Thread: 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.

Substitute and solve simultaneously for c and k. eg. for the first question you have

and

so you have y=5 and y=1

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 . So c=5. Now use that and the 2nd point. Divide by 5, then take the natural log of both sides, . So now you have k.

so k= ln(1/5)
where do I go now

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.

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

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?

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

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).