Determine the intervals on which the function is concave up and concave down
f(theta) = 21theta + 21Sin^2(theta),
[0,pi]
I have tried to implicitly differentiate but I am getting nowhere,... can anyone please help?
f(Ø) = 21Ø + 21(sinØ)^2
f ' (Ø) = 21 + 42sinØcosØ = 21 + 21sin (2Ø)
f '' (Ø) = 21 cos(2Ø) (2)
= 42 cos (2Ø)
the period of cos 2Ø is π , and our domain is [0,π]
so we have one complete cosine curve to consider.
Remember that a function is concave up if its second derivative is positive, and negative if .....
42 cos(2Ø) is positive from 0 to π/4 and then again from 3π/4 to π
so it is concave up for 0 < Ø < π/4 OR 3π/4 < Ø < π
and concave down between π/4 and 3π/4
why implicit?
f(x) = 21x + 21sin^2 x
f' = 21 + 21 sin 2x
f'' = 42 cos 2x
f''=0 at x=pi/4,3pi/4
f''>0 for x in (0,pi/4)
f''<0 for x in (pi/4,3pi/4)
f''>0 for x in (3pi/4,pi)
so now you know the cioncavity.
To determine the intervals on which the function is concave up and concave down, we need to find the second derivative of the given function and analyze its sign.
Let's start by finding the first derivative of the function. To do that, we'll differentiate each term separately:
f(theta) = 21theta + 21Sin^2(theta)
f'(theta) = (d/dtheta)(21theta) + (d/dtheta)(21Sin^2(theta))
The derivative of 21theta with respect to theta is simply 21, and the derivative of 21Sin^2(theta) requires the chain rule.
The chain rule states that if we have a composition of functions, such as Sin^2(theta), the derivative is the derivative of the outer function (Sin^2(theta)) multiplied by the derivative of the inner function (Sin(theta)):
(d/dtheta)(21Sin^2(theta)) = 21 * (d/dtheta)(Sin^2(theta)) = 21 * (2Sin(theta) * (d/dtheta)(Sin(theta)))
Now, we need to calculate (d/dtheta)(Sin(theta)). The derivative of Sin(theta) with respect to theta is simply Cos(theta):
(d/dtheta)(Sin(theta)) = Cos(theta)
Therefore, we can substitute this result back into the expression for the derivative:
f'(theta) = 21 + 42Sin(theta)Cos(theta)
Next, let's find the second derivative of the function by differentiating the first derivative with respect to theta:
f''(theta) = (d/dtheta)(21 + 42Sin(theta)Cos(theta))
To differentiate 21 with respect to theta, we get 0 since it's a constant term. The derivative of 42Sin(theta)Cos(theta) requires the product rule and the chain rule.
The product rule states that if we have a product of functions, such as Sin(theta)Cos(theta), the derivative is the first function (Sin(theta)) multiplied by the derivative of the second function (Cos(theta)), plus the second function (Cos(theta)) multiplied by the derivative of the first function (Sin(theta)):
(d/dtheta)(42Sin(theta)Cos(theta)) = 42 * (Sin(theta) * (d/dtheta)(Cos(theta))) + 42 * (Cos(theta) * (d/dtheta)(Sin(theta)))
The derivative of Cos(theta) with respect to theta is -Sin(theta), and the derivative of Sin(theta) with respect to theta is Cos(theta). So we can substitute these results back into the expression:
f''(theta) = 0 + 42 * (Sin(theta) * (-Sin(theta))) + 42 * (Cos(theta) * Cos(theta))
Simplifying further, we get:
f''(theta) = -42Sin^2(theta) + 42Cos^2(theta)
Now, we need to determine the sign of the second derivative to identify the intervals of concavity.
Since -42Sin^2(theta) and 42Cos^2(theta) are both multiplied by positive constants, the sign of the second derivative is entirely determined by the signs of Sin^2(theta) and Cos^2(theta).
Remember that Sin^2(theta) and Cos^2(theta) are both trigonometric functions squared, so they will always be positive or zero. Therefore, the sign of -42Sin^2(theta) and 42Cos^2(theta) depends on whether the value is zero or positive.
For -42Sin^2(theta) to be positive, Sin^2(theta) must be zero. This occurs when Sin(theta) = 0, which happens at theta = 0, pi, 2pi, etc.
For 42Cos^2(theta) to be positive, Cos^2(theta) must be zero. This occurs when Cos(theta) = 0, which happens at theta = pi/2, 3pi/2, etc.
By analyzing the intervals between these points, we can determine the concavity of the function. For concave up regions, the second derivative should be positive (f''(theta) > 0), and for concave down regions, the second derivative should be negative (f''(theta) < 0).
Considering the interval [0, pi], we have the following points:
0, pi/2, pi
Analyzing the intervals between these points, we find:
- Between 0 and pi/2:
At theta = 0, f''(theta) = -42Sin^2(0) + 42Cos^2(0) = 0 + 42 = 42 (positive)
At theta = pi/2, f''(theta) = -42Sin^2(pi/2) + 42Cos^2(pi/2) = -42 + 0 = -42 (negative)
Since the second derivative changes sign from positive to negative, the function is concave up on the interval [0, pi/2] and concave down on the interval (pi/2, pi].
Therefore, the intervals on which the function is concave up are [0, pi/2], and the intervals on which the function is concave down are (pi/2, pi].