P(t)= 100 sin((pi t)/2)+500 Calculate the maximum value for t. I have derived the function to P'= 50pi cos((pi t)/2)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

P(t)= 100 sin((pi t)/2)+500 Calculate the maximum value for t. I have derived the function to P'= 50pi cos((pi t)/2)

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Let P' equal 0 and solve for t. 0 = 50pi cos(pi t / 2) cos(pi t / 2) = 0 pi t / 2 = arccos 0 = pi / 2 + 2kpi t = 1 + 4k; where k is any integer
ok, i havnt learnt arccos but the textbook says "Take the first two values to see which is minimum" pi t/2 = pi/2 or 3pi/2 t= 1,3 I dont understand this part.
Well, the text is telling you to see which one produces a maximum value. Either t = 1 or t = 3 produces a maximum value. If you put t = 1 into the original equation P(1) = 100sin(pi (1)/2) + 500, you will get a maximum P = 600. If you put t = 3 into the original equation P(3) = 100sin(pi (3)/2) + 500, you will get a minimum P = 400. The t = 1 is clearly the maximum value. Is this Calc 1 / AB? I'd imagine you would have learned how to calculate arc cosines in trigonometry/precalculus.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Uh, i live in Australia so i would imagine that the subjects would be different. But yeah, i want to know how the textbook got t=1,3?
Oh, I'm sorry. Anyway, I guess the values t = 1 and t = 3 are the first values that would come to mind that would produce a maximum/minimum. There's nothing really special about them though. If you plug any integer k into t = 1 + 4k to get a number t. That number t will produce a maximum value. Likewise, if you plug any integer k into t = 3 + 4k to get a number t, that number t will produce a minimum value.
Oh so, for any trig max/min problems the t= 1,3 would most likely be max/ win?
It depends on what is inside the sine. For example, if P = 100sin(t) + 500, then, the maxima would occur at t = pi / 2 + 2pi; the minima would occur at t = 3pi / 2 + 2pi. It could also change if the sine was a cosine instead. It takes a fundamental understanding of trigonometry to see the relationship. But no, t = 1, 3 are not always the max and min t values...
oh right, guess i need to brush up on trig!
we cant find the max value of t . it will infinite as domain of this function is (-infinite, infinite) . we need not to find out maxima or minima of p(t)
Of course we can't find the max value of t. I assumed we were talking about relative extrema.

Not the answer you are looking for?

Search for more explanations.

Ask your own question