<p>I am stumped on this one problem in a practice AP Calc AB exam my teacher gave my class today:</p>
<p>The weight in pounds of a certain bear cub t months after birth is given by *w<a href=“%5Bi%5Dt%5B/i%5D”>/i</a>. If *w<a href=“2”>/i</a> = 36, *w<a href=“7”>/i</a> = 84, and dw/dt was proportional to the cub’s weight for the first 15 months of his life, how much did the cub weigh when he was 11 months old?</p>
<p>(A) 125 pounds
(B) 135 pounds
(C) 145 pounds
(D) 155 pounds
(E) 165 pounds</p>
<p>I thought about using a differential equation w/ something like…</p>
<p>dw/dt = kw
integral (dw/w) = integral (k) dt
ln w = kt + C
w = e^(kt + C)
but I don’t know what to do from here, and I don’t know if that work is even correct for this problem… Any help would be greatly appreciated!</p>
<p>This is how I did it, though I could be wrong with finding the k variable. Anywho continuing from where you left off:</p>
<p>w = e^(kt + C)
w = Ce^(kt)</p>
<p>Comparing the givens:</p>
<p>36 = Ce^(2k)
84 = Ce^(7k)</p>
<p>Solving for C:</p>
<p>C = 36 / (e^(2k))
C = 84 / (e^(7k))</p>
<p>Then using a graphing calc, solve for k:</p>
<p>k = 0.1695</p>
<p>Plug into one of the equations solving for C and you get:</p>
<p>C = 25.65</p>
<p>Then you have the equation:</p>
<p>w = 25.65<em>e^(0.1695</em>t)</p>
<p>Plug 11 into the equation and you get:</p>
<p>w = 165 lbs.</p>
<p>(E)</p>
<p>I’m sure I could be wrong somewhere in there; this is my first CC math solution yay!</p>
<p>Excellent! You did just fine, lol. Many thanks :)</p>
<p>It’s basically just that an exponent, eventually, is going to grow MUCH faster than a variable to some constant number.</p>
<p>For example, between x^2 and 2^x, it’s obvious that as x->infinity, 2^x will grow much faster.</p>
<p>I’m not sure if (A) is suppose to be x^(18-5x), or as you wrote it. If it’s x^(18-5x), then your choices are narrowed down to (A) and (E), the only two with variable exponents. But, since (A) would be (18-5x), the exponent is obviously becoming increasingly negative and the function would approach 0. So, therefore, (E) is your answer…</p>
<p>(A) y = x^18 - 5x
(B) y = 5x^2
(C) y = ln (x^2)
(D) y = (ln x)^2
(E) y = e^(0.01x)</p>
<p>EDIT:</p>
<p>Whoops… did you figure out why that was the answer? You deleted your post. :p</p>