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View Full Version : [High School AP Calc] a few problems

insanefreddy926
11-30-2009, 07:29 PM
yo here's a few take home questions I got. All help is appreciated. :]

1. An object moves along the x-axis with initial position x(0)=2. The velocity of the object at time t >= 0 is given by v(t)=sin((pi/3)t). What is the acceleration of the object at time t=4? What is the position of the object at time t=4? What is the total distance traveled by the object over the time interval 0 <= t <= 4?

I got:

a(4) = -pi/6
x(4) = (4pi + 9)/(2pi)
total distance = 15/(2pi)

2. Let f(x)= x^3 + px^2 + qx. Find the values of p and q so that f(-1)=-8 and f'(-1)=12. Find the value of p so that the graph of f changes concavity at x=2. Under what conditions of p and q will the graph of f be increasing everywhere?

for the first part I got p = -2 and q = 5
for the second part I got p = -6
for the last part I said whenever p is greater than or equal to 0 and q is greater than 0

3. Let f(x)= 4 - x^2. For 0 <= w <= 2, let A(w) be the area of the triangle formed by the coordinate axes and the line tangent to the graph of f at the point (w, 4 - w^2). Find A(1). For what value of w is A(w) a minimum?

I know what's going on here, but I'm having trouble coming up with the equation for A(w).

4. Consider the curve given by the equation y^3 - 3xy = 2. Find dy/dx. Write an equation for the line normal to the curve at the point (1,2). What is the concavity of the curve at that point?

edit: whoops I messed up, the dy/dx value I got is y/(y^2 - x) and the normal line is y - 2 = (-3/2)(x - 1)
don't know how to determine the concavity there. would it be from the second derivative?

lxDestinyxl
10-21-2010, 12:43 PM
For part 4, differentiate it again implicitly. If you differentiate it successfully, you will be able to replace your dy/dx's with the value of the derivative you got for your normal. Simplify, and you will get a positive or negative result, which will tell you of the curve's concavity at that point.

MrRubix
10-21-2010, 02:01 PM
A(w) is, as the problem says, formed by the axes and the tangent line to the parabola for 0<=w<=2 (these are your bounds because they keep things bound between the top of the parabola and the x axis).

Tangent line to f(x) is defined by the derivative of f(x), or f'(x) = -2x, which tells you the slope of the tangent at a given point.

General equation for tangent line here:
y = f(w) + f'(w)(x-w)
Or, here,
y = (4-w^2) - 2w(x-w)

x intercept (when y=0, plug into tangent equation above). This acts as our triangle's base.
x = (4+w^2)/2w

y intercept (when x=0, plug into tangent equation above). This acts as our triangle's height.
y = 4+w^2

The triangle's area here for A(1) is therefore 1/2*(b)(h) or, for w=1, (1/2)*(5/2)*(5) = 6.25.

Regarding the minimum triangle area, we know that A(w) = ((4+w^2)/2w)*(4+w^2))/2 = (4+w^2)^2/(4w)

So we need to find A'(w)=0, which means finding the derivative of A(w) -- which equals:

(4w(2(4+w^2)*2w) - 4*(4+w^2)^2)/(16w^2) = 0 via quotient rule

Simplified:
(w^2+4)*(3w^2-4) = 0
or
3w^4+8w^2-16=0
Which holds only when w = positive or negative (2/rad3)

Since w must be positive here, the triangle's area is minimized when w = 2/rad3

lxDestinyxl
10-22-2010, 08:08 AM
You were supposed to give a lead towards the answer, rather than completely solving it. Ah well.

MrRubix
10-22-2010, 09:09 AM
<--- does not give a ****

if the OP doesn't know how to solve it, then he's screwed anyway if it's on a test
i personally feel having the full answer is the best way to learn something. You can see all the steps and how things are applied.

All_That_Chaz
10-22-2010, 12:19 PM
As long as Rubix is doing math homework for people I have some crazy Actuarial P-Exam questions I don't get...

MrRubix
10-22-2010, 01:53 PM
Post away and I shall enlighten

MrRubix
10-22-2010, 01:58 PM
I just realized the OP's post was like a year ago

lmfao

All_That_Chaz
10-22-2010, 02:00 PM
Nah I can figure them out and I have a solutions manual anyway. I'm just freaking out about my midterm on Tuesday.

EDIT: yeah I think you might be a bit too late on that one.