dx/dt =1, x(0,s)=0, dy/dt=x, y(0,s) = s, du/dt=(y-1/2x^2)^2, u(0,s)=e^s
I did well at the beginning to get x(t,s) =t and y(t,s)=1/2t^2 + s, but got stuck with the du/dt part.
You can sub in x=t and y=1/2t^2 +s for x and y to get du/dt = s^2. But that's still three variables, and I can't see...
I seem to be getting an unsolvable integral here (integral calculator says it's an Ei function, which I've never seen). My thought was to use Bernoulli to make it linear and then integrating factors. Is that wrong? The basic idea is below:
P(x) 1, Q(x) = 1/2(1-1/x), n=-1, so use v=y^1-...
Hmm, yeah, maybe it was just the wording that is a little ambiguous. It says for a=2 AND b≠-1, the solution is inconsistent (makes sense, because then you get something like 0=1), for a≠2 AND b+4a^2-4a-7≠0, it is unique, and for a=2 AND b=-1 there are infinitely many solutions (also makes...
It makes sense that a=2 would cause problems because then we wouldn't have a matrix of full rank and we'd be unable to determine a value for w. But the key also says that when b+4a^2-4a-7≠0. Why is that an issue? For example, if a=1, that just says implies that w=0. Through back-subsitution...
But isn't y_2 showing a length? And why would they spend so much time going through Simpson's if that isn't applied here? It seems fishy they would give f''''(x), point to the idea that the two estimates are the same, and ask for "best estimate." Plus since the average of Simpson = the...
I'm confident in calculating the beginning values. Only question I had was whether in this graph y2 is assumed positive or negative. Obviously f(pi)= negative value, but is y2 negative or is it = |f(pi)|. Assuming the latter, it all makes sense and when I sum the two Simpson's Rule estimates...
I'm pretty confused here because after getting stuck on this problem, I tossed it into an integral calculator and it said the answer was 2 Si(x) + cos(x)/x + C. In intro calc we definitely haven't learned the Si(x) function or even gotten to any of the Taylor polynomial stuff yet.
I tried IBP...
Yes, $$\frac d {dx}\int_0^x x\sin(t) dt$$ is correct. Thanks, apologies on forgetting the dt. Is it OK to then just proceed with $$\frac d {dx}x\int_0^x sin(t) dt$$ and try for the product rule? It doesn't seem to specify if x is a function of t, but given how involved it gets when I don't...
I was told this problem could simply be solved with calc-1 techniques, so I'm tempted to say we could do d/dx(x∫(limits: 0,x) sin(t) dt. Then it's a simple product rule: d/dx (x) * ∫(0,x) sint dt + x * d/dx(∫ (0,x) sin (t) dt) = 1 - cos(x) + x*sin(x). However, I wonder if we have to allow that...
Haha, yeah, I definitely made a mistake there. λ=1/2 or -1/2. Thanks for the hint on adding the constraint, that makes sense and helps pull it together conceptually.
OK so, I have a few paths: λ=1/2 implies y=-z and x=1, so I get (1, -z, z), whereas λ=-1/2 implies y=z and x= -1, so I get (-1...
Attached is the question, and I'm working on part c. I'm maximizing f, and thus looking at F, because it's the gradient. My work was showing setting (1,-2z,-2y)=0, but I don't know what 1=0 implies for the x-component. Also, with Lagrange I don't get anything that creates definitive values...
Oops, grad(f)=0. All the work is correct consistent, I just must have held the shift button after hitting the parenthesis. In any case, the the derivatives are set equal to zero, which is I set (1,-2z,-2y)=0, then also did the Lagrange stuff.
I found that f= x -2yz. To maximize f, I can first inspect the solutions to grad(F)=0. z=y=0 pops out, but I'm not sure what to do with the x-component equaling 1. Do we just include (x,0,0) as a solution? I think the problem wants specifics though, based on what I've seen previously from...
Thanks, I will try to look it over. I was pretty careful, but evidently made an error along the way. Perhaps I exchanged a u with a v somewhere. I don't think I could spot the double angle identity, but I'll work through solving the df/du equation again and check my substitutions. Thanks for...
The -2yz term could be positive or negative, depending on which way you go in the y- and z-directions. Which intuitively feels like a saddle point, but I could be wrong there.
Thanks, this makes more sense. I got cos(u)=0, sin(v)=0, sinu=-sqrt(2)*cot(v) from the df/du equation. Then for df/dv I get cot(v)=0, cos(v)=1
-The cot(v)=0 is promising, because that works for v=pi/2, 3pi/2. I sub back in to then find sin(u)=0, so u= 0, pi.
-Similarly, cos(u)=0 gives u=...
Thanks so much for your reply. So all the undefined points are just ignored, and I should only consider the ones that have discrete values for u and v?
Also, the weird thing that happens when I do this is I get a lot of undefined values. For example, I'll get sin(v)=0, which implies v=0. So then when I substitute back in to the equation sin(u)=-sqrt(2)*cot(v), I get an error, since cot(0) is undefined. So while the z-coordinate is defined, x...
Thanks for the guidance.
For (0,0,0), I made the second partial derivatives matrix and substituted, and obviously get an R^3 zero matrix. This just means this one is indeterminate?
Also, I went ahead and parameterized and derived, but I want to make sure that my logic is correct.
f-x^2-2yz=...
The beginning is straight forward and I found f=x^2-2yz, which satisfies grad(f)=F. Then I calculated W= f(x,y,z)-f(0,1,1) since it's conservative.
I get stuck when trying to find the max and mins. Given grad(f)=0 at extrema, we can see (0,0,0) is a point. On the boundary, I have to...
I agree on the formula, but not sure where I went wrong. Since it's arc length and the path is parameterized as c(t)= (t, 2e^t, e^(2t)) and c'(t)= (1, 2e^(t), 2e^(2t)) = (dx/dt, dy/dt, dz/dt). So squaring all terms and summing up gives: 1 + 4e^(2t) + 4e^(4t). So then I took the square root...
Problem: See Attachment. Parts (a) & (b) are clear, but my confusion arises in (c)-- I feel like there is a much simpler form. While technically my answer is correct, there must be something I'm missing.
I parameterized the curve C=(t, e^2t, e^2t) and got c'(t)=(1,2e^2t,2e^2t), which should...