# F(x) = ce^kt

• Nov 3rd 2008, 03:58 PM
theflyingcow
F(x) = ce^kt
Find the function f(t) = ce^kt that passes through the given points in the plane.

a) (0,5) and (1,1)

b) (2,4) and (3,7)

Okay i don't know how the teacher did this in class today. Teach me the ways.
• Nov 3rd 2008, 04:03 PM
Substitute and solve simultaneously for c and k. eg. for the first question you have
$\displaystyle 5 = ce^{k(0)}$
and
$\displaystyle 1 = ce^{k(1)}$
• Nov 3rd 2008, 04:05 PM
theflyingcow
so you have y=5 and y=1
• Nov 3rd 2008, 04:05 PM
Jameson
Quote:

Originally Posted by theflyingcow
Find the function f(t) = ce^kt that passes through the given points in the plane.

a) (0,5) and (1,1)

b) (2,4) and (3,7)

Okay i don't know how the teacher did this in class today. Teach me the ways.

a) Think of these two points in the form of (t, f(t)). So plugging in the first point we get $\displaystyle 5=ce^{k*0}=ce^0 = c(1) = c$. So c=5. Now use that and the 2nd point. $\displaystyle 1=5e^{k*1}$ Divide by 5, then take the natural log of both sides, $\displaystyle \ln( \frac{1}{5} )= k\ln(e)=k$. So now you have k.
• Nov 3rd 2008, 05:05 PM
theflyingcow
so k= ln(1/5)
where do I go now
• Nov 3rd 2008, 05:12 PM
Jameson
Ok you need to think about your function, F(t), and what every letter represents. I solved for c and k. They are both constants defined by the two points in part (a). For part (b) they will change. e is a constant number 2.71... which is Euler's constant. If you are unfamiliar with this, you should look it up. Now you wrote in the title of this thread, F(x), not F(t). Think about this. If F is a function dependent on x, why are there no x's in the equation? If x isn't the variable, what is? t is. Or you need to change it to F(x)=ce^{kx}. Either way, but you need to be consistent.

So, you have c and k (constants) and e (always a constant), so this problem is completed.
• Nov 3rd 2008, 05:22 PM
theflyingcow
but we use two points and the overall k = (ln 1/5) / 1.
• Nov 3rd 2008, 05:26 PM
Jameson
Quote:

Originally Posted by theflyingcow
but we use two points and the overall k = (ln 1/5) / 1.

I don't get this post.

Yes, two points are needed to SOLVE for c and k, which are constants. What do you mean by the overall k?
• Nov 3rd 2008, 05:27 PM
theflyingcow
the k for the second point sorry
so the function is y = 5e^((ln 1/5) / 1)*t)
• Nov 3rd 2008, 05:32 PM
Jameson
Quote:

Originally Posted by theflyingcow
the k for the second point sorry
so the function is y = 5e^((ln 1/5) / 1)*t)

Yes, but you don't need to write anything /1 . It's understood that anything divided by one is itself and it's always omitted.

I will say you'll need to use the method that badgerigar described in his post for part (b).