Download Math 1220-3 Mock Exam 3

Math 1220-3
Mock Exam 3
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1. (6 pts) One way to find a root of an equation is the bisection method. Describe how this method
works.
Solution
The first thing to do is to pick numbers a0 , b0 close to the root r of the function f such that f (a0 )f (b0 ) <
0. Then you want to find the average value of these two numbers, call this value c. Now define numbers
a1 and b1 by a1 = a0 and b0 = c if f (a0 )f (c) < 0 or a1 = c and b0 = b1 if f (b0 )f (c) < 0. Now repeat
this process until you get as close to the root as you desire.
2. (5pts) Another way to find a root of an equation is by Newton’s method. Describe how this method
works. You may use a diagram to help with your explanation.
Solution
First pick a x1 close to the root you are trying to find. Then find the equation of the tangent line to
the graph at the point whose x coordinate is x1 . Now find when this line intersects the x-axis, and
call this value x2 . Now repeat the process as you did for x1 , to find x3 . Now keep on repeating this
process untill you get as close to the root as you desire.
X3
X2
X1
3. (6pts) Describe how Euler’s method works in approximating the solution of the differential equation
y 0 = f (x, y).
Solution
Just read page 505 in the book for a detailed describtion of how the method works.
4. (5pts) Find the slope of the tangent to the graph of r = 3 + 3 cos θ at the point of the graph where
θ = π6 .
Solution
The slope of the tangent line is given by the equation
f (θ) cos θ + f 0 (θ) sin θ
dy
= 0
dx
f (θ) cos θ − f (θ) sin θ
This implies that the slope of the tangent line is given by the equation
(3 + 3 cos θ) cos θ − (3 sin θ) sin θ
dy
=
dx
−(3 sin θ) cos θ − (3 + 3 cos θ) sin θ
Now just substitute θ = π/6 into the equation to obtain the slope.
5. (6 pts) Find the points of intersection between the graphs r = 5 sin θ and r = 2 + sin θ.
Solution
To do this we need to equate the two equations and then solve for θ. We have 5 sin θ = 2 + sin θ. This
implies that 4 sin θ = 2, which tells us that sin θ = 21 . So θ = π6 , 5π
6 . So the points of intersection are
(5/2, π/6) and (5/2, 5π/6).
6. (5 pts) Find the area of the region enclosed by the graph r = 4 cos 3θ.
Solution
We need to find the value of the integral
1/2
Z
π/6
(4 cos 3θ)2 dθ.
0
This will give us one sixth of the area. So the total area is
3
Z
π/6
16 cos2 3θ)dθ
=
3
Z
0
0
=
=
π/6
8(cos 6θ + 1)dθ
π/6
24 [sin 6θ/6 + θ]0
4π.
7. (6pts) Write the equation r = b cos θ in terms of cartesian coordinates.
Solution
If we multiply both sides of the equation by r we obtain the equation r2 = br cos θ. Remember if we
want to go from polar coordinates to cartesian coordinates we have to know how they are related. They
are related by the equations r2 = x2 + y 2 and r cos θ = x. Hence the equation in cartesian coordinates
becomes x2 + y 2 = bx.
8. (8 pts) Find the equation of the tangent line to the equation x2 + y 2 = 169 at the point (5, 12).
Solution
dy
dy
dy
First we use implicit differentation to find dx
. Then 2x + 2y dx
= 0. Hence dx
= −x
y , provided y 6= 0.
So at the point (5, 12), the slope of the tangent line is −5/12. Hence the equation of the tangent line
is y − 12 = (−5/12)(x − 5).
9. (5 pts) Sketch the graph of the equation
Solution
x2
16
−
y2
4
= 1, indicating vertices, foci, and asymptotes.
y = -1/2 x
y = 1/2 x
Y
( -4, 0 )
( 4, 0 )
X
The √
vertices are (4, √
0) and (−4, 0). The foci are (c, 0) and (c, 0), where c2 = a2 + b2 . Hence the foci
are ( 20, 0) and (− 20, 0).
10. (12 pts) Find the Taylor polynomial of order 3 based at 1 for the function ex . Use your answer to
approximate e0.8 .
Solution
Let f (x) = ex , then the Taylor polynomial of order 3 based at 1 is
P3 (x) = f (1) + f 0 (1)(x − 1) +
f 000 (1)
f 00 (1)
(x − 1)2 +
(x − 1)3 ,
2
6
which is P3 (x) = e + e(x − 1) + 2e (x − 1)2 + 6e (x − 1)3 . To approximate e0.8 , we simply substitute
x = 0.8 into the Taylor polynomial.
11. (6pts) Solve the differential equation
d2 y
dx2
dy
+ 3 dx
= ex .
