<p>I was wondering if I could get some help with the some AP calculus problems which I have to do for summer work. If you know how to do any of these, even one, a reply would be greatly appreciated. </p>
<p>*Write an equation in slope-intercept form of the line with parametric equations x = 2t+1 and y = -3t-4</p>
<p>*Write the equation for the inverse of y = arccot x. Then graph the function and its inverse.</p>
<p>*Identify the equation of the line tangent to the graph of
(x^2/25) + (y^2/4) = 1 at (3, 1.6)<br>
A. 3x+10y-25=0 B. 15x-20y-13=0 C. 3x-10y+25=0 D. 20x-15y-36=0</p>
<p>*Write the polar equation 4r = 8/cos θ in rectangular form.
A. y = 2 B. x = 2 C. x = 1/2 D. y = 1/2</p>
<p>*Which equation represents a spiral of Archimedes?
A. r = 3 θ B. r = 3 + 3 sin θ C. r = 3 cos 2θ D. r = 3 + 2 cos θ</p>
<p>*Find the rectangular coordinates of the point whose polar coordinates are
(20, 140º). Round to the nearest hundredth.</p>
<p>*Find the quotient [cos (5π/12) + i sin (5π/12)] / [2{cos (π/12) + i sin (π/12)}]. Then write the result in rectangular form.</p>
<p>*Express 4[cos (π/6)+ i sin (π/6)] in rectangular form.</p>
<p>*Determine the eccentricity of the conic section represented by
9x^2 - 16x^2 = 144.</p>
<p>Thanks!!</p>
<p>for #1, solve for t from the x= equation and plug it into the y= equation</p>
<p>remember, slope intercept form is y=mx+b</p>
<h1>2 arccot x is the inverse of cot x, and since the inverse of the inverse of a function is just the function, cot x is the function you seek. Just throw them both on a graphing calculator and graph away. They should both be symmetric with respect to the line y = x.</h1>
<h1>3 "Tangent" is a signal for slope, which signals a derivative. Implicitly differentiate the equation and you get</h1>
<p>.08x + .5yy' = 0.
Now since a tangent line intersects the graph at the point you want to find the tangent line of, the x and y values are the same, so plug in 3 for x and 1.6 for y and solve for y', giving you
.24 + .8y' = 0, so y' = -.3.
Now you have the slope which is -.3, and so the equation should look like y = mx + b, so we have
y = -.3x + b.
Again the y and x value won't change so we plug in above x and y and solve for b, giving us
1.6 = -.3(3) + b
Solve for b, and we get b = 2.5
Now our tangent line looks like this
y = -.3x + 2.5
Now it's simply a matter of manipulating the choices to get above equation. Let's try A
3x+10y-25=0 10y=-3x+25 y = -.3x + 2.5
Well, that was convenient, looks like A is the answer.</p>
<p>thanks a ton, both of you.</p>
<p>can any1 help with any others?</p>
<ol>
<li>Remember for polar (i'll use a instead of theta because I'm illiterate)</li>
</ol>
<p>x = rcosa
y = rsina</p>
<p>In 4r = 8/cosa, bring the cos a to the other side to get 4rcosa = 8
4x = 8, x = 2. B</p>
<ol>
<li>I actually have no clue what that is, but using elimination...
B is a cardiod
C is a rose
D is a limacon (right? but something like that)</li>
</ol>
<p>So, it has to be A.</p>
<ol>
<li>Turn it into cis form, which is much easier to read.
(cis is basically an abbrev. for cos + isin)</li>
</ol>
<p>[cis(5pi/12)]/[2cis(pi/12)]</p>
<p>Divide the "r"s and subtract the angles, so you end up with
(1/2)cis(5pi/12 - pi/12) = .5cis(pi/3) = (1/2)cos(pi/3) + (i/2)sin(pi/3)
= 1/4 + i*sqrt(3)/4</p>
<p>Last One.</p>
<p>Eccentricity is c/a</p>
<p>In standard form, the equation is x^2/16 - y^2/9 = 1
a = 4, b = 3
c^2 = a^2 + b^2, c = 5
Eccentricity is 5/4</p>
<p><em>Write the polar equation 4r = 8/cos θ in rectangular form.
