<p>The first 3 terms are 4/3, 8/9, and 16/27. This is a geometric series, with r=2/3. Using the formula Sum = a/(1-r), you get:
Sum = (4/3)/(1-2/3)
= (4/3)/(1/3)
=4</p>
<p>lol another:</p>
<p>the length of a curve from x =1 to x=4 is give by integral from 1 to 4(sqrt(1=9x^2))dx. if the curve contains the point (1,6), which of hte following could be an equation for this curve?
a) y = 3+ 3x^2
b) y = 5 + x^3
c) y = 6 + x^3
d) y = 6-x^3
e) y = 16/5 + x + 9/5x^5</p>
<p>why is the answer b? i'm getting a</p>
<p>FRQ time. :) This is from PR 2002-2003, BC Calculus Exam #1. </p>
<p>Let f and g be functions that are differentiable throughout their domains and that have the following properties:</p>
<p>(i) f(x+y) = f(x)g(y) + g(x)f(y)
(ii) lim (a -> 0) f(a) = 0
(iii) lim(h -> 0) (g(h) - 1)/h = 0
(iv) f '(0) = 1</p>
<p>(a) Use L'Hopital's rule to show that lim (a -> 0) f(a)/a = 1.
(b) Use the definition of the derivative to show that f '(x) = g(x).
(c) Find ∫g(x)/f(x)dx.</p>
<p>I kept getting 4 for my answer too. However, my answer sheet says the answer should be 6, which was confusing me.</p>
<p>
[quote]
4(sqrt(1*=*9x^2))dx
[/quote]
</p>
<p>Is that - or +? :D</p>
<p>I have the answer key in front of me. It says that its 4. how wierd is that.</p>
<p>lol....i have another:</p>
<p>a curve C is defined by the parameteric equations x = t^2 - 4t + 1 and y = t^3. what is the equation of the line tangent to the graph of C at the point (-3,8)</p>
<p>i keep getting y = 8, but my answer key says it is x = -3</p>
<p>^i was able to get x =-3 but it involved dividing by 0 then then multiplying be zero...lol</p>
<p>
[quote]
ol....i have another:</p>
<p>a curve C is defined by the parameteric equations x = t^2 - 4t + 1 and y = t^3. what is the equation of the line tangent to the graph of C at the point (-3,8)</p>
<p>i keep getting y = 8, but my answer key says it is x = -3
[/quote]
Funny, I got y=x+11</p>
<p>Its because when t=2, dx/dt is undefined, meaning the curve has to be vertical. Since it's vertical it has to be x=something. It also has to touch the point (-3,8), so x=-3.</p>
<p>I meant, you are using (dy/dt)/(dx/dt). So when dx/dt is 0, the derivative is undefined.</p>
<p>guys, do we just have to memorize one or two formulas for the logistics differential equaiton?
My teacher hasn't covered it in class.</p>
<p>Anyone have any more problems?</p>
<p>Oops sorry noober, misread your post. Thought you only wrote x^x(lnx)</p>
<p>Can we do a logistics problem? Any logistics problem? Or just a more detailed explanation?</p>
<p>I don't understand how to deal with absolute value. For example to find the integral of |x| from -1 to 3 ... ?</p>
<p>Think about the graph. The range [-1, 3] includes 0. We cannot find the area under 0 in the absolute value function. However, the |x| behaves like y = -x from (-∞, 0) and like y = x from (0, ∞).</p>
<p>tomorrow baby! ya'll better start gettin crunk!</p>
<p>Quick Questions:</p>
<p>How do you find the area in a curve in polar?
Arc length in Regular? Parametric? Polar?
What are the power series for e to the x, sin x, cosx, and other ones we need to know?</p>
<ol>
<li><p>Int a to b of ½r²d(theta)</p></li>
<li><p>Regular: Int a to b of √(1+(dy/dx)²)dx
Parametric: Int a to b of √((dx/dt)²+(dy/dt)²)dx
Don't worry about it for polar.</p></li>
</ol>