Applications of the Initial Value Problem y' = f(x, y); y(c) = d 123
SOLUTIONa. Differentiating the equation for the semi-circle, we find- x
y'(x) = V9='X2' g - x2
So the arc length is{ 2 {2 ~
s = 11 ,/1 + (y')2 dx = 11 V 1 + ~ dxg - x^2 + x^2 1
2
--~-dx= 3dx
g - x2 1 -Jg - x2.
We numerically calculated an approximate value for this arc length usingSOLVEIVP by setting J(x, y) = 3/-)9 - x^2 , by inputing [1, 2] for the interval
of integration, and by inputing the initial condition y(l) = 0. The program
output showedf 2 3dx
11 V9=X2 ~ 1.169672.
Thus, the desired arc length, s , is approximately equal to 1.169672.
b. The surface area of the solid generated by revolving the given arc of the
semi-circle about the y-axis is
Sy= 27r f
2xJl + (y')^2 dx = 67f f
2
~·
11 11 9 - x^2We calculated an approximate value for the integral f 1
2
xdx/-J9 - x^2 usingSOLVEIVP by setting J(x, y) = x/-J9 - x^2 , by inputing [1, 2] for the interval
of integration, and inputing the initial condition y(l) = 0. From the program
output, we found
f 2 xdx
11 V9=X2 ~ .5923588.
Hence, the surface area S ~ 67r(.5923588) = 11.16570.
c. The volume of the solid generated by revolving the given region about the
y-axis is
Vy= 27f 1
2
x[y(x) - v's] dx = 27f 12
x[V9 - x^2 - v's] dx.Setting J(x, y) = x [-J9 - x^2 -v's], inputing [1, 2] as the interval of integration,
and inputing the initial condition y(l) = 0, SOLVEIVP numerically computed
1
2
x [V9 - x^2 - v's]dx ~ 0.4615936.