Download Chapter 6 Practice Test Group 1 with solutions Chapter 6

Chapter 6 test
6.1:
Find the general solution to the exact differential equation:
1.
Dy
𝐷𝑥
= sin x − e−x + 8x 3
2.
Du
𝐷𝑥
= (sec 2 x 5 )(5x 4 )
Find anti derivative and use the reverse power rule
[y= -cos x + e-x + 2x4 +C]
[u= tan(x5) + C]
Solve the initial value problem explicitly:
3.
Du
𝐷𝑥
= 7x 6 − 3x 2 + 5
u = 1 when x = 1
[u= x7 – x3 + 5x - 4]
4.
Dx
𝐷𝑡
1
1
t
t2
= −
1
+6
[x= ln t + + 6t – 7]
t
x = 0 when t = 1
Solve the initial value problem using the fundamental theorem. The
answer should have a definite integral.
5. F ′ (x) = sin(x 2 ) and F(1) = 5
1
5 = ∫1 sin(𝑡 2 ) 𝑑𝑡 + 𝐶
1
∫1 sin(𝑡 2 ) 𝑑𝑡= 0
5 =C
𝑥
[𝑦 = ∫1 sin(𝑡 2 ) 𝑑𝑡 + 5]
Construct a slope field for the differential equations. Use a graph with
the window [-1, 1] by [-1, 2] with twelve lattice points. Use slope
analysis.
6.
Dy
𝐷𝑥
= 2x + y
7.
Dy
𝐷𝑥
=x
Use Euler’s Method, and find the percentage error.
8. Find y when x= 1.3 and X= 0.1
𝑑𝑦
=𝑦−𝑥
𝑑𝑥
(x,y)
(1,2)
(1.1, 2.1)
(1.2, 2.2)
[2.3]
1
1
1
Dy
𝐷𝑥
= y − x (2, 1)
X
Y= 𝑑𝑦
(𝑥)
𝑑𝑥
(X+X, Y+Y)
.1
.1
.1
.1
.1
.1
(1.1, 2.1)
(1.2, 2.2)
(1.3, 2.3)
9. Find y when x= 1.7 and X= - 0.1
Dy
𝐷𝑥
= x − y (2, 2)
𝑑𝑦
=𝑥−𝑦
𝑑𝑥
(x,y)
(2,2)
(1.9, 2)
(1.8, 2.01)
[2.031]
0
-.1
-.21
X
Y= 𝑑𝑦
(𝑥)
𝑑𝑥
(X+X, Y+Y)
-.1
-.1
-.1
0
.01
.021
(1.9, 2)
(1.8, 2.01)
(1.7, 2.031)
6.2
Integration with u-substitution
10. ∫
[
1
14
𝑑𝑥
(8𝑥−1)3
(𝑥 2 + 2)7 + 𝐶]
1
11. ∫ 𝑠𝑒𝑐 2 (8𝑥 )𝑑𝑥
3
[
1
24
(tan(8𝑥 )) + 𝐶]
12. ∫ 𝑥(𝑥 2 − 1)5 𝑑𝑥
[
(𝑥 2 −1)6
12
+ 𝐶]
13. ∫ 3 sin(1 − 3𝑥 ) 𝑑𝑥
[ cos(1 - 3x) + C ]
6.3
Find the indefinite integral
14. ∫ 𝑦 ln 𝑦 𝑑𝑦
[
𝑦2
2
ln 𝑦 −
𝑦2
4
+ 𝐶]
Solve the initial value problem
15.
𝑑𝑢
𝑑𝑥
= 𝑥 𝑠𝑒𝑐 2 (𝑥) (0, 1)
[𝑢 = 𝑥 𝑡𝑎𝑛𝑥 + ln(𝑐𝑜𝑠𝑥 ) + 1 ]
Use parts and solve for the unknown integral
16. ∫ 𝑒 𝑥 𝑠𝑖𝑛𝑥 𝑑𝑥
[
𝑒 𝑥 𝑠𝑖𝑛𝑥− 𝑒 𝑥 𝑐𝑜𝑠𝑥
2
+ 𝐶]
Use tabular integration to find the antiderivative
17. ∫ 𝑥 4 𝑒 −𝑥 𝑑𝑥
[(−𝑥 4 − 4𝑥 3 − 12𝑥 2 − 24𝑥 − 24)𝑒 −𝑥 + 𝐶]
Solve the differential equation
18.
𝑑𝑦
𝑑𝑥
𝑥2
[(
4
= 𝑥 2 𝑒 4𝑥
𝑥
1
8
32
− +
) 𝑒 4𝑥 + 𝐶]
6.4
Use separation of variables
19.
𝑑𝑦
𝑑𝑥
=𝑦+2
𝑑𝑦
∫ 𝑦+2 = ∫ 𝑑𝑥
Ln[y+2] = x+ C
Ln[2+2] = 0+C
Ln[4] = C
Ln[y+2] = x + ln [4]
eln(y+2)= e(x+ln 4)
y+2= ex
[𝑦 = 4𝑒 𝑥 − 2]
20.
𝑑𝑦
𝑑𝑥
= −2𝑥𝑦 2 (1, 0.25)
[𝑦 = (𝑥 2 + 3)−1 ]
21. Radium 226 decays at a rate proportional to the quantity present.
Its half-life is 1612 years. How long will it take foe one quarter of the
given quantity to decay?
Y= ½ y0
½ y0= y0ekt
½ y0= y0ek(1612)
½= ek(1612)
1
𝑙𝑛2
1612
=
1612𝑘
1612
K= -.00043
Decay ¼ then we are left with ¾ y0
¾ y0= y0e(.00043)t
Ln ¾ = ln e(.00043)t
Ln ¾ = -.00043t
[t=669]
22. Let K= 4 and A= 6 What is the exponential solution and is it a growth
or decay graph?
Y=Aekt
Sub in numbers and k ≥ 0 so equation is positive.
[6e4t and growth]
23. If your money compounds 7% quarterly, how long will it take to
double?
A(t)= 2A0
2𝐴0
𝐴0
=
𝐴0 (1+
.07 4𝑡
)
4
𝐴0
2=(1+ .07/4)4t
Ln 2 =ln(1+ .07/4) x 4t
[t=10]
24. A hard-boiled egg at 98⁰C is put in a pan under running 18⁰C water
to cool. After 5 minutes, the egg’s temperature is found to be 38⁰C.
How much longer will it take the egg to reach 20⁰C?
Ts=18 T0= 98
T-18 = (98-18)e-kt
T= 38 when t=5
38= 18 + 80e-5k
e-5k= ¼
-5k= ln ¼ = -ln4
K=1/5 ln4
T= 18+ 80e-(0.2ln 4)t
Y= 20 y= T= 18+ 80e-(0.2ln 4)t
t= 13.3 it has already been 5 minutes so 13.3-5 = 8.3
[About 8 more minutes]