what is the derivative of:
V(t) = 130sin5t + 18
3 answers
V '(t) = 650cos(5t)
thank you, now do you think you can help me find where the maxiumum and minimum values are for this function and where they occur?
for the actual function not the derivative
for the actual function not the derivative
The max and/or min of a function occur when the derivative is zero.
that's why you were probably asked to find the derivative.
so
650 cos(5t) = 0
divide by 650
cos 5t = 0
the cosine curve is zero at π/2 and 3π/2
so 5t = π/2 or 5t = 3π/2
t = π/10 or t = 3π/10
you also know that the period of cos 5t is 2π/5
so adding or subtracting 2π/5 to any an answer will produce a new answer.
so t = π/10, π/2 , 9π/10, 13π/10 ..... or 3π/10 , 7π/10, 11π/10 ....
subbing any of these into the original function will yield either 148 or -112
try it, make sure your calculator is set to radians.
that's why you were probably asked to find the derivative.
so
650 cos(5t) = 0
divide by 650
cos 5t = 0
the cosine curve is zero at π/2 and 3π/2
so 5t = π/2 or 5t = 3π/2
t = π/10 or t = 3π/10
you also know that the period of cos 5t is 2π/5
so adding or subtracting 2π/5 to any an answer will produce a new answer.
so t = π/10, π/2 , 9π/10, 13π/10 ..... or 3π/10 , 7π/10, 11π/10 ....
subbing any of these into the original function will yield either 148 or -112
try it, make sure your calculator is set to radians.