A. y = 2 B. x = 2 C. x = 1/2 D. y = 1/2
The answer for this is B. You could do the algebraic way and possibly make a huge error or you can graph it on your calculator which I did :). However, this is how you do it algabriacly (sp)
4r=8/cosθ
r=2/cosθ
1=2/rcosθ *note</em> rcosθ = x
1=2/x
1/2=1/x
2=x</p>
<p>*Find the rectangular coordinates of the point whose polar coordinates are
(20, 140º). Round to the nearest hundredth.
I got (-3.96, 19.60). The angle is 140, so it's in the 2nd quadrant. the hypotneuse (aka r) is 20. So, 20sin140= y coordinate and 20cos140= x coordinate.</p>
<p>*Find the quotient [cos (5π/12) + i sin (5π/12)] / [2{cos (π/12) + i sin (π/12)}]. Then write the result in rectangular form.
First subtract (pi)/12 from 5(pi)/12.That's (Pi)/3. so the quotient is [cos (pi/3) + i sin (pi/3)]. Now we need to switch it into rectangular. Change it to (1/2 + i (radical 3)/2). And I'm pretty sure that's your answer.</p>
<p>Swim, you're forgetting that 2 in front of the cis(pi/12).</p>
<p>thanks to everybody...</p>
<p>-swim2daend and jenkster... u guys both did number 7 and the two answers slightly vary.. can either of u find a mistake in the other's?</p>
<p>-i'm going to be a senior goin into ap calc... i took honors precalc last year and we didnt do any of this stuff... so i dont know any of those names jenkster pointed out for the spiral of archimedes question... so im just gonna put A unless anybody knows otherwise?</p>
<p>again, thanks a ton to every1</p>
<p>i see that jenkster replied regarding number 7 before i did.</p>
<p>I take it this is Calc BC?</p>
<h1>4 If you took physics, you must remember that you can split a force into components. So force F would have a vertical component of Fsin(θ) and the horizontal component would be Fcos(θ). The same thing applies to polar equations. Just think of "r" as a vector and its components are x and y, so the horizontal component, x, would be x=rcos(θ) and the vertical component would be y=rsin(θ). Manipulating the equation shown, we get</h1>
<p>rcos(θ)=2
Since I just said x=rcos(θ), we just substitute that for x and we get x=2, which is B</p>
<h1>5</h1>
<p>A graphing calculator question. Just go to polar mode and graph each one, and use Zoomfit if you can't see it. If it looks like a spiral, you've got it.
It looks like A is the only one that fits.</p>
<p>I realize that the answers to these are already given, but I just thought I'd give you another perspective on how to solve these</p>
<p>ya my number 7 is wrong. I forgot about the 2.</p>
<p>haha thanks so much guys... jenkster especially. </p>
<p>i guess that only leaves:</p>
<p>*Express 4[cos (π/6)+ i sin (π/6)] in rectangular form.</p>
<p>and i don't seem to know this one either although it seems easy:</p>
<p>*Which series is divergent?
A. 1 + (1/2^3) + (1/3^3) + (1/4^3) + ...
B. (1/5) + (1/5^2) + (1/5^3) + ...
C. (4/5) + (4/6) + (4/7) + (4/8) + ...
D. 2 + (2/3) + (2/9) + (2/27) + ...</p>
<p>No problem. Hey, I learned something too, haha. For the turn into rectangular form, all you have to do is evaluate the cos and sin. Like, what does cos(pi/6) equal? sqrt(3)/2. Plug that in for cos(pi/6).</p>
<p>Ooo, diverging series is sooooooo not necessary for pre-calc. You'll learn all about it in BC.</p>
<p>But just to satisfy your curiosity, the answer is C, because it is a harmonic series.</p>
<p>yeah, not sure why they want you to do that problem before you learn about series... </p>
<p>anyway, a harmonic series a/n is divergent. </p>
<p>man, i am already starting to forget all those dumb convergence tests.... everything is fading. I hope I don't forget EVERYTHING by the time I take multivariable next year.... crap.</p>
<p>or wait...instead of us doing the work for you why dont you give a try at the problems yourself?</p>