dy/dx = x cos x^2
dy = x cos x^2 dx
let z = x^2
dz = 2 x dx so x dx = dz/2
so we have
dy = (1/2) cos z dz
y = (1/2)sin z + c
when x = 0, z = 0
-3 = (1/2)(0) + c
c = -3
so
y = (1/2) sin z - 3
when x = pi, z = pi^2
y = (1/2) sin(pi^2) - 3
= -.430-3
=-3.43 I do not agree either
If dy/dx = x cos x^2 and y = −3 when x = 0, when x = pi, y = .
A. −3.215
B. sqrt(2)
C. 1.647
D. 6
E. 3pi
All of the potential solutions I've come up with don't satisfy any of the given values.
3 answers
-.215-3
-3.215
-3.215
Ah-ha! As I was working on this earlier, somehow I got to sin(pi^2) - 3, too. But it gave me the same -3.43. Now that the 1/2 is in there, it makes much more sense.
Complicated math messes with one's head; you know how to do all the fancy stuff, but then the basic arithmetic flies out the window! lol
Complicated math messes with one's head; you know how to do all the fancy stuff, but then the basic arithmetic flies out the window! lol