Recent content by crims0ned

I am a mechanical engineering major that has been working in the fire protection industry and going to school in the evening for about two years. My ultimate goal when I decided to go back to college was to become a licensed professional engineer as fast as possible. But last summer the...

Yeah I think I got it now, I set the gradient equal to the normal of that other plane. \nabla f \left(x_0,y_0,z_0\right)= u=<1,2,3> then I set the partials equal to that normal and I get f_x=2x_0; f_y=-1; f_z=2z_0 so... \nabla f \left<2x_0,-1,2z_0\right>=<1,2,3> then my only problem is my...

Isn't the definition of a critical point a point on f(x) that occurs at x0 if and only if either f '(x0) is zero or the derivative doesn't exist at that point?

You're right it shouldn't be an inflection point because the original functions tangent line doesn't at x=2, but possibly it's asking for all critical points on the second derivative? Is this some kind of online homework program?

okay, so we have f(x)=\frac{(x^2-3)}{(x-2)} and f''(x)=\frac{2(x-2)}{(x-2)^4} Even though x=2 is a vertical asymptote it is still a critical point on the second derivative

f(x)=(x^2-3)(x-2) is a polynomial and therefore it's domain should be all real number and this should carry on to it's derivatives as well. f(x)=(x^2-3)(x-2) = x^3-2x^2-3x+6 right? So there shouldn't be a quotient at all.

Your function is f(x)=(x^2-3)(x-2) right? Maybe if you foil it out first before you take the derivative?

At what point on the paraboloid y=x^2+z^2 is the tangent plane parallel to the plane x+2y+3z=1 ? Tangent plane equation is... Fx(X,Y,Z,)(x-X)+Fy(X,Y,Z)(y-Y)+Fz(X,Y,Z)(z-Z)=0; for x^2+z^2-y=0 My attempt at the problem... First I found the unit normal for the plane I'm trying to...

Maybe the rate of change in the kinetic energies of the inlets and outlet per minute or second? K.E. = \frac{1}{2}mv^2

once i got to \int_{-1}^{t-1} \sqrt{u^2+1}du i trig subbed and got \int sec^3(\theta) d\theta but can i change the limits along with the variable and still solve for s ?

there seems to be only an i and j component

the original problem is r(t)=<cos(t)-tcos(t), sin(t)-tsin(t)> ; t=o

Yes, that's it, just the magnitude of dr/dt. Are you familiar with these types of problems? Because all of the ones I've worked pier to this and everything in my calculus book always out really clean.

the equation is this s= \int \sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2} dt evaluated from some starting t to t

we've been using s reparametrizations just as kind of random parametrization i guess. pretty much it seems to be finding the arc-length and evaluating it from some number (t sub not) to an indefinite t. And what ever we come up with we then solve for t is terms of s and substitute it back into...