Recent content by Meadman23

I end up with { [ (z/t) + Δz] - (z/t) } / Δz then Δz / Δz then 1then taking the limit I get 1Or am I supposed to do {[(z + Δz)/t] - (z/t)} / Δz[(z/t) + (Δz/t) -(z/t)] / Δz(Δz/t) / Δz1/t How does this get me to the partial derivative with respect to z and how would I know that doing...

delta z is in the numerator and in the denominator on the LHS. I'm not sure what you're trying to say.

? But wouldn't that give me 0/0 if I take the limit directly?

I'm mostly trying to understand how the expression in brackets becomes a partial derivative. I know I end up with 0/0 if I replace all deltaz with 0, so that leads me to believe L'hopitals rule is used here. I just don't see how I could take L hopitals when I have two variables..

I'm just not getting it. ' I see that the angle between r and x is phi, thus in using the formula, r (dot) x = (1)(1) cos(phi). Then I see the angle between r and y is (90-phi), thus in using the formula, r (dot) y = (1)(1)cos(90-phi) = sin (phi). I then see the angle between phi and x is...

Its the attached picture. I'm not seeing why when using the dot product, they start using sin.

Just like the title says, would that technically be true? I know the cross product is normal to the plane of the two vectors being crossed, which would make it z. However, since the angle between two vectors is 0, sin (0) = 0...

Homework Statement Shown in attachment Homework Equations The Attempt at a Solution I'm trying to analyze the circuit in the attached picture. This is a step response with a 3V input or 3u(t). What I've done so far is: 1. convert all of the components to the s-domain. R = R, L = sL, C =...

The image attached is of an underdamped RLC step response. I know that I can find the damped frequency of the response by first finding the period of the wave, and manipulating the period such that I can do 2*pi*f. If I'm looking at this waveform and the only info I know about it is this...

Okay, let me see if I get this right. 1. The slope of (1) is [4-0]/[4-0] = 1 2. Using this slope and the point (4,4) and plugging it into the y - y0 = m(x - x0) equation yields : y - 4 = 1(x - 4) -> y - 4 = x -4 -> y = x 3. From this, my equation for (2) is then shown to be y = x?

I figured that's the way I could solve for the function of this waveform. That is, I could find the function of the part of the wave 1. from x = 0 to x = 4 2. from x = 4 to x = 5 3. from x = 5 to x = 6 Then, after I've deduced the correct functions that define each of the 3...

Homework Statement If I have a waveform that looks like it does in the attached picture, how can I figure out the y-int at the point where the line curves like triangle? The Attempt at a Solution This is confusing to me because I can figure out the slope of the triangle, but I...

I might be reaching here but let's see. I have: u = h^2 + r^2 and r^2 u - r^2 = (h^2 + r^2) - r^2 = h^2 Thus, r^2 vanishes and h^2 is no longer of a function of r. Thus, I can factor it out of my integral. Am I seeing this right?

I mean, h^2 = u - r^2 which means to me that the value of h^2 is dependent on the value of r. How could it not be?

Okay, thanks guys. I split the fraction up into two. ∫u/(2*u^(3/2)) - ∫h^2/(2*u^(3/2) I wasn't able to get it to come out right until I considered the h^2 as a constant and factored it out along with the 1/2. 1/2∫u/(u^(3/2)) - h^2/2 ∫1/(2*u^(3/2) Why can I consider h^2 as a constant...