Recent content by mickellowery

Alright so with the correct equations would the proper integral for the circumference be: \int\sqrt{(-3sin(t))^2 +(2cos(t))^2}dt And then for the area enclosed by the ellipse would I use \int(3cos(t)-2sin(t))2 evaluated from 0 to \Pi?

Oh geez I just noticed a typo in the original problem. It should be x=3cos(t) y=2sin(t) not y=sin(t) sorry about that.

OK the way I did it I actually used \int\sqrt{1+dy/dt/dx/dt} and that's how I came up with the -\frac{2}{3}cot(t)2 I had -\frac{2}{3}\frac{cos(t)}{sin(t)} and I simplified it to cot(t). Would this even be the right formula to use for the circumference? I thought that arc length would be the...

Homework Statement Consider the ellipse given by the parametric equation x=3cos(t) y=sin(t) 0\leqt\leq2\Pi. Set up an integral that gives the circumference of the ellipse. Also find the area enclosed by the ellipse. Homework Equations \int\sqrt{1+(dy/dx)^2}dt The Attempt at a...

Homework Statement \int\frac{e^xcos(log_7(e^x+9))}{(e^x+9)ln(7)}dx Homework Equations The Attempt at a Solution Let u= (ex+9) du= exdx New integral \int\frac{cos(log_7(u))}{(u)ln(7)}du This is where I got l little lost. Should I let log7(u)=\frac{ln(u)}{ln(7)}? Or is this...

Sorry, the problem says to simplify the expression. The final answer is supposed to be \frac{x}{\sqrt{1+x^2}}

Homework Statement sin(tan-1(x)) Homework Equations The Attempt at a Solution y=tan-1(x) tan(y)=x sec2(y)= 1+tan2(y) sec(y)=\sqrt{1+x^2} This is where I'm getting stuck. I know that I have to say that the sin(y)= whatever, but I'm not sure how to tie the sin sec and tan...

Homework Statement Two identical particles, each having a charge of +q are fixed in space and separated by a distance d. A third particle with charge -Q is free to move and lies initially at rest on the perpendicular bisector of the two fixed charges a distance x from the midpoint between the...

Excellent! Thanks so much for all your help again Doc. You're my hero!

Alright I think I've got it Doc. Would the second one be 2d2= 340(1.92+t1 ? I'm still thinking about the third one a little, but would it have bounced off both walls?

OK so the first echo which has no time attached to it will be off the close wall and the second echo which is 1.92s is off the far wall. The third echo which is 1.47s should be off the close wall again.

I think what I'm having trouble seeing is whether or not all of these echoes are bouncing off of the closer wall. I see that it should be 2d_1= 340t_1 but then would it be 2d_1= 340t_2?

OK so it's d1+d2= (340)t1+(340)t2?

OK I have been trying to come up with an expression for time, but I'm having trouble. Would it just be: t1= \frac{340}{d1} t2= \frac{340}{d2}?

Homework Statement A cowboy stands on horizontal ground between two parallel vertical cliffs. He is not midway between the cliffs. He fires a shot and hears it echoes. The second echo arrives 1.92s after the first and 1.47s before the third. Consider only the sound traveling parallel to the...