9. I got 3x^2(x)^(x^2)(lnx)
12. I moved the dy and dx onto opposite sides and ended up with (1/(x^2-x^3+1))^.5
14. The formula I used was the antiderivative of (10/4)x from 10 to 14, which is 110.
15. I don't see how e^(-x) converges, since, as the particle approaches infinity, I got 1, which by nth term would mean it diverges.
20. This is confusing since, by the information given, you can deduct two different answers for radius
22. Since the numerator is an additional equation, maybe the n! crosses out the numerator, but I don't see how the relationship of the function with e.
23. Integrating it, the function looks like sinx-x+(x^3/3) from pi to zero, which does not equal to pi.
29. Looks like the derivative of cos(x+h), but where did the 2 in -sinx come from?
33. When setting the ln function to infinity, I gor pi/1/1/x, which when appraoching infinity, should equal to zero, which would make the answer (a). Did anyone else get that?
36. My answer slightly differed, with 3x^2(x^2+2)^(1/2). Did anyone else get that?
44. The answer makes it so that x(t), not a(t), tells the time when the v(t) is increasing, but why?
Lastly,
45. If n is approaching infinity, then wouldn't the answer be a)0, since (1/n) approaching infinity is zero and sin(0)=0?
3 comments:
9. you have to ln of both sides - so it would be lny=lnx^(x^3) which equals lny=(x^3)lnx - then take the derivative of both sides. you would get (1/y)dy/dx=(3x^2)lnx+(x^3/x) then simplify.
14. the formula is W=Fd, where W=work, F=force, and d=distance
15. you aren't using convergence tests. if the limit equals a number, then it converges.
20. you're given dv/dt and V. by using the volume formula you can find r, there is only one possible answer (r=3) then take the derivative of the volume formula, plug in dv/dt and r to find dr/dt. then go ahead and try to find the rate of change of surface area.
this is all i did so far.
for #29 i used l'hopitale's rule so it became limit as h approaches zero of sinx+sinx which equals 2 sinx
for #33 i don't understand why you would do anything with ln.. you just substitute (1+cos2x) with 2cos^2x and integrate it
for #36, the original equation is (x^2 + 2)^(3/2); the derivative is (3/2)(x^2 + 2)^(1/2)(2x) with the chain rule.. it simplifies to 3x(x^2 + 2)^(1/2) which is answer B
for #44, it's asking when the velocity is increasing so you would look at the acceleration vector and find where it equals zero and find where it is positive.. it should work out fine
I answered how to do number 14 in Maria's post, that should help you get the answer.
For 15, like Priya said, you're using indefinite limits (is that what it's called?). So for example, in I. you make the equation the limit as b approaches infinity from the left of the integral of e^-xdx from 0 to b. From integrating and then finding the limit, you should get 1, which means it converges. You do this for II. and III. as well, but you use the limit as a approaches 0 from the right (II. shouldn't exist and III. should equal 2) so the answer is E.
And by typing that info for 33, I think you mean number 32. Yeah you do e^(limit as x approaches 0 of ln(1+sinpix)^1/x). So you first need to find the exponent part (not the e yet) and since you ln'd the equation it becomes ln(1+sinpix)/x. Both the top and the bottom equal 0, so you use L'Hopital's rule and get the limit as x approaches 0 of picospix/(1+sinpix). Plugging in 0 for x, you get pi/1=pi. Then you have to make sure to make that the exponent of e, so the answer is e^pi, which is D.
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