What value for function y=3cos(t-pi) + 2 gives an instantaneous rate of change of 0?
a) 0
b) pi/2
c) pi/3
d) pi/4
pi/3
dy/dx = -3 sin(t-π)
=0
sin(t-π) = 0
t-π = 0 , or t-π = π or t-π = 2π
t = ... -3π, -2π, -π , 0 , π, 2π , 3π . .
Answer is 0, right?
To find the value that gives an instantaneous rate of change of 0, we need to find the value of t that satisfies the equation dy/dt = 0.
Let's differentiate y with respect to t using the chain rule:
dy/dt = -3sin(t-pi)
For dy/dt to be 0, sin(t-pi) must be 0, because multiplying by -3 does not change the fact that dy/dt is 0.
The equation sin(t-pi) = 0 has solutions for t when t-pi is equal to an integer multiple of pi. In other words:
t-pi = n*pi
Solving for t, we get:
t = n*pi + pi
Now we need to find the values of n that correspond to the multiple choice options.
a) n = 0, t = 0 + pi = pi
b) n = 1/2, t = 1/2*pi + pi = 3/2*pi
c) n = 1/3, t = 1/3*pi + pi = 4/3*pi
d) n = 1/4, t = 1/4*pi + pi = 5/4*pi
From the given options, the value that gives an instantaneous rate of change of 0 is option b) pi/2.