Question

plz help ! =] find the second derivative (d^2y/dx^2) for the function y=a(1-cost), x=a(t-sint) ... i don't understand how to apply the chain rule exactly ..

Answer 1

write ' for d/dt. Then
x' = a(1-cos t)
y' = a.sin t
Thus
\[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{\sin t}{1 - \cos t} \]
Now
\[ \frac{d^2y}{dx^2} = \frac{d \ }{dx} (dy/dx) = \frac{dt}{dx} \frac{d \ }{dt} (dy/dx) \]

Now both terms on the right, you can evaluate.

Answer 2

how did you know to do the last step ? can you refer me to a formula or something ?
Thank you !!!

Answer 3

By the chain rule ... you have
\[ \frac{d \ }{dx} (dy/dx) \]
Now \( dy/dx \) itself is a function of t, call it \( f(t) \). Hence
\[ \frac{d \ }{dx} f(t) = \frac{dt}{dx} \frac{df}{dt} \] Fortunatley, it is also true that
\[ \frac{dt}{dx} = \frac{1}{dx/dt} \]

Answer 4

thanks alot !!! that was the part that i missed =]