<p>Hey guys,
I don’t know if this has been asked before, but…</p>
<p>does anyone know how accurate the PR Practice tests are for BC calc?</p>
<p>thank you.</p>
<p><a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools;
<p>question 2 for example, using algebra, how do you solve it?</p>
<p>Could somebody please explain the Lagrange Error Bound? An example is 6(d) from 2008B: <a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>how important are memorizing inverse trig functions?</p>
<p>I would say that memorizing inverse trig functions is fairly important.</p>
<p>the Lagrange Error Bound is simply an extention of Taylor/Mclaurin polynomials. Say you found the value of f(x) to the nth degree. To find error, you find the value of f(x)(x-a)/(n!) for the (n+1) degree. </p>
<p>So you just calculate the value that is next up.</p>
<p>can anyone answer my question about the accuracy of PR tests?</p>
<p>I think only the integrals of inverse trigs are important (you mean arctan, arcsec, and arcsin right? Not secu, cotu, cscu?)</p>
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<p>a) The formula for the area enclosed by a polar curve is ∫((r^2)/2)dθ, with appropriate endpoints. Therefore, in this case, since r = θ + sin2θ and the curve goes from 0 to pi, we must evaluate the integral ∫(((θ + sin2θ)^2))/2)dθ with endpoints 0 and pi. ***It’s worth mentioning that sometimes multiple-choice on the AP test will use symmetry to alter the endpoints of the integral in relationship to the 1/2 used in the formula. This doesn’t apply here since the figure in this problem is asymmetrical, but if the figure were, say, a circle, you could evaluate the integral ∫(θ^2)dθ from to 0 to pi.</p>
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<p>Not sure if this is the Lagrange error bound, but here’s how I did this problem:</p>
<p>The question is basically asking you to find the point in the series ln(5/4) at which the error is <1/100. Now, since in this series each term gets smaller as the series continues AND the series is alternating, we know that the absolute value of each term is larger than all the terms summed together that follow it. (i.e. if I have 1/4 - 1/8 + 1/16…, 1/8 > 1/16…) Therefore all we need to do is find the first term of ln(5/4) that has an absolute value less than 1/100. This is (1/3)(1/2)^6, or 1/192, which is certainly < 1/100.</p>
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<p>You mean derivatives? For what it’s worth, all of the trig (inverse and not inverse) derivatives are on my school’s “stuff to memorize” list, which is not a very big list. I would know them.</p>
<p>a curve is described by the parametric equations x = t^2 + 2t and y = t^3 + t^2.An equation of the line tangent to the curve at the point determined by t=1 is?</p>
<p>supposedly the answer is 5x + 4y = 7</p>
<p>how do you get this?</p>
<p>hey everyone, does anyone know how to do #18 on the BC part of the course description? <a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board, page 35</p>
<p>also if anyone has a link for a tutorial for Riemann Sums in general, that would be greatly appreciated!</p>
<p>dy/dx = (dy/dt) / (dx/dt)</p>
<p>I think that’s your problem. It’s pretty easy once you take it from there.</p>
<p>I’m pretty sure you take the dx/dt and dy/dt, and then dy/dx for the slope. Then plug in t=1 into the x and y equations to get your coordinates, and use those to get a value for the slope. Then put all that together into point-slope form. </p>
<p>Second one, differentiate implicitly.</p>
<p>Can somebody explain to me the concavity rules? I’ve seen a lot of these types of problems on the FRQ’s, and can’t seem to get it correctly. (I.e. 2nd derivative test)</p>
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<p>When the test asks about concavity, think 2nd derivative. When the 2nd derivative is positive, the graph is concave up. When the 2nd derivative is negative, the graph is concave down. (Think about this; if the 2nd derivative is positive, then the 1st derivative is increasing; doesn’t this look “concave up”?). Also, a point of inflection is when a graph’s concavity changes AKA when the 2nd derivative goes from positive to negative or from negative to positive.</p>
<h1>18.</h1>
<p>You want the Reimann Sum version of the integral of the cross-sectional area, which is</p>
<p>int(from 0 to 2) (1+3x)dx</p>
<p>this is equivalent to the Reimann Sum formula of</p>
<p>SUM(1 to n) f(x<em>i) * delta</em>x</p>
<p>f(x<em>i) = 1 + 3 (a + (b-a)*i / n), delta</em>x = (b-a) / n, a = 0 and b = 2 (because those are the bounds of integration), so</p>
<p>SUM(1 to n) [ 1 + 3 (0 + 2*i / n) ] * (2 - 0) / n</p>
<p>= SUM(1 to n) [1 + 6i/n] * (2/n)</p>
<p>You need to know the basic formula (above) and that x<em>i = a + delta</em>x * i, and also that delta_x = (b-a) / n</p>
<p>ChemE14 - I just get 5/4 when I do that… this problem is really ruining my day</p>
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<p>I meant, using algebra, how do you solve for the limits of intergration?</p>
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<p>Well, as the others stated, the key here is knowing that the slope of a tangent line for a parametric curve is (dy/dt)/(dx/dt). Now here we know that dy/dt = 3t^2 + 2t and that dx/dt = 2t + 2. That gives us dy/dx = (3t^2 + 2t)/(2t+2). At t = 1, that equals 5/4. This gives us the slope of the tangent line, 5/4 (as reflected in the answer you provided.) Now we also have to consider the constant for the equation of the line as a whole, since the slope is only part of this problem. Our equation is unfinished and currently y = (5/4)x + C (c is just some constant we haven’t found yet). To find this we plug 1 into the values for x and y the problem gave, giving us the point (3,2). Plug those values into our unfinished equation to get 2 = 15/4 + C. Therefore C = -7/4, and our equation is y = (5/4)x - 7/4. Multiply both sides by 4, to get 4y = 5x - 7, which was I think the answer you gave.</p>
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<p>My mistake. The problem tells you that the domain for the function is 0 <= θ <= pi. Verify this by looking at the graph; the function only goes from θ = 0 to θ = pi. Therefore, the limits are 0 and pi.</p>
<p>briangt - Okay, so now you have the slope of your tangent line. Now, to get your equation for the tangent line, you need the point at t=1. To get this, simply plug t=1 into the x(t) and y(t) equation. Then use point slope formula and you’ll get it.</p>
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<p>Yes, but how do yuo do without looking at the graph is what I want to know.</p>