What does k indicate for y=e^(-kt)

  • Thread starter brandy
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In summary, k as a percentage indicates the portion of temperature difference that is equal to the rate of change of temperature with respect to time, or the steepness of the curve when plotting exponential functions.
  • #1
brandy
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Homework Statement


for T=e^(-kt)
explain what k as a percentage indicates.

2. Relevant information
T=temperature in degrees
t=time in minutes

The Attempt at a Solution


i really don't know, i was thinking something like the percentage of rate of decay and i fiddled around with the a f(t+1) - f(t) and then taking that away from f(t+2) - f(t+1) etc and dividing them and such but i still don't know what k indicates.
please help.
 
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  • #3
From the equation you just specified it seems clear that you are working with rate of decay problems:
[tex] T = e^{(-kt)}[/tex]
To truly understand rate of decay you generally need a little knowledge of calculus but I will try to explain this for you in terms of precalculus.

For all temperature decay problems, the rate of change of temperature depends on the amount of temperature difference currently present:

Imagine a very hot object placed in cold water. At first the object cools quickly when the temperature difference between the object and the water is large. As the temperature between the object and the water start to become closer the object cools at a slower rate. (Meaning that change in temperature divided by change in time is less.) K is simply a constant that describes the rate at which the object cools based on the difference in temperature from the object and its surrounding environment.

I see that you were attempting to find the percentage of temperature lost after a certain amount of time when you tried [f(t +1) - f(t)] / [f(t + 2) - f(t + 1)] but that number will not be consistent for every t because the equation is exponential and not linear. Since you probably haven't taken calculus yet I don't believe I can explain the meaning of k any deeper but don't worry because you're probably not expected to deeply understand the problem.

If you want to continue to understand the meaning of the above equation I recommend graphing it on a graphing calculator and experimenting with different values of k and seeing how that changes the equation.
 
  • #4
i actually have done caclulus.
what i meant when i said the stuff about f(t+1) is i was just fiddleing with numbers to see if i can get some meaning.
so i tried:
f(t+1) - f(t)

i also tried, (f(t+2) - f(t+1)) - (f(t+1) - f(t))

i also tried, f(t+1)/f(t)

i know about the rate of decay and that the decay percentage is 100*e^-k
and i know the rate of decay is -Ak*e^-kt

but i still can't express in words what k as a percenrage indicates.
i can tell you what e^-k indicates
but not k
this is my dilema
 
  • #5
K indicates the portion(percentage?) of temperature difference that is equal to the rate of change of temperature with respect to time:

[tex]\frac{dT}{dt} = -k(T_O - T_A)[/tex]

[tex]dT = -kTdt[/tex]

[tex]\frac{dT}{T} = -kdt[/tex]

[tex]\int \frac{dt}{T} = \int -kdt[/tex]

[tex]\ln{(T)} = -kt[/tex]

[tex]T = e^{(-kt)}[/tex]

I hope my words could help. Your're right; it is a difficult thing to word.
 
  • #6
oh yes. why didnt i think of that!
its soooo simple
oh well
i guess we all have our moments
 
  • #7
No problem. I didn't think of it either. I had to dust off the old calculus book and look at the proof.
 
  • #8
k can be imagined as the steepness of the curve if you plot exponential functions.
 

FAQ: What does k indicate for y=e^(-kt)

What does k indicate for y=e^(-kt)?

K indicates the rate of change or decay of the function. It is the constant that determines how quickly the value of y decreases as t increases.

How does k affect the graph of y=e^(-kt)?

K affects the steepness of the graph. A larger k value results in a steeper curve, while a smaller k value results in a more gradual decrease.

What happens to the graph when k is negative?

When k is negative, the graph of y=e^(-kt) will be reflected across the y-axis, resulting in a decreasing exponential curve instead of an increasing one.

What is the relationship between k and the half-life of y=e^(-kt)?

The half-life of y=e^(-kt) is equal to ln(2)/k. This means that as k increases, the half-life decreases, and vice versa.

Can k be zero or negative in the function y=e^(-kt)?

No, k must be a positive number in this function. A value of zero or a negative value would result in division by zero, which is undefined.

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