2009 AP Calculus BC Test Answers

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thatguy47

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Here's a link to the AP free-response test everyone took on Wednesday. This is for BC only.
http://apcentral.collegeboard.com/apc/p ... lus_bc.pdf

Can anyone solve number 3 or number 6 correctly?
Here's my answers:
1.
a) -.1 m/s^2
b) 1.35
c) t=2
d) Caren
2.
a) 980
b) t=1.363
c)
d)21.052
3.
a)
b)
c)
d)
4.
a) 4.25
b)
c)
5.
a) -3
b) 8
c)
d)
6.
a)
b)
c)
d) one point?
 
If we integrate the given dx/dt and dy/dt we get

\(\displaystyle x=.8t, \;\ y=\frac{18}{5}t-\frac{49}{10}t^{2}\)

By setting dy/dt=0 and solving for t=18/49 sec and subbing back into y, we get:

\(\displaystyle y=\frac{18}{5}(18/49)-\frac{49}{10}(18/49)^{2}=\frac{162}{245}\approx .661\)

That is the distance from her shoulders to the top when she starts down, the high point.

Twice that plus 11.4 gives the total distance traveled till she hits the water.

\(\displaystyle 2(\frac{162}{245})+\frac{57}{5}=\frac{3117}{245}\approx 12.722 \;\ m\)

The total time can be found from this.

I hope I am on the right track. My projectile physics is a little rusty.
 
Some people are getting:
3.
a)12.722 m
b)1.936 seconds
c)1.5488m

I didn't attempt to solve this problem so I'm not sure if these answers are right or not.
 
I think those values are mixed up. 12.722 should be for part c.

To find the total time till hitting the water solve \(\displaystyle 3.6t-4.9t^{2}=-11.4\)

t=1.936 seconds.

To find the max x distance, sub that time into x=.8t

.8(1.936)=1.5488 m.

I agree with those solutions. They seem correct.

To find the angle use tan and the y and x found previously.
 
Ya, they have:
3.
a) 12.06 m
b) 1.936 s
c) 12.94 m
d) 1.51 radians

Here's an updated list of answers:
http://talk.collegeconfidential.com/ap- ... swers.html
1.
a) -.1 m/s^2
b) Distance traveled by Caren; 1.8 m
c) t=2
d) Caren
2.
a) 980 people
b) 1.363 s
c ) 387.5 hours
d) .776 hours
3.
a) 12.06 m
b) 1.936 s
c) 12.94 m
d) 1.51 radians
4.
a) 4.25
b)
c)
5.
a) -3
b) 8
c) 18
d)
6.
a) e^(x-1)^2 = 1 + (x - 1)^2 + (x-1)^4/2! + (x - 1)^6/3! + ... + (x-1)^2n/n! + ...
b)f(x) = 1 + (x-1)^2/2! + (x-1)^4/3! + (x-1)^6/4! + ... + (x-1)^n/(n+1)! <- i tested that value
c) -infinity < x < infinity (converges for all x)
d) f'(x) = x - 1 + (x-1)^2/2! + ...
f''(x) = 1 + ....
no points of inflection b/c all the powers of (x-1) are even and the constant is 1, so it f''(x) is always greater than 0
 
I was calculating some wrong things. I was thinking horizontal distance.

Part a: \(\displaystyle 3.6(.367)-4.9(.367)^{2}=.661\)

add this to 11.4 and get 12.06 m
 


I got a somewhat different result for the first part of exercise (3).

I considered the diver's shoulders as a point moving through space.

x'(t) gives horizontal velocity, and y'(t) gives vertical velocity.

The antiderivative for x'(t) is 0.8t + C1.

The antiderviative for y'(t) is -4.9t^2 + 3.6t + C2.

The point's path through space is parabolic. Functions x(t) and y(t) represent a parameterization of this path.

x(t) = 0.8t + C1

y(t) = -4.9t^2 + 3.6t + C2

If we introduce an XY-coordinate system, then the X-coordinate of the point at time t is x(t), and the Y-coordinate of the point at time t is y(t).

