y(t)=(3/2)*sin(12*pi*t) y't=(3/2)*cos(12*pi*t) Is that correct?
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Originally Posted by No Logic Sense y(t)=(3/2)*sin(12*pi*t) y't=(3/2)*cos(12*pi*t) Is that correct? Close... You forgot to apply the chain rule! You need to multiply to your answer. --Chris
Hello, Originally Posted by No Logic Sense y(t)=(3/2)*sin(12*pi*t) y't=(3/2)*cos(12*pi*t) Is that correct? Nope. You forgot to use the chain rule Here, and Basically, you'll have to multiply by the derivative of 12 pi t, with respect to t
Okay, let me try again: y't=(3/2)*cos(t) Is that correct?
No wait, I can do it! If 12 * pi * t = u and sin(u) = y(t) du/dy = -sin(u) dt/dy = t y'(t)=(3/2) * t * -sin(12*pi*t) Is that now correct?
Originally Posted by No Logic Sense No wait, I can do it! If 12 * pi * t = u and sin(u) = y(t) du/dy = -sin(u) dt/dy = t y'(t)=(3/2) * t * -sin(12*pi*t) Is that now correct? Hmmm... There is a confusion... Pardon me if it is my fault Get back to my formula. So the formula give :
Oh yeah, you're right. No don't worry it wasn't your fault. It has just been ages since I did the whole diffential thing so I thought 12 * pi would dissapear, but of course it's t. Thanks for the answers.
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