Recent content by shft600

Homework Statement Consider the population model with dx/dt=rx(1-x)-ax/(1+x). Compute the normalized forward sensitivity indices of the (positive) internal equilibrium with respect to r and a. The default parameter values are r=4, a=3.Homework Equations SI(x,t)=(∂x/∂t)*(t/u)The Attempt at a...

That would be it! I knew it was something simple, thanks!

Homework Statement Simply enough, solve the following using U-Substitution. dy/dx = y2-2y+1 and the answer key provides y = 1+(x+C)-12. The attempt at a solution I know how to do this, I thought. But I'm just spinning my wheels and need somebody to show me my error... I tried moving...

I'm trying to study for my final on friday from past exams and their keys, and one question is the integral of 3x+3x I get through it all fine, but then according to the key, the derivative of 33x is just (ln3)(3x) so here's my question, I thought that by chain rule, it should end up being...

never mind, that ended up being much easier than I expected

excellent, I actually just found that mistake as I refreshed this page but now I've taken a look at the next question and am even more confused... I = int{ sin7x cos-3/5x dx would this be an improper integral type question or can I use the same ideas on the last question?

anybody? I haven't made any progress on this last integral yet... I found something online that claimed a rule that cos(3x)=4cos3(x)-3sin(x), so I manipulated it into cos3(2x)=(1/4)(cos(6x)+3sin(2x)), and then by integrating the right side of that I got (1/24)(sin(6x)-3/8(cos(2x))), which is...

I can get up to the point that =(1/8) [ x - (3/2)sin(2x) + (3/2)( x + (1/4)sin(4x)) - int(cos3(2x)dx ] but here I get stuck again on the new integral again, I tried int{ cos3(2x) int{ cos2(2x) cos(2x) dx int{ (1-sin2(2x)) (cos(2x)) dx, u=sin(2x), du=2cos(2x) then, 1/2 int{ (1-u2)(du)...

Evaluate the following integral: I=int{sin6(x)dx This form of question has shown up on both midterms, and is on my final assignment of the semester and most likely the final exam as well, and I still haven't figured out how to solve them. Could somebody point me in the right direction? Should...

ooh, my bad... so its equal to arcsin(x-1)+C !

by that, it should be -(x2-2x) = -((x-1)2-1), or a=1, b=-1, c=-1 so then, making u=x-1, du=1, it becomes int{ du/sqrt(-u2-a2) and that falls into the arccosh formula as: cosh-1(-u/a)+c or cosh-1(-(x-1)/1)+c cosh-1(1-x)+c, and that should be the final answer for the indefinate integral right...

hm.. I'm trying that but still stuck, I get int: 1/sqrt(x(2-x))dx, I can move the root up to the top easily enough by multiplying by sqrt(x(2-x))/sqrt(x(2-x)), but that doesn't help much. I found the formula relating a/sqrt(b+x2) to (a/sqrt(b))arctan(x/sqrt(b))+C, but that still doesn't...

Homework Statement Evaluate the following integral: (int)1/sqrt(2x-x2)dx Homework Equations d(arcsinx)/dx = 1/sqrt(1-x2) The Attempt at a Solution I just need a good start in the right direction on this one so I can at least try it. I know the arcsin d/dx as above, but how do I...

Hm.. I'm not entirely sure, I just posted the formula straight off my formula sheet. The first part is the work of moving the varying amount of fluid, and the second just moving it out through the open space in the dish. The force is constant for the (T-h) (10-5cm), but we need to take a...

so then is it just the integral of pi(ey)2? In which case V(y)= Pi((1/2)(e2y)) from 0-5, and a constant Pi((1/2)(e2(5))) from 5-10cm?