Finding Range of Temperature (T) as t Approaches Infinity

  • Thread starter Thread starter glid02
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary
SUMMARY

The discussion centers on determining the range of temperature (T) as time (t) approaches infinity using the equation T = 15/1.36(0.6*sin(t) - cos(t)) + 100 - 23.971/e^(0.6t). As t approaches infinity, the term 23.971/e^(0.6t) approaches zero, simplifying the equation to T = (15/1.36)(0.6*sin(t) - cos(t)) + 100. The range of temperatures is found by calculating the maximum and minimum values of the expression (15/1.36)(0.6*sin(t) - cos(t)) added to 100.

PREREQUISITES
  • Understanding of trigonometric functions, specifically sine and cosine.
  • Familiarity with limits and behavior of exponential functions as they approach infinity.
  • Basic knowledge of algebraic manipulation of equations.
  • Experience with temperature modeling in mathematical contexts.
NEXT STEPS
  • Calculate the maximum and minimum values of the expression (15/1.36)(0.6*sin(t) - cos(t)).
  • Explore the properties of sine and cosine functions over their periodic intervals.
  • Study the behavior of exponential decay in the context of limits.
  • Investigate temperature modeling techniques in differential equations.
USEFUL FOR

Mathematicians, physics students, and engineers interested in temperature modeling and the behavior of functions as they approach limits.

glid02
Messages
54
Reaction score
0
I already found this equation:
T= 15/1.36(.6*sin(t)-cos(t))+100-23.971/e^(.6t)

Now I'm supposed to find the range of temperatures (T) as t approaches infinity.

I tried 100+(15/1.36*.6*sin(90))
and 100-(15/1.36*cos(0))

but that's not right. What am I missing?

Thanks.
 
Physics news on Phys.org
If I got it right your equation looks like this:
[tex]\frac{15}{1.36}[0.6sin(t)-cos(t)]+100-\frac{23.971}{e^{0.6t}}[/tex]

Since [tex]\frac{23.971}{e^{0.6t}}[/tex] goes to [tex]0[/tex] when t goes to [tex]\infty[/tex] you should calculate the maximum and the minimum of the expression [tex]\frac{15}{1.36}[0.6sin(t)-cos(t)]+100[/tex] and this will be your range of temperatures.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
Replies
3
Views
2K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
8
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K