By placing the origin such that the point's location at t = 0 is (0,Y) and the point's location at the surface of the water is (X,0), then C1 = 0 and C2 = 11.4.

x(t) = 0.8t

y(t) = -4.9t^2 + 3.6t + 11.4

When the vertical velocity is zero, the point is at it's maximum vertical displacement in the positive direction.

y'(t) = 0 leads to t = 0.3673.

x(0.3673) = 0.2939

y(0.3673) = 12.0612

So, I get (0.3, 12.1) for the XY-coordinates of the parabola's vertex.

When I remove the parameter t, I get the following quadratic function:

Y(X) = -7.65625X^2 + 4.5X + 11.4

The maximum here is also 12.0612, so I'm thinking that the answer to part(a) is approximately 12.1 meters, and the answer for part (b) is approximately 1.94 seconds.

For part (c), I'm reading the question to ask for the distance of the point's trajectory along the parabola, not the total vertical displacement.

I may work on parts (c) and (d) after I get back from dinner. 8-)

MY EDITS: Fixed typographical error, and then came back to fix BBCode syntax error. And back again, to fix a grammatical error (sigh).
 
galactus said:
I was calculating some wrong things … [now I] get 12.06 m
Click to expand...


Yay. This matches my result.

I know there's a calculus formula for the distance along a function's graph, but I need to look it up.

For part (c), I'm thinking that I can use this method with my function Y(X) above from X = 0 to X = 1.5490 to get the distance traveled by the point along the curve from t = 0 to t = 1.94.

 
Yes, part c is the total distance traveled by the diver.

It takes .367 seconds to get to the top of the curve. So, 3.6(.367)-4.9(.367)^2=.661 m

double that and add 11.4. We get 12.72 m

I get 12.06 m for part a and 1.936 sec. for part b.

for part d, I am getting about 1.49 rad.
 
Here's an updated list of the answers. Do these all look right?

1a) -.1 miles/min^2
1b) 1.8 miles, total distance travelled on time interval in miles
1c) t=2, velocity changes sign
1d) Caren 1.4 miles < Larry 1.6 miles

2a) 980 people
2b) t=1.363, second derivative test
2c) 387.5 hours (first fundamental theorem)
2d) .776 hours

3a) dy/dt = 0 @ t=.3673, so 12.061m
3b) A = 1.936 seconds
3c) s = 12.946m
3d) arctan(dy/dt/dx/dt) = 1.519 radians

4a) 4.25
4b) 2+4(x+1)-6(x+1)^2
4c) y = -4/e^(1/3) * e^(-x^3/3) + 6

5a) -3
5b) 8
5c) 18
5d) y+2 = 3(x-5) tangent line; y-3 = 5/3(x-8) secant line, just plugin and state since the graph is always concave down, tangent line over approximates, secant line underapproximates

6a) 1 + (x-2)^2 + (x-1)^4 / 2 + (x-1)^6 / 3! + ... + (x-1)^(2n) / n!
6b) 1 + (x-1)^2 / 2 + (x-1)^4 / 3! + (x-1)^6 / 4! + ... + (x-1)^(2n) / (n+1)!
6c) -inf < x < inf
6d) no points of inflection since original function is an always positive even function, so second derivative is an always positive even function, indicating concavity never changes. if you show the second derivative, you'll see that it's always >=1.
 
galactus said:
Yes, part c is the total distance traveled by the diver. This wording still seems ambiguous, to me.

It takes .367 seconds to get to the top of the curve. So, 3.6(.367)-4.9(.367)^2=.661 m

double that and add 11.4. We get 12.72 m …
Click to expand...


I read part (c) as asking for the total distance that the point moves along the parabolic trajectory from t = 0 to t = 1.94.

It looks, to me, like the result above is the total of the absolute values of the two vertical displacements.

I get approximately 12.9 meters (along the curve).

I had to look up the formula for the length of a curve. I found two formulas; one for y as a function of x, and one for both x and y as functions of t. They both give me the same result.

\(\displaystyle L \; = \; \int_a^b \sqrt{1 \; + \; \left( \frac{dy}{dx} \right)^2} \, \; dx\)

\(\displaystyle L \; = \; \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 \; + \; \left( \frac{dy}{dt} \right)^2} \, \; dt\)

 
thatguy47 said:
… 3d) arctan(dy/dt/dx/dt) = 1.519 radians … I think you mean arctan([dy/dt]/[dx/dt]).
Click to expand...


