Calculus: Optimization, Differentiation, Integration, and Linear Algebra

Calculus
Institution Name
Student Name
Date

Q1
a. Writing an expression
A=wl
Where A is the area, w equals to width and l equals to length
L=2 w +l
L is the length of the fencing wire
Now 2 w+l=200
b. Expressing area in terms of width
A=wl
L=200−2 w
A=w (200−2 w)
−2 w2+200 w
c. Using differentiation to optimize A
dA
dw =−4 w +200
At the optimum point
dA
dw =0
−4 w+ 200=0
−4 w=−200
4 w=200
w=50 m
At w=50 m
l=200− ( 2∗50 ) =100 m
The maximum area
A=wl
50∗100=500 m2
Q2 Differentiate
a. f ( x ) =−e
x
3 sin (6 x)

Using the product rule
duv =u' v+ v ' u
u=−ex / x
v=sin (6 x)
u'= du
dx
¿−1
3 e
x
3
v'= dv
dx
d /dx (6 x )=6
Applying the chain rule
d
dx ( sin 6 x )=6 cos (6 x )
Now we replace thevalues in the formula for the product rule
f ' ( x ) = [ −1
3 e
x
3 ∗sin ( 6 x ) ] +[6 cos ( 6 x ) ∗−e
x
3 ]
¿−1
3 e
x
3 sin ( 6 x ) −6 e
x
3 cos (6 x )
b. g ( x )= x3 + 4
lnx
Using the quotient rule
d ( u
v )=[ u' v−u v'
v2 ]
u=x3 + 4
v=lnx
u'=3 x
v'= 1
x

g' ( x ) =
3 xlnx− ( x3+ 4 ) ( 1
x )
( lnx )2
c. h ( t )= (ln ( 3 t ) +2 t
1
2 )2
3
using the chain rule
h' ( t ) ={ 2
3 (ln ( 3 t ) +2 t
1
2 )(−1 /3 )}∗{( 3∗1
3 t )+t
−1
2 }
¿ {2
3 (ln ( 3 t )+ 2t
1
2 )−1
3
}{ 1
t + t
−1
2 }
Q3 Indefinite integral
f ( x )=x ( 1
2 )(x ¿¿ 2+2)(x +3)¿
Removing the brackets
(x
5
2 + 2 x
1
2 )( x+3)
x
7
2 + 3 x
5
2 +2 x
3
2 + 6 x
1
2
` Now using the addition/sum rule and the power rule we integrate the polynomial
∫ x
7
2 +3 x
5
2 +2 x
3
2 + 6 x
1
2
¿ x
9
2
9
2
+ 3 x
7
2
7
2
+ 2 x
5
2
5
2
+ 6 x
3
2
3
2
x
9
2
9
2
+ x
7
2
6
7
+ x
5
2
4
5
+ x
3
2
4
Q4 Integration by substitution
a. ∫ sinx
( 5−3 cosx )3 dx
u=5−3 cosx