Solution
dy
x
First of all let u = dx
. Then we rewrite the equation as du
dx + 3u = e . Now this is a First order
differential equation, which we can solve. Now multiply both sides of the equation by the integrating
3x
4x
factor to obtain the differential equation e3x du
dx + 3e u = e . The solution to this equation is
x
−3x
x
−3x
u = 1/4e + Ce
. This implies that y = (1/4)e + C1 e
+ C2 .
12. (5pts) A spring with a spring constant k of 5 pounds per foot is loaded with a 10-pound weight and
allowed to reach equilibrium. It is then raised 1 foot and released. What are the equation of motion,
the amplitude, and the period? Neglect friction.
Solution
2
10 d x
First of all we write down the differential equation. The differential equation is given by 32
dt2 = −5x.
2
d x
This can be rewritten in the form dt2 + 16x = 0. The general solution to this differential equation is
x = C1 cos 4t + C2 sin 4t. Now we use the initial conditions to find C1 and C2 . The initial conditions
are x(0) = 1 and x0 (0) = 0. These imply that C1 = 1 and C2 = 0. Hence the equation of motion is
x(t) = cos 4t. Hence the amplitude of motion is 1 inch and the period is π/2.
2
13. (7pts)
√ The slope of the tangent to the parabola x = −14y at a certain point on the parabola is
−2 7/7. Find the coordinates of that point.
Solution
dy
The equation of the curve is y = −x2 /14. Hence dx
= −x/7. So if the slope of the tangent to the
√
√
parabola at a certain point (x, y) is −2 7/7, then we must have −x/7 = −2 7/7, which implies that
√
x = 2 7. Now we plug this value into the equation of the curve to find the y coordinate.
Now we
√
obtain y = −28/14 = −2. Hence the coordinates of the point we are interested in is (2 7, −2).
14. (7pts) Solve the differential equation y 00 + y 0 − 2y = x2 .
Solution
First of all we have to find the general solution yg to the homogeneous equation y 00 + y 0 − 2y = 0. The
auxiliary equation to this is r2 + r − 2 = 0. The solutions to this equation are 1 and -2. So the general
solution yg is C1 ex + C2 e−2x . Now to find the particular solution. Try a particular solution of the form
yp = Ax2 +Bx+C. If we substitute this back into the equation we get 2A+2Ax+B−2(Ax2 +Bx+C) =
x2 . Equating coefficients we see that A = −1/2 , B = −1/2 and C = −3/4. Hence the solution to the
differential equation is C1 ex + C2 e−2x + −1/2x2 − 1/2x − 3/4.
15. (7pts) A spring with a spring constant k of 20 pounds per foot is loaded with a 10-pound weight
and allowed to reach equilibrium. It is then displaced 1 foot downward and released. If the weight
experiences a retarding force in pounds equal to one-tenth the velocity, find the equation of motion.
Solution
2
10 d x
1 dx
The differential equation that models this situation is 32
dt2 = − 10 dt − 20x. This can be rewritten as
2
25 dx
d x
dt2 + 8 dt + 64x = 0. To solve this differential equation we need to write down the auxiliary equation.
The auxiliary equation is r2 + 25
8 r + 64 = 0. The solutions to this equation are
√
−4 28 51i
±
25
25
Hence the general solution to the differential equation is
√
√
x(t) = C1 e(−4/25)t cos((28 51/25)t) + C2 e(−4/25)t sin((28 51/25)t)
Using the initial conditions x(0) = 1 and x0 (0) = 0, we find that C1 = 1 and C2 = 4/25.
16. (8pts) Solve the differential equation y 00 + y 0 − 2y = sin x.
Solution
First of all we have to solve the homogeneous equation y 00 +y 0 −2y = 0. The auxiliary equation to this is
r2 + r − 2 = 0. The solutions to this equation are 1 and -2. So the general solution yg is C1 ex + C2 e−2x .
Now to find the particular solution. Try a particular soultion of the from yp = A sin x + B cos x. When
we substitute this into the equation we get −A sin x − B cos x + A cos x − B sin x − 2A sin x − 2B cos x =
sin x. Comparing coefficients we find that A = −3/10 and B = −1/10. Hence the solution to the
differential equation is C1 ex + C2 e−2x − (3/10) sin x − (1/10) cos x.
17. (7pts) Solve the differential equation y 00 − 6y + 13y = 0.
Solution
2
To do this we first write down the auxiliary
equation.
√
√ The auxiliary equation is r − 6r + 13 = 0. The
6± (36−52)
6± (−16)
solutions to this equation are r =
=
. In complex form the solutions are 3 ± 2i.
2
2
Hence the general solution to the differential equation is given by y = C1 e3x cos 2x + C2 e3x sin 2x.