calculus – Evaluate the length of the curve $r(t)=(sqrt{t}, t, t^2)$

Question: Find the length of the curve $r(t)=(sqrt{t}, t, t^2)$ for $1<=t<=4$.

My attempted solution: I’ve computed the derivative of the vector function and has already set up the integral:

$$int_1^4 sqrt{frac{1}{4t}+1+4t^2} dt$$
I’m unsure if this integral is set up correctly because I couldn’t seem to integrate it even with software like Symbolab and Wolfram Alpha.

I’d appreciate it if someone could show me how this is done. Thanks in advance!

calculus and analysis – How can i write in Mathematica that 2*x is an integer?

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calculus and analysis – Manipulation of a symbolic matrix expression

I have the following code:

$Assumptions = Element(B | F | A, Matrices({d, d}, Reals))
$Assumptions =   Element(Id | C | C0 , Matrices({d, d}, Reals, Symmetric({1, 2}))); 
$Assumptions = Element(t, Reals);
W = ((ScriptCapitalC) (C - Id)) ** (C - Id) + ((ScriptCapitalD) (C - C0)) ** (C - C0) /.  C -> (TensorTranspose(F + t* A, {1, 2}) ** (F + t* A))
D(W, {t, 2})
D(W, {t, 2}) /. t -> 0

which provides me the answer:

enter image description here

I would rather expect that

  1. 0 ** F will be replaced by 0 or completely removed from output
  2. F ** F should actualy be F^T * *F (F is not a symmetric matrix)
  3. A ** A should be replaced by A^ T** (A is not a symmetric matrix)

where ‘T’ denotes the matrix transpose operator. Can you help?

calculus – When doing a change of variables integration, what does a jacobian (determinant) of 0 tell us?

Given a change of variables $x = f(u, v), y = g(u, v)$ if we get that $partial(x, y)/partial(u, v) = det (begin{matrix} partial x/partial u & partial x/partial v \ partial y/partial u & partial y / partial v end{matrix}) = 0$. What does that tell us?

Idea: in single variable calculus, a determinant of $0$ tells us that the function is constant. So is the idea similar here? Naively asking the two variable function to be constant seems to be obviously false, so I’m wondering if it means something weaker such as being constant on a surface/area instead? And in the specific case of change of variables, how do we intepret this?

Simple Physics and Calculus in Mathematica V4

I think the only real problem keeping you from using the more modern Quantities framework is the fact that Integrate(a0, t, t) and a0 Integrate(1, t, t) are not equivalent if one or more accelerations in a0 are zero. In the first case we get {0 m/s^2, -4.9 t^2 m/s^2} and in the second case we get {t^2 (0 m/s^2), t^2 (-4.9 m/s^2)}. Since we’re integrating constants, we can pull the constants out of the integral and put them in front. I’ve done it for both v0 and a0, though it was only absolutely necessary for a0. