Recent content by H2instinct

Luckily the easy way to do this also involves more learning. It's still not making sense btw.

If you are saying to multiply them like this... RHS = \left((a_{x} \cdot c_{x} \cdot b) + (a_{y} \cdot c_{y} \cdot b) + (a_{z} \cdot c_{z} \cdot b) \right) - \left((b_{x} \cdot c_{x} \cdot a) + (b_{y} \cdot c_{y} \cdot a) + (b_{z} \cdot c_{z} \cdot a)\right) That really...

Ya. Just don't understand from here... RHS = \left(a_{x} \cdot c_{x} + a_{y} \cdot c_{y} + a_{z} \cdot c_{z}\right) b - \left(b_{x} \cdot c_{x} + b_{y} \cdot c_{y} + b_{z} \cdot c_{z}\right) a to the solution.

You are correct... I confused regular multiplication with dot product. But what exactly is the parentheses multiplication of vectors? I though that was the dot product. Still confused...

I am not fully understanding the dot product multiplication rule because how I think it is supposed to be multiplied to show component form just keeps coming out as I have it shown above. Also, I fixed the signs of the LHS, I believe that is now correct. This is where I get stuck: RHS =...

Homework Statement Prove, by writing out in component form, that \left(a \times b \right) \times c \equiv \left(a \bullet c\right) b - \left(b \bullet c\right) aand deduce the result, \left(a \times b\right) \times c \neq a \times \left(b \times c\right), that the operation of forming the...

Complex Numbers Proof Multiplying the top and bottom by the complex conj. of the bottom: \frac{a+ib}{c+id} * \frac{c-id}{c-id} Gives me: \frac{(ac+bd) - i(ad-bc)}{c^{2}+d^{2}} In form x+iy it is: \frac{(ac+bd)}{c^{2}+d^{2}} + (\frac{(bc-ad)}{c^{2}+d^{2}})*i This is where...

Ah Ok, I am getting how you got the boundaries then. And right, it obviously wouldn't work for a parametrized function that is only in the 1st and 3rd quadrant, etc.

Yes it did work, thanks for the help, I will remember to change it to the (0 , \frac{\pi}{2}) quadrant when it calls for abs. value.

Homework Statement Homework Equations The hint. The Attempt at a Solution So using the hint I took the derivative of each of the parametrized functions. \frac{dx}{d\theta} = -3sin(\theta)*cos^{2}(\theta)*a \frac{dy}{d\theta} = 3cos(\theta)*sin^{2}(\theta)*a Then I plugged them into this...

P.S. I really love this forum, everyone is so helpful!

Ok I will try and explain what I did: After getting those substitutions I applied them all to the original equation and got out: (1/3)v(-2/3)* Dv/Dx + (1/3)v(1/3) = (1/(3ex))*v(-2/3) Multiplied by the inverse of dy/dv to get: dv/dx + v = 1/(ex) This is where I got confused, the...

Homework Statement 3y^2*y'+y^3= e^-x The Attempt at a Solution I am using Bernoulli's equation to substitute V in and I keep coming out with Y^3= e^x(-e^-x+c) my V = y^3 Y = v^(1/3) and dy/dv=(1/3)v^(-2/3) I peeked to see if I was correct, the right answer is supposed to...

Thanks, I understood it much better from that.

Thank you for the help, but I still don't see the proof from (1.3.5...(2n-1))/2^n == (2n)!/((4^n)(n!)) It's fine. I'm going to move on and get help from a classmate when the time comes. Thanks for the help though!