Recent content by Laney5

Im not sure if this is the right section to post this question.. Calculate the differential of the map f:R^3 -> R^2 , (x ,y ,z)->(xy3 + x2z , x3y2z) at (1 ,2 ,3) in the direction (1 ,-1 , 4) I know how to get the differential of the map (finding the jacobian matrix) but the only part i am...

Im trying to figure out how to do this question. This is an example in the book i have. I am not sure how they got the answer. Here is the example from the book: Find the Transition Matrix P from the basis B={t+1, 2t, t-1} to B'={4t^{2}-6t, 2t^{2}-2, 4t} for the space R[t]. A little...

u=2lnz-1 du=2(\frac{1}{z})dz (\frac{1}{2})du=(\frac{1}{z})dx (\frac{1}{u}).(\frac{1}{2})du=(\frac{1}{x})dx \frac{1}{2}\int\frac{1}{u}du=\int\frac{1}{x}dx (\frac{1}{2})lnu=lnx + C ln(u)^{1/2} - lnx = C ln[\frac{(u)^{1/2}}{x}] = C u=2lnz-1 ln[\frac{(2lnz-1)^{1/2}}{x}] = C...

Ya, its that part alright! Do i write it like \frac{1}{u} \int \frac{dz}{z}? \frac{1}{u} \int \frac{dz}{z} = Integral(\frac{dx}{x}) \frac{1}{u} (lnz) = lnx + C \frac{1}{2lnz - 1} (lnz) = lnx + C Is this right?

Im looking for a general solution to the following equation but i can't seem to get an answer. 3xdy = (ln(y^{6}) - 6lnx)ydx I've got this far anyway.. \frac{3x}{y} dy = (ln(y^{6})-ln(x^{6})) dx \frac{dy}{dx} = \frac{y}{3x} . 6ln\frac{y}{x} \frac{dy}{dx} = \frac{y}{x} ...