what is the derivative of:
V(t) = 130sin5t + 18
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.
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
To find the derivative of the function V(t) = 130sin(5t) + 18, we can use the chain rule. The chain rule states that if we have a function of the form f(g(t)), then its derivative is given by f'(g(t)) * g'(t).
In this case, g(t) = 5t and f(x) = 130sin(x) + 18.
First, let's find the derivative of f(x) = 130sin(x) + 18. The derivative of sin(x) is cos(x), so the derivative of 130sin(x) is 130cos(x). Since the derivative of a constant (18 in this case) is 0, the derivative of f(x) = 130sin(x) + 18 is f'(x) = 130cos(x).
Next, let's find the derivative of g(t) = 5t. The derivative of t with respect to t is 1, so the derivative of g(t) is g'(t) = 5.
Now, we can apply the chain rule. The derivative of V(t) = f(g(t)) is given by f'(g(t)) * g'(t).
f'(g(t)) = 130cos(5t) and g'(t) = 5.
Therefore, the derivative of V(t) = 130sin(5t) + 18 is V'(t) = 130cos(5t) * 5.
Simplifying that expression, we have:
V'(t) = 650cos(5t)