Now, everyone knows the basic method to find the instantaneous rate of change of an equation ; that is, to find the difference between 2 points nearest to the point required a rate of change, divided by the difference in x-values of the respective y-values chosen. SIMPLE. What about trigonometric functions? Well, I have to say that is JUST THE SAME. Like 100% the same.
There is only a few things you need to note:
- Given a trigonometric graph with/out transformation, the turning points of the graph give an instantaneous rate of ZERO!!!
- Given a plain trigonometric function, and asked to find the instantaneous rate when a < x < b,graph the function and use the Tangent Operation to find the instantaneous rate of change simply subsititute the a and b into the function to find the y-values. Then use the traditional way to find the instantaneous rate of change.
- It really doesnt matter how complicated or how simple the function is, because your only need two distinct points to find the instanatneous rate of change
Take a look at this video to understand some detailed examples of this topic, if you are still unsure or how to find instantaneous rate of change. Remember to practise!
There is only a few things you need to note:
- Given a trigonometric graph with/out transformation, the turning points of the graph give an instantaneous rate of ZERO!!!
- Given a plain trigonometric function, and asked to find the instantaneous rate when a < x < b,
- It really doesnt matter how complicated or how simple the function is, because your only need two distinct points to find the instanatneous rate of change
Take a look at this video to understand some detailed examples of this topic, if you are still unsure or how to find instantaneous rate of change. Remember to practise!
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