Trig, how long is the graph under y=0

  • Thread starter Thread starter BadatPhysicsguy
  • Start date Start date
  • Tags Tags
    Graph Trig
AI Thread Summary
The discussion focuses on determining how long the temperature, represented by the function f(x) = 20 + 25*sin(0.85x), remains below zero during the first 10 hours. Participants suggest using a graphing calculator to find the values of x where f(x) = 0, which indicates the intervals where the temperature is negative. One user initially miscalculated the period and attempted to estimate the duration incorrectly. It is emphasized that algebraic methods can be used instead of relying solely on the calculator. Ultimately, the solution involves finding the specific x-values within the given range to accurately determine the time the temperature is negative.
BadatPhysicsguy
Messages
39
Reaction score
0

Homework Statement


f(x)=20+25*sin(0.85x)
x = number of hours from start. f(x) = temperature.
For how long is the temperature negative (under 0) during the first 10 hours? We haven't learned how to derive/integrate trig equations so that is out of the question.

Homework Equations

The Attempt at a Solution


Most people with similar problems have said they used their graph calculator to solve these kinds of questions. I tried figuring out the period (7.39) and took the half of that + the remaining 2.61 but during those there is no more under the graph. So the answer would seem to be around 3.7 but it is only 1.5. How do I use my graph calculator (TI-84) to solve this?
 
Physics news on Phys.org
BadatPhysicsguy said:

Homework Statement


f(x)=20+25*sin(0.85x)
x = number of hours from start. f(x) = temperature.
For how long is the temperature negative (under 0) during the first 10 hours? We haven't learned how to derive/integrate trig equations so that is out of the question.

Homework Equations

The Attempt at a Solution


Most people with similar problems have said they used their graph calculator to solve these kinds of questions. I tried figuring out the period (7.39) and took the half of that + the remaining 2.61 but during those there is no more under the graph. So the answer would seem to be around 3.7 but it is only 1.5. How do I use my graph calculator (TI-84) to solve this?
You could set f(x) = 0 and solve the resulting equation. There are two values of x in the interval 0 ≤ x ≤ 10 for which f(x) = 0. Find these values and you'll have the interval where the temperature is negative.
 
BadatPhysicsguy said:

Homework Statement


f(x)=20+25*sin(0.85x)
x = number of hours from start. f(x) = temperature.
For how long is the temperature negative (under 0) during the first 10 hours? We haven't learned how to derive/integrate trig equations so that is out of the question.

You don't need calculus to solve this problem. You just need to find out for 0≤x≤10 where f(x) < 0. You can use a little algebra to clean things up a bit.

The Attempt at a Solution


Most people with similar problems have said they used their graph calculator to solve these kinds of questions. I tried figuring out the period (7.39) and took the half of that + the remaining 2.61 but during those there is no more under the graph. So the answer would seem to be around 3.7 but it is only 1.5. How do I use my graph calculator (TI-84) to solve this?

Instead of using your calculator to think for you, try working the problem out by analyzing it. You're not going to school to learn how to work a calculator, but how to learn to analyze and solve problems.
 
  • Like
Likes Mark44
I'd like to thank you Mark44 and SteamKing. Stupid of me to miss that!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top