2009 AP Calculus BC Test Answers

[mmm4444bot]

I need to revisit 1(a) because it appears from your post that I probably got it wrong.

I'm okay. Somehow, I got it into my head that the units in my answer were meters per minute^2. (How dense! Caren is not a turtle!)

It's -0.1 miles/min^2, and that converts to -8.8 ft/sec^2.

(Whew. BTW, I did it the "easy" way. :wink: )

 

[BigGlenntheHeavy]

Problem 1(d). We know Larry lives 1.6 miles from school, but doesn't Carin also live 1.6 miles from school as after going .2 miles to school from her home, she had to go back the .2 miles to get her calculus book, hence they both live the same distance from school: 1.6 miles.

 

[mmm4444bot]

I calculated Caren's distance traveled over the 12 minutes as 1.8 miles, so, after subtracting the 0.4 miles from returning home and back, I ended up with 1.4 miles.

 

[galactus]

Has anyone tackled #6?. The Taylor series for \(\displaystyle e^{(x-1)^{2}}\) centered at x=1 is:

\(\displaystyle 1+(x-1)^{2}+\frac{(x-1)^{4}}{2}+\frac{(x-1)^{4}}{6}+...................\)

It's general series is \(\displaystyle \sum_{k=0}^{\infty}\frac{(x-1)^{2k}}{k!}\)

For \(\displaystyle f=\frac{e^{(x-1)^{2}}-1}{(x-1)^{2}}\) is:

\(\displaystyle 1+\frac{(x-1)^{2}}{2}+\frac{(x-1)^{4}}{6}+.....................\)

\(\displaystyle \sum_{k=1}^{\infty}\frac{(x-1)^{2k}}{k!}\).

The ratio test gives \(\displaystyle \frac{(k+1)!}{(x-1)^{2(k+1)}}\cdot\frac{(x-1)^{2k}}{k!}=\frac{k+1}{(x-1)^{2}}\)

\(\displaystyle \lim_{k\to \infty}\frac{k+1}{(x-1)^{2}}=\infty\)

Therefore, the series diverges. Consequently, the radius of convergence is R=0