Initial Value - Calculus - Solved Quiz, Exercises of Calculus

Key points of this past exam are: Initial Value, Problem, Solution, Method, Two Steps, Initial Point, Region Bounded

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MATH 106A,B - CALCULUS II WINTER 2008
QUIZ 3
NAME:
Show ALL your work CAREFULLY.
(a) Consider the initial value problem
dy
dx = 1 + x2ywith y(0) = 2.
Estimate the value y(2) (when x= 2) of the solution using Euler’s method with two steps with initial point
(0,2). DO THIS BY HAND and show all your steps.
Since we use two steps from x= 0 to x= 2, the step size is x= 1 with x0= 0 and x1= 1. Euler’s
method yields y1=y0+f(x0, y0)·xwhere f(x, y) = 1+x2y. It follows that y1= 2+[1+(0)2(2)](1) = 3.
Repeating the method again, we get y2=y1+f(x1, y1)·x= 3+[1+(1)2(3)](1) = 7. Thus, y(2) 7.
(b) Find the exact area of the region bounded by the curve x=y2+ 1, the line x= 1, and the line y= 1.
The area of the desired bounded region is given by
A=Z1
0
(y2+ 1) 1dy =y3
3¯
¯
¯
1
0=1
3.
If we integrate with respect to x, the area takes the form R2
11x1dx.
1
12
(c) Set up (but DO NOT evaluate) a definite integral representing the exact volume of the solid obtained
by rotating the region in (b) around the y-axis.
If we consider horizontal slices, the volume is represented by
Z1
0
[π(y2+ 1)2π(1)2]dy =πZ1
0
y4+ 2y2dy.
or, integrating with respect to x, by
Z2
1
2πx(1 x1) dx.
Date: January 28, 2008.
1

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MATH 106A,B - CALCULUS II WINTER 2008

QUIZ 3

NAME:

Show ALL your work CAREFULLY.

(a) Consider the initial value problem

dy dx

= 1 + x^2 y with y(0) = 2.

Estimate the value y(2) (when x = 2) of the solution using Euler’s method with two steps with initial point (0, 2). DO THIS BY HAND and show all your steps.

Since we use two steps from x = 0 to x = 2, the step size is ∆x = 1 with x 0 = 0 and x 1 = 1. Euler’s method yields y 1 = y 0 +f (x 0 , y 0 )·∆x where f (x, y) = 1+x^2 y. It follows that y 1 = 2+1+(0)^2 (2) = 3. Repeating the method again, we get y 2 = y 1 +f (x 1 , y 1 )·∆x = 3+1+(1)^2 (3) = 7. Thus, y(2) ≈ 7.

(b) Find the exact area of the region bounded by the curve x = y^2 + 1, the line x = 1, and the line y = 1.

The area of the desired bounded region is given by

A =

0

(y^2 + 1) − 1 dy =

y^3 3

1 0

If we integrate with respect to x, the area takes the form

x − 1 dx.

1

(^1 )

(c) Set up (but DO NOT evaluate) a definite integral representing the exact volume of the solid obtained by rotating the region in (b) around the y-axis.

If we consider horizontal slices, the volume is represented by ∫ (^1)

0

[π(y^2 + 1)^2 − π(1)^2 ] dy = π

0

y^4 + 2y^2 dy.

or, integrating with respect to x, by ∫ (^2)

1

2 πx(1 −

x − 1) dx.

Date: January 28, 2008. 1