Multiple Choice

[kyzoo]

...I didn't bother to write down my multiple choice answers =X so i dunno

[VeryCrazyEdu.]

pretty sure C is -k/V which when graphed on the calc gives a very similar graph (i think this is what it was)

[taiga]

O, for the slope field question, what did everyone get? I'm really doubtful of my answer =X

I got dV/dt = -kv^2

I was thinking
~ Magnitude of slope goes up as V goes up, hence dV/dt cannot be inversely proportional to V
~ Magnitude of slope is negative for V > 0, and since k>0, hence must be neagtive sign in front of the expression

As V gets bigger, K becomes lesser

Change in t doesent effect gradient.

And it is defined at some certain points (cant remember)

but yeah, i got the same answer

[Stormer]

pretty sure C is -k/V which when graphed on the calc gives a very similar graph (i think this is what it was)

I can't really remember. I remember having that answer but for some reason I also remember having C Or B...OR D.  Lol I dunno

[kyzoo]

pretty sure C is -k/V which when graphed on the calc gives a very similar graph (i think this is what it was)

0.o But the graph represented by the slope field is V(t), not dV/dt

[VeryCrazyEdu.]

pretty sure C is -k/V which when graphed on the calc gives a very similar graph (i think this is what it was)

0.o But the graph represented by the slope field is V(t), not dV/dt

graph it as a differential equation in ur calc now....like -1/x or even -10/x   ...in the calc u put in the gradient function.

If you stepped back and looked at the shape of the graph in the question it looked like -ln(v) ...this differentiated is -1/v (with the dilation of k of course)

[kyzoo]

pretty sure C is -k/V which when graphed on the calc gives a very similar graph (i think this is what it was)

0.o But the graph represented by the slope field is V(t), not dV/dt

graph it as a differential equation in ur calc now....like -1/x or even -10/x   ...in the calc u put in the gradient function.

If you stepped back and looked at the shape of the graph in the question it looked like -ln(v) ...this differentiated is -1/v (with the dilation of k of course)

=/ But V is on the y-axis. So the graph of t = - ln(V) is like the graph of y = -ln(x) rotated by 90 degrees

[tram]

i got a few different anwsers comapred to the OP

6: b
11: e
15: e
18: d
21: c

[m@tty]

I had a few different answers:

7. B

15. E

21. C

O, for the slope field question, what did everyone get? I'm really doubtful of my answer =X

I got dV/dt = -kv^2

I was thinking
~ Magnitude of slope goes up as V goes up, hence dV/dt cannot be inversely proportional to V
~ Magnitude of slope is negative for V > 0, and since k>0, hence must be neagtive sign in front of the expression

I chose this, most logical option. Though I was dubious in doing so.. For some reason it something didn't seem right.

[VeryCrazyEdu.]

pretty sure C is -k/V which when graphed on the calc gives a very similar graph (i think this is what it was)

OHHH NOOOO  lol just realised my mistake....i think :\

0.o But the graph represented by the slope field is V(t), not dV/dt

graph it as a differential equation in ur calc now....like -1/x or even -10/x   ...in the calc u put in the gradient function.

If you stepped back and looked at the shape of the graph in the question it looked like -ln(v) ...this differentiated is -1/v (with the dilation of k of course)

=/ But V is on the y-axis. So the graph of t = - ln(V) is like the graph of y = -ln(x) rotated by 90 degrees

[darkphoenix]

Yeah i got -kv^2, cant really remember how i got it though =.=

[m@tty]

When V is close to zero, the gradient shown is close to zero.

When V increases, so does the slope.

Therefore V is proportional to dV/dt.

Also, for positive V a negative slope is given. Thus there must be a negative sign in expression.

[3Xamz]

I put -kV^2 for that too.

* It had be, as we moved across the graph (Horizontally), The gradients were the same so this meant dV/dt is not effected by t.
* Lower the V, the less the gradient became, so dV/dt was proportional to V
* Slope was negative, and k was stated to be positive, so there had to be a negative sign in there.

Which led me to pick dV/dt = -kV^2

I think two/three options were immediately eliminated cos they either HAD "t" or DIDNT have "V".
The other two, I think it was always positive and the other V was in the denominator.

[98.40_for_sure]

i think i picked -k/V or something, because it was undef at 0 hahah shit another -1

[3Xamz]

i think i picked -k/V or something, because it was undef at 0 hahah shit another -1

I think its cos V=0, so dV/dt = 0
The slope fields would be parallel to the x-axis so we can't see it printed