Hello,
I'm supposed to do the following questions for the equation
y=cos(pi/3)x for intervals [0,6]
a) For what values of x does the instantaneous rate of change appear to equal 0?
b) For what value(s) of x does the instantaneous rate of change appear to reach a maximum value? a minimum value?
Im guessing a) is x=3 and x=6?
I have no clue how to do b) since I know the derivative is 1/2? So not sure how to find max/min of instantaneous rate of change...
Any help is appreciated, thanks!
why are you guessing? I hope you actually did some kind of calculation.
since cos(kx) has period 2pi/k,
cos(pi/3 x) has period 6
the rate of change is y' = -pi/3 sin(pi/3 x)
so, it is zero at pi/3 x = 0 or pi or 2pi; that is, x = 0,3,6
sin(pi/3 x) is at a max when pi/3 x = pi/2, or x = 3/2
min at x = 5/2
Note that these values are midway between the zeros.
Don't forget your general knowledge of the shape of sine curves. Online graphing sites can help.
To find the values of x for which the instantaneous rate of change appears to be equal to zero, we need to find the critical points of the function. Critical points occur where the derivative of the function is equal to zero or does not exist.
Let's start with part a:
a) To find the instantaneous rate of change, we need to take the derivative of the function y = cos(pi/3)x. The derivative of y with respect to x can be found using the chain rule as follows:
dy/dx = -(pi/3)sin(pi/3)x
To determine when the instantaneous rate of change appears to be equal to zero, we need to solve the above equation for dy/dx = 0.
-(pi/3)sin(pi/3)x = 0
Since the sine function is zero at x = n*pi for any integer n, we have:
(pi/3)x = n*pi
x = (3/3)n
So x can take the values x = n, where n is an integer. In the interval [0, 6], the values of x that satisfy this condition are 0, 1, 2, 3, 4, 5, 6. Therefore, x = 0, 1, 2, 3, 4, 5, 6 are the values for which the instantaneous rate of change appears to equal 0.
For part b, to find the values of x for which the instantaneous rate of change appears to reach a maximum or a minimum, we need to find the critical points of the derivative function.
b) The derivative of y = cos(pi/3)x is -(pi/3)sin(pi/3)x as we calculated in part a. To find the maximum and minimum values, we need to find the critical points of the derivative.
To find the critical points, we need to set the derivative equal to zero:
-(pi/3)sin(pi/3)x = 0
sin(pi/3)x = 0
Since the sine function is zero at x = n*pi for any integer n, we have:
(pi/3)x = n*pi
x = (3/3)n
So x can take the values x = n, where n is an integer. In the interval [0, 6], the values of x that satisfy this condition are 0, 1, 2, 3, 4, 5, 6. Therefore, x = 0, 1, 2, 3, 4, 5, 6 are the values for which the instantaneous rate of change appears to reach a maximum or minimum.