PLEASE HELP ME WITH THIS CALCULUS QUESTION!!! EASY 10 POINTS!!!
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PLEASE HELP ME WITH THIS CALCULUS QUESTION!!! EASY 10 POINTS!!!

[From: ] [author: ] [Date: 11-04-30] [Hit: ]
ln 3.5 = 2k + ln 2 (replace y (weight of puppy) with 3.5 pounds and t (2 months).)So then ln 3.5 - ln 2 = 2kk = [ln (3.5/2)]/2 = .......

So when y = 2, t = 0. (Initial weight of puppy is 2 pounds)

So ln 2 = c.

So when t = 2, ln 3.5 = 2k + ln 2 (replace y (weight of puppy) with 3.5 pounds and t (2 months).)

So then ln 3.5 - ln 2 = 2k

k = [ln (3.5/2)]/2 = ...(hell if I know xD)

So when t = 3, ln y = 1.5*[ln(3.5/2)] + ln 2

So to find y at this point (for 3 months) all you have to do is let e^(RHS) = y.

And y should be 4.6 (approximately) ^^.

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Weight increasing at a rate proportional to its weight: dw/dt = kw

Separate the variables:

dw/w = k dt..............integrate

ln w = kt + C..............write exponential form

w = e^(kt + C) = e^C(e^kt)...........when t = 0, w = 2, so e^C = 2

w = 2e^(kt)........when t = 2, w = 3.5. Use this to solve for k

3.5 = 2e^(2k)........e^(2k) = 1.75........2k = ln(1.75).....k=(ln(1.75))/2 = ~.2798

w(3) = 2 e^((.2798)(3)) = ~ 4.63 pounds

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wouldn't it help you if you googled it to find a VERY THROUGH EXPLANATION


TRY IT ALTEAST please
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