Find the absolute minimum and absolute maximum values of f on the given interval.
f(t) = 16 cos t + 8 sin 2t,
[0, π/2]
nope max is actually 12sqrt(3)
f'(t) = -16sin(t) + 16cos(t)
f'=0 when sin(t)=cos(t)
t = π/4
f(0) = 16
f(π/4) = 16/√2 + 8 = 8(1+√2) > 16
f(π/2) = 0
So, it looks like
max = 8(1+√2)
min = 0
Let's check the graph:
http://www.wolframalpha.com/input/?i=16+cos+t+%2B+8+sin+2t+for+0+%3C%3D+t+%3C%3D+pi%2F2
Well, finding the absolute minimum and maximum values of a function requires a bit of calculation. But let me put on my clown hat and help you through it with a smile!
First, we need to find the critical values by taking the derivative of f(t). Using some trigonometric identities, we get:
f'(t) = -16sin(t) + 16cos(2t)
Now, let's set f'(t) equal to zero to find the critical points. Hmm, looking at this equation, I must say, finding zeros of trigonometric functions can sometimes be a bit "trig-icky"! But we'll manage!
16cos(2t) - 16sin(t) = 0
Next, let's solve this equation for t. We can factor out 16 to simplify things:
16(cos(2t) - sin(t)) = 0
The only way for this equation to be equal to zero is if either cos(2t) - sin(t) = 0 or 16 = 0. But wait a minute – 16 equals zero? That's not possible! So we can safely ignore 16 = 0.
Now, let's focus on cos(2t) - sin(t) = 0. To keep things light-hearted, let's give it a name: "Silly Trig Equation"! Solving the Silly Trig Equation can be a bit silly indeed!
cos(2t) - sin(t) = 0
We need to find the values of t that satisfy this equation. Unfortunately, there's no quick "trick" to solve it, so we'll have to put on our serious math hats for a moment for some algebraic calculations.
By using the Pythagorean identity (cos^2(x) + sin^2(x) = 1), we can rewrite the equation as:
cos(2t) - √(1 - cos^2(t)) = 0
Rearranging this equation:
cos(2t) = √(1 - cos^2(t))
Now, let's square both sides:
cos^2(2t) = 1 - cos^2(t)
Expanding cos^2(2t):
(1 - 2sin^2(t))^2 = 1 - cos^2(t)
We can simplify further:
1 - 4sin^2(t) + 4sin^4(t) = 1 - cos^2(t)
Rearranging and combining like terms:
cos^2(t) - 4sin^2(t) + 4sin^4(t) = 0
Now, this equation is getting really silly! But don't worry, we're close to finding the critical points.
Using another trigonometric identity (cos^2(x) = 1 - sin^2(x)), we can rewrite the equation as:
(1 - sin^2(t)) - 4sin^2(t) + 4sin^4(t) = 0
Simplifying:
1 - sin^2(t) - 4sin^2(t) + 4sin^4(t) = 0
Combining like terms:
1 - 5sin^2(t) + 4sin^4(t) = 0
Now, we have the equation 4sin^4(t) - 5sin^2(t) + 1 = 0. To solve this equation, we can use a substitution, say z = sin^2(t). Then:
4z^2 - 5z + 1 = 0
Factoring:
(4z - 1)(z - 1) = 0
Solving for z:
z = 1/4 or z = 1
Now we'll solve for sin(t) using z = sin^2(t):
sin^2(t) = 1/4 or sin^2(t) = 1
Taking the square root of both sides:
sin(t) = ±1/2 or sin(t) = ±1
Now we need to solve for t. From sin(t) = 1/2, we get two solutions: t = π/6 and t = 5π/6. From sin(t) = -1/2, we also get two solutions: t = 7π/6 and t = 11π/6. Finally, from sin(t) = 1, we get t = π/2, and from sin(t) = -1, we also get t = 3π/2.
Okay, now we have all the critical points: t = π/6, 5π/6, 7π/6, 11π/6, π/2, 3π/2.
To find which of these are within the given interval [0, π/2], we see that π/6, 5π/6 are within the interval, while 7π/6, 11π/6, π/2, and 3π/2 are not. So the critical points within the interval are t = π/6 and t = 5π/6.
To find the absolute minimum and maximum values, we need to evaluate f(t) at the critical points and the endpoints of the interval.
f(0) = 16cos(0) + 8sin(2*0)
= 16(1) + 8(0)
= 16
f(π/2) = 16cos(π/2) + 8sin(2*π/2)
= 16(0) + 8sin(π)
= 0
f(π/6) = 16cos(π/6) + 8sin(2*π/6)
= 16(√3/2) + 8(1)
= 12√3 + 8
f(5π/6) = 16cos(5π/6) + 8sin(2*5π/6)
= 16(-√3/2) + 8(-1)
= -12√3 - 8
The absolute minimum value is the minimum of these values, which is -12√3 - 8, and the absolute maximum value is the maximum of these values, which is 16.
So, the absolute minimum value is -12√3 - 8 and the absolute maximum value is 16.
