ) =4 π cos ( 2 πt ) =¿∧f ( t ) ' =−sin (t)¿

Math Assignment 4
dv
dt =2 sin(2 πt ).−sin (t)+ cos (t ). 4 π cos (2 πt )
¿−2 sin(2 πt) sin (t )+4 π cos ( 2 πt ) cos (t )
Question 3 (b)
dv
dt =−2 sin(2 πt )sin (t)+4 π cos (2 πt ) cos (t )
When t=1 second , dv
dt =−2 sin(2 π ) sin (1)+4 π cos ( 2 π ) cos (1)
¿ 0+ 4 π × 0.5403=2.1612 π
Rate of change=6.7896
Question 4 (a)
i=10 v2 (e40 v)
di
dv =10 d
dv v2 ( e40 v )
We apply the product rule: ( f ( t ) . g ( t ) ) '=g ( t ) . f ( t ) '+ f ( t ) . g ( t ) '
di
dv =g ( t ) . f ( t )' + f (t ) . g ( t )'
g ( t )=v2∧g ( t ) ' = d
dv ( v2 )=2 v
f ( t ) =e40 v ∧f ( t ) '= d
dv e40 v=40 e40 v
di
dv =10 ( v2 ( 40 e40 v ) +e40 v ( 2 v ) )