AB Sol ch6 pt FT - Bellarmine College Preparatorywebs.bcp.org/sites/jmolina/Calc AB practice...
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Solutions to Chapter 6 Practice Test Part 1: No Calculators. Leave your answers un simplified. Evaluate each integral: [4 points each] a) 2 1 cos 2 dx x - ∫ b) 3 1 2 dx x x - ∫ 2 sec 2 xdx =- ∫ = ( ) 1 tan 2 2 x C - + = 3 1 2 x dx x dx - - = ∫ ∫ 2 ln x x C - + + c) 1 0 x e dx π ∫ = 1 0 1 x e π π = ( ) 1 1 e π π - d) /4 0 tan d π θ θ ∫ 4 0 sin cos x dx x π ⌠ ⌡ = let cos u x = , sin du xdx =- , sin du dx x = - . 4 sin sin 0 x du u x π - ⋅ ∫ = 4 4 0 0 1 ln ln cos x x du u x u π π = = =- =- =- ⌠ ⌡ = 2 ln 2 - Remark on solution to (d) above: ( ) cos sin d x x dx =- , hence the factor of – 1 above is manipulated so that the integral on part (d) is of the form ( ) ( ) f x f x dx ′ ∫ = ( ) ln f x C + . The technique of u – substitution will still work. Simply let cos u x = and go from there. 2) Find y(6) if 2 dy x dx y - = and (0) 5 y = . Show all steps! [6 points] 2 ydy xdx =- 2 ydy xdx ⇒ =- ∫ ∫ 2 2 2 x y C ⇒ =- + 2 2 x y C ⇒ =± - + . Now use initial condition of (0) 5 y = () 2 0 0 5 25 2 y C C ⇒ = =± - + ⇒ = 2 25 2 x y ⇒ =± - + ∴ ( ) 6 18 25 y =± - + = 7 ± _
Transcript of AB Sol ch6 pt FT - Bellarmine College Preparatorywebs.bcp.org/sites/jmolina/Calc AB practice...
Solutions to Chapter 6 Practice Test Part 1: No Calculators. Leave your answers un simplified. Evaluate each integral: [4 points each]
a) 2
1
cos 2dx
x
−∫ b)
3
1 2dx
x x
−
∫
2sec 2xdx= −∫ = ( )1
tan 22
x C− + = 31 2xdx x dx
−− =∫ ∫2ln x x C
−+ +
c)1
0
xe dx
π∫ =
1
0
1 xe
π
π= ( )
11eπ
π− d)
/ 4
0
tan d
π
θ θ∫
4
0
sin
cos
xdx
x
π
⌠⌡
= let cosu x= , sindu xdx= − ,
sin
dudx
x=
−.
4 sinsin
0
x duu x
π
−⋅∫ =
44
0 0
1ln ln cos
x
xdu u x
u
ππ=
== − = − = −⌠
⌡ =
2ln
2
−
Remark on solution to (d) above: ( )cos sind
x xdx
= − , hence the factor of – 1 above is manipulated so
that the integral on part (d) is of the form ( )( )
f x
f xdx
′
∫ = ( )ln f x C+ . The technique of
u – substitution will still work. Simply let cosu x= and go from there.
2) Find y(6) if 2
dy x
dx y
−= and (0) 5y = . Show all steps! [6 points]
2ydy xdx= − 2 ydy xdx⇒ = −∫ ∫2
2
2
xy C⇒ = − +
2
2
xy C⇒ = ± − + . Now use initial condition of
(0) 5y = ( )20
0 5 252
y C C⇒ = = ± − + ⇒ =2
252
xy⇒ = ± − + ∴ ( )6 18 25y = ± − + = 7±
_
Part 2: You may use a calculator. Round your decimal answers in the third place after the decimal point. 1) A bank account accrues interest at a rate of 2.35%, compounded continuously. [a and c @ 2 pts each, b @ 6 pts]
a) Write a differential equation that relates the balance (y) and time (t) wehre t is measured in years. [NOTE: If you cannot write the differential equation, go to your teacher for help. However, you will be deducted two points on the spot.]
The constant of proportionality is 0.0235, so 0.0235dy
ydt
= .
b) Use calculus to solve your differential equation from part (a) if the initial amount of money in
the bank account is $5000.00.
0.0235dy
ydt
= ⇒ 0.0235dy
dty
= 1 0.0235y
dy dt⇒ =∫ ∫ ln 0.0235y t C⇒ = +
.0235ty Ae⇒ = ( ) ( ).0235 00 5,000y Ae⇒ = =
�
5000A⇒ = ⇒ ( ) .02355000 ty t e⇒ =
c) Using the equation from part (b), when will the account be worth $100,000.00?
Solve for t in the equation ( ) .02355,000 100,000ty t e= = . You can use either algebra or graphical
methods by using your calculator. Algebraically: 0.023520 te= ⇒
1ln 20
0.0235t =
⇒ 127.478t ≈ years.
2) Consider the differential equation 22dy
xydx
= − . [part a @ 2 pts, part b @ 1 pt, part c @ 6 pts]
a) Sketch a slope field for this equation on the 12 points shown on the axes below.
b) While the slope field in part (a) is only drawn for 12 points, it is defined for an infinite number of points. Describe the set of points in the x-y plane where the slope will be positive.
We must investigate 0dy
dx>
22 0xy⇒ − >
⇒ 0x <
c) Find the particular solution to the given differential equation with 1
(1)4
f = .
22
dyxdx
y= − 2 2dy
yxdx⇒ = −∫ ∫ 21
x Cy
⇒ − = − +
( )2
1y
C x⇒ =
+. To find the initial condition ( )
( )2
1 11
4 1y
C⇒ = =
+3C⇒ = ⇒ ( ) 2
1
3f x
x=
+
