<p>bump......</p>
<p>K polar coordinate question time (GASP!):</p>
<p>find the area inside one petal of the rose curve r = 4cos(3[theta])</p>
<p>NOTE: Use your calculator for this one to get 3 dec approx, Doing it w/o a calculator is way too tedious.</p>
<p>Area = integral 0.5r^2 d[theta] from A to B
= 2* integral 0.5(4cos(3theta))^2 dtheta from 0 to pi /4
= 16 integral cos^2(3theta) dtheta from 0 to pi/4
= whatever the equals after being evaluated</p>
<p>right?</p>
<p>Very close. The last interval should be pi/6, not pi/4 then you are all good. (because at pi/6, the petal reaches its end)</p>
<p>oh right, because it's 3theta, not 2theta.</p>
<p>This is probably incorrect:</p>
<p>(1/2)∫(-π/6, π/6) (4cos(3ϑ))^2 dϑ
= (1/2)∫(-π/6, π/6) (16cos^2(3ϑ)) dϑ
= 8 ∫(-π/6, π/6) (cos^2(3ϑ)) dϑ
= 8 ∫ ((cos(6ϑ)+1)/2) dϑ < - - - I'll put in (-π/6, π/6) at the end.
= 4 ∫ (cos(6ϑ)+1) dϑ
= 4 ∫ (cos(6ϑ) dϑ + 4 ∫ 1 dϑ
= 4 ((1/6)sin(6ϑ)) + [4(ϑ)]
= (2/3)<em>sin(6ϑ) + (4ϑ) | (-π/6, π/6)
= [(2/3)</em>sin(π) + (2π/3)] - [(-2/3)<em>sin(-π) + (-2π/3)]
= (2/3) + (2π/3) - [(2/3)</em>sin(π) - (2π/3)]
= (2/3) + (2π/3) - [(2/3) - (2π/3)]
= (4π/3).</p>
<p>Chaos, you are nearly perfect, you just made one silly mistake at the end, it should be simply 4pi/3 [because sin(pi)= 0]. I doubt ETS will force you to do such a problem, sorry for the tedious problem.</p>
<p>Right. Oops. :D</p>
<p>Yes, that problem wasn't very fun. I need the practice, though. >_<</p>
<p>Last one before I go to sleep:</p>
<p>∫x*cos(x)dx</p>
<p>Ehhh. Integration by parts.</p>
<p>Let
v = sin(x)
dv = cos(x) dx
u = x
du = dx</p>
<p>= uv - ∫vdu
= x<em>sin(x) - ∫sin(x)dx
= x</em>sin(x) + cos(x) + C.</p>
<p>a' = cos(x)
a = sin(x)
b = x
b' = 1</p>
<p>∫x*cos(x)dx = xsin(x) - ∫sin(x) dx
= xsin(x) +cos(x) + C</p>
<p>Post more problems. :cool:</p>
<p>A free response I need help on:</p>
<p>A metal wire of length 8 cm is heated at one end. The table below gives selected values of the temperature T(x), in degrees Celsius, of the wire x cm from the heated end. The function T is decreasing and twice differentiable.</p>
<h2>Distance| 0 | 1 | 5 | 6 | 8</h2>
<p>Temp | 100 | 93 | 70 | 62 | 55</p>
<p>Are the data in the table consistent with the assert that T''(x) [second derivative] > 0 for every x in the interval 0 < x < 8? Explain your answer.</p>
<p>Good luck on the BC test everyone. It should be a breeze. Don't sweat it, and have fun. :D</p>
<p>I'm glad that I only need a 3.</p>
<p>Here is a problem I posted in the other BC thread</p>
<p>Okay: Logistic growth problem!</p>
<p>dP/dt = (P/5)(1-P/12)</p>
<p>a) If p(0) = 3, what is the limit of P(t) as t goes to infinity?</p>
<p>b) If p(0) = 20, what is the limit of P(t) as t goes to infinity?</p>
<p>lim t--> infinity means that it's asking for the carrying capacity.</p>
<p>I feel so inadequate. I could barely do any of these questions. :(</p>
<p>K here is another (simple) one: d/dx ∫4t^4 + 5 (from 2 to x^2) dt</p>
<p>here is one right off one old FRQ</p>
<p>given that x(0) = 5 and v(t) = sin (13-(t^2)) what is the particle's position at x = 3</p>
<p>calc active</p>