Hope all this math clowning around didn't make your head spin! Just remember, sometimes the math can get a little tricky, but with patience, a smile, and maybe a clown nose, you can solve any problem!
To find the absolute minimum and absolute maximum values of f(t) = 16 cos(t) + 8 sin(2t) on the interval [0, π/2], we need to find the critical points and evaluate the function at those points as well as the endpoints.
Step 1: Find the derivative of f(t):
f'(t) = -16 sin(t) + 16 cos(2t)
Step 2: Set f'(t) equal to zero to find the critical points:
-16 sin(t) + 16 cos(2t) = 0
Step 3: Solve the equation for critical points.
Divide both sides by 16: sin(t) - cos(2t) = 0
Using trigonometric identities, we can rewrite cos(2t) as 1 - 2sin^2(t): sin(t) - 1 + 2sin^2(t) = 0
Rearranging the equation: 2sin^2(t) + sin(t) - 1 = 0
Factoring the quadratic equation: (2sin(t) - 1)(sin(t) + 1) = 0
Setting each factor equal to zero:
2sin(t) - 1 = 0 or sin(t) + 1 = 0
Solving for t:
2sin(t) - 1 = 0
sin(t) = 1/2
t = π/6 or t = 5π/6
sin(t) + 1 = 0
sin(t) = -1
t = 3π/2
Therefore, the critical points are t = π/6, t = 5π/6, and t = 3π/2.
Step 4: Evaluate f(t) at the critical points and endpoints:
f(0) = 16 cos(0) + 8 sin(2*0) = 16
f(π/2) = 16 cos(π/2) + 8 sin(2π/2) = 8
f(π/6) = 16 cos(π/6) + 8 sin(2π/6) = 16(√3/2) + 8(1) = 12√3 + 8
f(5π/6) = 16 cos(5π/6) + 8 sin(2*5π/6) = 16(-√3/2) + 8(-1) = -12√3 - 8
f(3π/2) = 16 cos(3π/2) + 8 sin(2*3π/2) = 16(0) + 8(-1) = -8
Step 5: Compare the values to determine the absolute minimum and absolute maximum:
The values are:
f(0) = 16
f(π/2) = 8
f(π/6) = 12√3 + 8
f(5π/6) = -12√3 - 8
f(3π/2) = -8
The absolute minimum value is -12√3 - 8 which occurs at t = 5π/6, and the absolute maximum value is 16 which occurs at t = 0.
Therefore, the absolute minimum value is -12√3 - 8, and the absolute maximum value is 16 on the interval [0, π/2].
To find the absolute minimum and absolute maximum values of a function on a given interval, we can follow these steps:
Step 1: Find the critical points of the function by taking the derivative and setting it equal to zero.
Step 2: Evaluate the function at the critical points and at the endpoints of the interval.
Step 3: Compare the values obtained in steps 1 and 2 to determine the absolute minimum and maximum values.
Let's apply these steps to the function f(t) = 16 cos t + 8 sin 2t on the interval [0, π/2]:
Step 1: Find the critical points.
Taking the derivative of f(t) with respect to t:
f'(t) = -16 sin t + 16 cos 2t
Setting f'(t) = 0:
-16 sin t + 16 cos 2t = 0
Simplifying the equation:
2 cos 2t = sin t
Using the double-angle identity cos 2t = 1 - 2sin^2(t), we can rewrite the equation as:
2(1 - 2sin^2(t)) = sin t
Expanding and rearranging the equation:
4sin^2(t) + sin(t) - 2 = 0
This is a quadratic equation in terms of sin(t). Solving it, we get:
(4sin(t) - 1)(sin(t) + 2) = 0
Setting each factor equal to zero:
4sin(t) - 1 = 0 --> sin(t) = 1/4
sin(t) + 2 = 0 --> sin(t) = -2 (extraneous solution)
So, the critical point is sin(t) = 1/4.
Step 2: Evaluate the function at the critical points and endpoints.
Evaluate f(t) at sin(t) = 1/4:
f(t) = 16cos(t) + 8sin(2t)
= 16cos(t) + 8 * 2sin(t)cos(t) (using double-angle identity)
Using sin(t) = 1/4, we can determine that cos(t) = √(1 - sin^2(t)) = √(1 - (1/4)^2) = √15/4
Plugging in the values:
f(t) = 16 √15/4 + 8 * 2 * 1/4 * √15/4
= 4√15 + 2√15
= 6√15
Evaluate f(t) at the endpoints:
f(0) = 16cos(0) + 8sin(2 * 0)
= 16 * 1 + 8 * 0
= 16
f(π/2) = 16cos(π/2) + 8sin(2 * π/2)
= 16 * 0 + 8 * 0
= 0
Step 3: Compare the values.
From the above calculations, we have:
f(0) = 16
f(π/2) = 0
f(t) = 6√15 at sin(t) = 1/4
Comparing these values, we see that:
The absolute maximum value of f(t) on the interval [0, π/2] is 16, which occurs at t = 0.
The absolute minimum value of f(t) on the interval [0, π/2] is 0, which occurs at t = π/2.