InquilineKea

Hm so I'm having difficulty in answering this question. This is what I have so far: 1. The problem statement, all variables and given/known data Consider a neuron with resting potential of -65 mV and threshold of -55 mV. It receives two synaptic inputs with similar synaptic conductances, one with reversal potential of -10 mV and the other with reversal potential of -58 mV. Draw the predicted postsynaptic response (change in membrane potential) to stimulation of each synapse alone, and then to simultaneous stimulation of both synapses. Briefly explain what’s going on and why the results might at first be confusing. 3. The attempt at a solution So the neuron will reach an action potential at -55 mV. So if the neuron gains 10 mV, it will depolarize and reach an action potential. Now, how will the synaptic inputs affect the neuron? Input 1 might have a reversal potential of -10 mV, but that doesn't say how much current it sends to the neuron, nor does it say the proportion of total positive/negative ions in the input compared to that of the neuron. So that confuses me. How do we add potentials? Do we just take some average of the resting potential of the neuron with the reversal potential of the inputs? Which would be sort of like adding up concentrations or ratios. Or do we add-55 to -10 and -55 to -58? Ions sum up additively (and don't involve taking ratios) but I highly doubt the question wants me to say that since then there would just be hyperpolarization and no action potential for either input. Thanks!

consider the function [math]\frac{1}{\epsilon^2 + z^2}[/math] So we know that there are two poles, one at [math]z = i \epsilon[/math], one at [math]z = - i \epsilon[/math]. So when this function never hits 0 on the real line, how do the singularities affect its behavior on the line? Okay, so poles are a subclass of singularities. I think that [math]z = i \epsilon[/math] and [math]z = - i \epsilon[/math] are poles - I may be wrong here. The question is - how do complex singularities (complex poles in this case) affect a function's behavior when the function is plotted on the real line?

Let C be the closed curve that goes along straight lines from (0,0,0) to (1,0,1) to (1,1,1) to (0,1,0) and back to (0,0,0). Let F be the vector field [tex]F = (x^2 + y^2 + z^2) i + (xy + xz + yz) j + (x + y + z)k. Find \int F \cdot dr[/tex] By Stokes Theorem, I know that I can transform this into a curl (instructions say don't evaluate the line integral directly). So I know I get a form of square that goes CCW. Projected on the xz-plane, it is simply a square (but the y-coordinate of the right end is 1). I think the plane is described by x = z, since there is no dependence on y. So x-z = 0 Anyways, I take the curl of F to get [tex]F \cdot dr = (1-x-y) i + (1-2z) j + (y+z-2y) k.[/tex] As we know, [tex]\int\int \nabla \times \vec{F}\cdot \vec{n}dA = \int\int F \cdot dS = \int\int -P \frac{\partial g}{\partial x} - Q \frac{\partial g}{\partial y} + R[/tex] where [tex]\nabla \times F(x,y,z) = P i + Q j + R k[/tex] and [tex] g = f(x,y) = z[/tex] The equation of my plane should be z = f(x,y) = x, right? (the plane is f, and determined by "Let C be the closed curve that goes along straight lines from (0,0,0) to (1,0,1) to (1,1,1) to (0,1,0) and back to (0,0,0).") in that case, [tex]\frac{\partial g}{\partial x} = 1, \frac{\partial g}{\partial y} = 0[/tex] So correcting my arithmetic.. [tex]F \cdot \nabla f = -(1 - x - y) + y + z - 2y dA[/tex] = [tex]\int_0^1 \int_0^1 -1 + x + z [/tex] We take z = x. Is this an appropriate choice? If so [tex]\int_0^1 \int_0^1 2x - 1 dx dy = \int_0^1 x^2 - x (1,0) = \int_0^1 0 ....[/tex] what am I doing wrong here? argh, does this forum support latex?

Ah, I see. thanks! So a strict free body diagram analysis would not work, since there are two times for the free body diagram, and the time lag causes the naive interpretation of the free body diagram (the interpretation I made above) to fail. And the reason would be that I am a biological agent and I can adjust myself after I push against the chair, such that part of the "equal and opposite reaction" would be canceled, right?

Hm okay. But how would this shift in the COM affect the free body diagram?

Hm, but if i myself do not move, then the force pairs that I exert on the chair (both of them) must cancel, must they not? If so, then there should be no acceleration. But I can accelerate the chair by pushing against it (and nothing else)

So I am a person sitting on a chair that has wheels. Now, the question is - can I push the chair when I am sitting on the chair and have nothing to push against? [other than the chair]? Note that I have no contact whatsoever with the floor. From experience, I know I can. So why? Whenever i try to push against the chair, according to Newton's 3rd Law, each force produces an equal and opposite force. So as I push against the chair, I exert a force, and it exerts an equal and opposite force on me. But I move along with the chair, so I must exert a second force on the chair, which is in the opposite direction as the direction of acceleration of the chair. (a) <--- (man on chair) ---->(chair on man) (b) --->(man's bottom on chair) <---(chair on man) such that man does not move with respect to the chair's reference frame. What about the outside world though? I know that the chair accelerates, so there must be a net force to the left. The guy is in the middle, and he tries to push the chair to the left. I see three forces exerted. (a) is the man pushing chair left, the chair edge pushes the man right. Then the man's bottom and the chair have to push each other (b). Man pushes chair right, chair pushes man left. And what if you replace the man with say, a motor? The difference here would be that the motor would probably be a lot more attached to the bottom of the chair than the man would be. So the question is - is the force in (a) greater than the force in (b) and why? Since if the force in (a) is greater than that in (b), then the chair will experience acceleration to the left, which is apparently what I experienced