6.7 Rates Of Change In Trigonometric Functions
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Determine the average rate of change of the function
y = 4 cos (x) + 3 for 0 < x < π/3
Step 1: Determine the y-values for when x = 0 and when x = π/3
y = 4 cos (0) + 3 y = 4 cos ( π/3) + 3
= 4(1) + 3 = 4 (1/2) +3
= 7 = 2 + 3
= 5
m = (y2 - y1) ÷ (x2 -x1)
= (5 - 7) ÷ (π/3- 0)
= -2 ÷ π/3
= - 1.91
Therefore the average rate of change on this interval is -1.91
Step 2) Using the slope formula, calculate the average rate of change. (NOTE: It’s easier to convert the angle from radians to degrees before continuing.)
The average rate of change of trigonometric functions are found by plugging in the x-values into the equation and determining the y -values. After having obtained both coordinates, simply use the slope formula: m=(y2 - y1)÷(x2 - x1). The resulting m value is the average rate of change of this function over that interval.
The instantaneous rate of change of trigonometric functions are found by using the slope formula with coordinates that come from x-values that are slightly higher and lower than the x-value in question by a fraction.
Check out our tutorial below to provide an in depth example of how to determine the average and instantaneous rate of change of trigonometric functions.
The rates of change of trigonometric functions are determined by using similar methods and strategies used when dealing with other functions.
Don’t remember? That’s alright, let’s briefly review them!
Example 1
Step1) Determine the y-values for when t= 8.001 and t= 7.999. (NOTE: Use a calculator to evaluate). Be sure to keep several (if not all) decimal places.
y = 3 sin 6(7.999) + 11 y = 3 sin 6(8.001) + 11
= 13.22922425 = 13.22964468
m = (y2 - y1) ÷ (x2 -x1)
= (13.22964468 -13.22922425) ÷ (8.001-7.999)
= (0.000420428 ÷ 0.002)
= 0.21
Therefore the instantaneous rate of change on this interval is 0.21.
Step 2) Using the slope formula, calculate the instantaneous rate of change.
Determine the instantaneous rate of change for the following function when t=8: y = 3 sin 6(t) + 11.
Sample Questions
Level 1/2
1. Determine the average rate of change of the function y =6 cos 2(x-π/3) + 5 for the interval π/3 x 2π/3.
2. How can we tell if the average rate of change of a sinusoidal function is positive or negative for an interval before actually solving for average rate of change? Explain.
Level 3
Determine the instantaneous rate of change at t = 14 for the following function: H(t) = 4 sin 5 (t) -10.
Level 4
4. The temperature in Toronto, Ontario behaved abnormally this spring. The temperature saw an increase from 0°C to 15°C and another decrease from 15°C to 0°. This cycle repeated every 12 hours. Express the temperature as a function of time and solve for the instantaneous rate of change at t = 22.
Key Concepts/Tips
Determining average rate of change or instantaneous rate of change is no different from calculating with other functions. The same strategies are used for other types of functions as well. The tangent lines occur at the maximum and minimum points of a function, due to its periodic nature the instantaneous rate of change is 0 in many areas, this is also where the tangent lines slope is zero. In a sinusoidal function the slope of the tangent lies between the maximum and minimum values, where the slope is at its greatest is where the point lies halfway between the min and max. You can sketch a tengent line to get a rough idea what values will coorespond to each area as well as use two points on the graph to find the slope of the secant line. You can calculate instantaneous rate of change by calculating two points very close together.
Example 2