As a confirmation, I calculated the slope manually from actual (x, y) coordinates obtained from using the time at entry (t2) and a "split second" prior to entry (t1).

t1 = 1.936255599492454682221413
t2 = 1.936255599492454682221414

I got approximately 1.52 radians, also. (Diver's entry almost vertical, since Pi/2 ? 1.57.)

Good job. 8-)

 
Yes, I got the same afterall. I had a booboo in my calculations. I also agree with your arc length. I was calculating the vertical distance and not the length along the curve.

\(\displaystyle {\theta}=tan^{-1}\left(\frac{\frac{18}{5}-\frac{49}{5}\left(\frac{\sqrt{5910}+18}{49}\right)}{\frac{4}{5}}\right)=tan^{-1}\left(\frac{\sqrt{5910}}{4}\right)\approx 1.51881171662 \;\ \text{radians}\)

Or about \(\displaystyle 87^{\circ} \;\ 1' \;\ 17.4''\)

Is that close to what you got, mmm?.
 
thatguy47 said:
… Do these all look right?

1a) -.1 miles/min^2
1b) 1.8 miles, total distance travelled on time interval in miles
1c) t=2, velocity changes sign
1d) Caren 1.4 miles < Larry 1.6 miles …
Click to expand...


These look correct, to me.

 
galactus said:
… Is that close to what you got, mmm?
Click to expand...


Close enough! :D

Actually, it's exactly what I got (rounded to your number of digits) using the xy-coordinates to determine the slope as described above:

m = -19.219131093782569857212315

arctan(-m) = 1.5188117166220119492749924

 
thatguy47 said:
… Do these all look right? …

… 2a) 980 people
2b) t=1.363, second derivative test
2c) 387.5 hours (first fundamental theorem)
2d) .776 hours …
Click to expand...


These all match my results.

 
I’d just like to (belatedly) point out that the diving problem was very poorly stated. Reference should have been made to the diver’s center of mass/gravity – not the shoulder position. A diver’s shoulders do not follow a parabolic trajectory. They start some height above the diver’s cg and, due to rotation during the dive, end up below the diver’s cg. Nitpicky, perhaps, but it bothered me... :)
 
On problem 1.(a).

a(t) = v '(t) = lim [v(t+h)-v(t)]/h as h->0.

t = 7.5 min = 450 sec.

a(450) = lim [v(450+h)-v(450)]/h as h ->0.

Let h = 1 sec, h is small, then a(450) = lim[v(451)-v(450)]/1 as h->0.

Hence a(7.5min) = a(450sec) = lim [ 1311.2ft/sec-1320ft/sec]/1sec as h ->0 = -8.8ft/sec².

By the way, what does BC mean?
 
wjm11 said:
… Nitpicky, perhaps, but it bothered me …
Click to expand...


I don't think it's nitpicky, at all.

I've seen the College Board make some fairly nitpicky comments over some student's answers (or forms of answers), in the past.

They should know better! (Ahem, and I should have noticed it, too.)

 
BigGlenntheHeavy said:
… By the way, what does BC mean?
Click to expand...


It's some sort of calculus topic descriptor.

There is a lot of information about the AB and BC topics covered at collegeboard.com.

Specifically, you can see the topic outlines for AB and BC on pages 7 through 13 in the following document.

http://apcentral.collegeboard.com/apc/p ... sedesc.pdf



Also, here's a blurb cut-and-pasted from their site:

In 1956, 386 students took what was then known as the AP Mathematics Exam. By 1969, still under the heading of AP Mathematics, it had become Calculus AB and Calculus BC. The Calculus BC exam covers the same differential and integral calculus topics that are included in the Calculus AB exam, plus additional topics in differential and integral calculus, and polynomial approximations and series. This is material that would be included in a two-semester calculus sequence at the college level. Because graphing calculator use is an integral part of the course, the exam contains questions that require students to use a graphing calculator.



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

 
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