Math Help - Hard problems-need severe help

1. Hard problems-need severe help

I've been working on practice problems over the summer in preparation for next years AP calc class, and these problems are beyond my knowledge. Any help is highly appreciated.

1. Sketch the polar curve r=2-4sinθ and determine the values of θ which generate the inner loop. Give these angles in exact radian measure.

2. A closed can in the shape of a right circular cylinder is designed to hold 12 oz. soda. If the height of the cylinder is x in., write an equation in terms of x to represent the amount of material needed to construct the closed can. (1 fl. oz. = 1.805 in^3) Find the minimum amount of material needed.

3. Solve over the given interval: sin(x) - cos(2x)=0, -pi/2 ≤ x ≤ pi

4. Two particles move in the x-y plane. For t≥0 the position of a particle A is x=(t-2)^2 and y=t-2 and the position of particle B is x=(3/2)t - 2 and y= (3/2)t - 4. At what time do the particles collide?

2. Originally Posted by KennyYang
I've been working on practice problems over the summer in preparation for next years AP calc class, and these problems are beyond my knowledge. Any help is highly appreciated.

1. Sketch the polar curve r=2-4sinθ and determine the values of θ which generate the inner loop. Give these angles in exact radian measure.
What have you done. If nothing else, just plot some points. Take $\theta = \frac{\pi}{2}$ and $\theta = 0$ and $\theta = \pi$ and see if you can get a sense of it. You calculator won't graph in polar coordiantes?

2. A closed can in the shape of a right circular cylinder is designed to hold 12 oz. soda. If the height of the cylinder is x in., write an equation in terms of x to represent the amount of material needed to construct the closed can. (1 fl. oz. = 1.805 in^3) Find the minimum amount of material needed.
$Volume = \pi\;r^{2}h$ That's about all you need.

3. Solve over the given interval: sin(x) - cos(2x)=0, -pi/2 ≤ x ≤ pi
This is a trig problem. $\cos(2x) = 1 - 2\sin^{2}(x)$ Put that in there and use your best quadratic equation skills.

4. Two particles move in the x-y plane. For t≥0 the position of a particle A is x=(t-2)^2 and y=t-2 and the position of particle B is x=(3/2)t - 2 and y= (3/2)t - 4. At what time do the particles collide?
First, do they EVER cross paths? Can you draw those parametric definition as given or would it be easier now to get rid of the "t"? After determining that they are at least in the same neighborhood, do they get there at the same time? You'll have to put the "t" back in to tell that. In any case, if they run into each other, the "x"s and "y"s will have to do it at the same time.

Solve these and see if you get the same solution

x: $(t-2)^{2} = \frac{3}{2}t - 2$

y: $t-2 = \frac{3}{2}t - 4$

Good Luck in AP Calc!

3. Thanks for the help TK! For #1 I tried graphing and tracing the points, and ended up with pi/6, pi/4, pi/3, pi/2, 12pi/5, 12pi/7, 3pi/2, 4pi/3, and 6pi/5. I'm pretty sure these are correct.
I'm still working on #2 as how to solve for x.
#3 was not hard once I realized I could use the quadratic formula, and ended up with x=-1 and x=1/2.
For #4 I ended up with the particles colliding at points (1,-1) and (4,2).
If anything looks wrong feel free to let me know. Thanks again for the assistance.

4. For #2:

The volume is ${\pi}r^{2}x=\frac{1083}{50}$....[1]

The surface area is $2{\pi}rx+2{\pi}r^{2}$.....[2]

Solve [1] for r and sub into [2]. Then [2] will be entirely in terms of x.

5. Originally Posted by KennyYang
Thanks for the help TK! For #1 I tried graphing and tracing the points, and ended up with pi/6, pi/4, pi/3, pi/2, 12pi/5, 12pi/7, 3pi/2, 4pi/3, and 6pi/5. I'm pretty sure these are correct.
I'm still working on #2 as how to solve for x.
#3 was not hard once I realized I could use the quadratic formula, and ended up with x=-1 and x=1/2.
For #4 I ended up with the particles colliding at points (1,-1) and (4,2).
If anything looks wrong feel free to let me know. Thanks again for the assistance.
#1 You may not be understanding the question. I think it wants ALL the angles that make up the inner loop. Let's see, (0,2), Okay, when do we first get to zero? $2-4\sin(\frac{\pi}{6}) = 0$. Keep going around until it hits zero again. $2-4\sin(\frac{5\pi}{6}) = 0$. So, ALL the angles are $\frac{\pi}{6} \le \theta \le \frac{5\pi}{6}$

#2 Let's see what you're doing. Have you found a derivative of a single variable? You WILL have to do that.

#3 I hope you did NOT get x = -1 or x = 1/2. Where did the sines go?

#4 So close!! You did not heed my warning. Their PATHS do indeed intersect at those locations, but they are NOT necessarily there at the same time. Check out both of those for "t" and see if you get the same value for each of our wandering points.

6. Hello,Kenny!

1. Sketch the polar curve $r \:=\:2-4\sin\theta$
and determine the values of $\theta$ which generate the inner loop.
We have a limacon with a loop.

The loop begins and ends when $r = 0$.

Solve: . $2 - 4\sin\theta\:=\:0\quad\Rightarrow\quad \sin\theta\,=\,\frac{1}{2}\quad\Rightarrow\quad\th eta \:=\:\frac{\pi}{6},\:\frac{5\pi}{6}$

If you make a sketch, you see the loop is generated for: . $\boxed{\;\frac{\pi}{6}\:\leq \:\theta \:\leq \:\frac{5\pi}{6}\;}$

7. Soroban and TK: ah thank you! I had slightly misunderstood what to do.
For #3: I had forgot about the sines. Plugging in -1 and 1/2 for them gave me 3pi/2 and pi/6. Thanks for the reminder =D.
For #4: Looking at the value of T showed me they only collide at (4,2).
Galactus, with your suggestion I ended up getting x by itself and then graphing the surface area equation to find 43.017 as the minimum amount of material needed.
I think that sums up pretty much everything, and all looks correct! Thanks everyone for your input. What a great community!