not gonna lie i've got no clue what you've done

you're supposed to get an equation in the form V = Ae^-kt

Then input the values, divide both simultaneous equations so A cancels out and solve k from there

you're supposed to get an equation in the form V = Ae^-kt

Then input the values, divide both simultaneous equations so A cancels out and solve k from there

(edited 1 year ago)

Original post by user8937264980

not gonna lie i've got no clue what you've done

you're supposed to get an equation in the form V = Ae^-kt

Then input the values, divide both simultaneous equations so A cancels out and solve k from there

you're supposed to get an equation in the form V = Ae^-kt

Then input the values, divide both simultaneous equations so A cancels out and solve k from there

why cant it be V=Ae^t +B

if you take logs of V = Ae^-kt you get logV = -kt + logA

this is in the form y=mx + c and will be the correct log graph

however if you take logs of V = Ae^t + B you get logV = logA + t + logB

this is not in y=mx + c and c would be logA + logB which would not be the correct initial value of the car. This also means the gradient would be 1 which is incorrect

this is in the form y=mx + c and will be the correct log graph

however if you take logs of V = Ae^t + B you get logV = logA + t + logB

this is not in y=mx + c and c would be logA + logB which would not be the correct initial value of the car. This also means the gradient would be 1 which is incorrect

Original post by leoishush

why cant it be V=Ae^t +B

correct me if i'm wrong but from what im seeing i think what your doing is trying to replace the k with a +B. I kinda get where your coming from with this idea (if this is what yuor thinking), however it wont work out as k is a constant that will be directly affecting t. By just putting + B what your effectively saying is that this constant is affecting the whole model directly, which it isn't.

Original post by leoishush

why cant it be V=Ae^t +B

Youve not posted the question, but Id guess from the description B=0 for the first couple of parts, though c) seems to want the decay to a fixed value. For the exponential part

* e^t is an increasing exponential as the exponent "kt" has k=1. Again, there must be something (data/description) that would lead you to believe k<0 so a decreasing exponential. You dont really need to assume the sign if youre solving for k, but you dont.

* The k scales time or equivalently you dont assume the base is e (so roughly 3). Your model increased by a factor of ~3 every time unit. Having a k means

e^(kt) = (e^k)^t

So the closer k is to zero, or equivalently e^k is to 1, the slower the increase/decrease in the growth/decay.

A simple reprsentation of radioactive decay is

n = Ae^(kt)

it goes to zero as k<0 and the value of k determines the half life (scales time). A is the initial ratioactivity level. Its similar here.

If you fitted your model to the data (t=0,t=1), youd have a model which exponetiall increased (not decayed) and went hurtling through zero after several time units and it would keep on getting more and more negative as A is negative. I guess the question has something about exponentail decay, going to zero, ... so your model wouldnt adhere to this. Your model has two parameters and there are two data points, so you can solve for your A and B. However, GIGO applies so if the initial model is wrong ....

(edited 1 year ago)

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