1. ## Rates of change

I need to find the derivative of;

h(t) = 62 + 60sin ((5t-1)pi / 2)

All help is appreciated.

xoxo Lex.

2. Originally Posted by Mercy99
I need to find the derivative of;

h(t) = 62 + 60sin ((5t-1)pi / 2)

All help is appreciated.

xoxo Lex.
Again, keep in mind that $\displaystyle \frac{d}{dt}\left[\sin(x)\right]=\cos(x)$.

Proof:

If $\displaystyle f(x)=\sin(x)$, then $\displaystyle f'(x)=\lim_{h\to{0}}\frac{f(x+h)-f(x)}{h}=\lim_{h\to{0}}\frac{\sin(x+h)-\sin(x)}{h}$

Note that $\displaystyle \sin(\varphi+\vartheta)=\sin(\varphi)\cos(\varthet a)+\cos(\varphi)\sin(\vartheta)$

Thus, $\displaystyle \lim_{h\to{0}}\frac{\sin(x+h)-\sin(x)}{h}=\lim_{h\to{0}}\frac{\sin(x)\cos(h)+\co s(x)\sin(h)-\sin(x)}{h}$

$\displaystyle \implies \lim_{h\to{0}}\frac{\sin(x)\cos(h)-\sin(x)}{h}+\lim_{h\to{0}}\frac{\cos(x)\sin(h)}{h}$

$\displaystyle \implies \lim_{h\to{0}}\sin(x)\frac{\cos(h)-1}{h}+\lim_{h\to{0}}\cos(x)\frac{\sin(h)}{h}$

We should know that $\displaystyle \lim_{h\to{0}}\frac{\cos(h)-1}{h}=0$ and $\displaystyle \lim_{h\to{0}}\frac{\sin(h)}{h}=1$

Thus, $\displaystyle \lim_{h\to{0}}\sin(x)\frac{\cos(h)-1}{h}+\lim_{h\to{0}}\cos(x)\frac{\sin(h)}{h}=\sin( x)\cdot 0+\cos(x)\cdot 1=\color{red}\boxed{\cos(x)}$

We need to apply the chain rule as well : $\displaystyle \frac{d}{dt}\left[f(g(t))\right]=f'(g(t))\cdot g'(t)$

Thus, $\displaystyle \frac{dh}{dt}=60\cos\left(\frac{(5t-1)\pi}{2}\right)\cdot \frac{5\pi}{2}=\color{red}\boxed{150\pi\cos\left(\ frac{(5t-1)\pi}{2}\right)}$

Does this stuff seem to make more sense now?

Feel free to ask more questions if you have them. I may not be on for much longer. There are others besides me who can answer your questions.

--Chris