# Math Help - Periodic Function - Instantaneous rate of change

1. ## Periodic Function - Instantaneous rate of change

Find the IRoC at the point (90 degrees, 5) on f(x) = 5sin(x)
I started with the different quotient formula:
IRoC = ( f(x+h) - f(x) ) / h , h-->0
=( 5 sin (90 +h) - 5 ) /h
= (5 sin h) / h

What do I do next?? I don't think I can factor out the h, even if I did what would sin take as an argument?

Thank you for the help.

Edit: Also, how would you apply the same formula with an exponential function like y = 100(0.85)^x , on x = 0 ??

2. ## Re: Periodic Function - Instantaneous rate of change

One of the first trig formulas you see in a Calculus book is " $\lim_{x\to 0}\frac{sin(x)}{x}= 1$". There are a number of ways of showing that, depending upon exactly how you define "sin(x)".

With $100(0.85^x)$, the difference quotient is $\frac{100(.85^{x+h})- 100(.85^x)}{h}$. Using the fact that $a^{x+ y}= a^xa^y$, we can write that as $100(.85^x)\frac{.85^h- 1}{h}$. So the derivative is $100(.85^x)\lim_{h\to 0}\frac{.85^x- 1}{h}$. Notice that we were able to take x completely out of the limit so the derivative is the original function, $100(.85^x)$, times some constant. How you take that limit to find the constant depends upon exactly how you want to handle exponentials but it comes out to be the natural logarithm of .85.

3. ## Re: Periodic Function - Instantaneous rate of change

We have not done Calculus yet, Is there a simple algebraic way of doing what you did?

4. ## Re: Periodic Function - Instantaneous rate of change

Originally Posted by sakonpure6
I started with the different quotient formula:
IRoC = ( f(x+h) - f(x) ) / h , h-->0
=( 5 sin (90 +h) - 5 ) /h
= (5 sin h) / h

What do I do next?? I don't think I can factor out the h, even if I did what would sin take as an argument?

Thank you for the help.

Edit: Also, how would you apply the same formula with an exponential function like y = 100(0.85)^x , on x = 0 ??
Well first of all, \displaystyle \begin{align*} 5\sin{ \left( 90^{\circ} + h \right) } - 5 \neq 5\sin{(h)} \end{align*}, rather

\displaystyle \begin{align*} 5\sin{ \left( 90^{\circ} + h \right) } - 5 &= 5\left[ \sin{ \left( 90^{\circ} \right) } \cos{(h)} + \cos{ \left( 90^{\circ} \right) } \sin{(h)} \right] - 5 \\ &= 5 \left[ 1\cos{(h)} + 0\sin{(h)} \right] - 5 \\ &= 5\cos{(h)} - 5 \\ &= 5 \left[ \cos{(h)} - 1 \right] \end{align*}

Now since you haven't done Calculus, you probably haven't had anything to do with limits yet, so just use your calculator to see what happens to \displaystyle \begin{align*} \frac{5 \left[ \cos{(h)} - 1 \right] }{h} \end{align*} when you plug in values of h that are very close to 0...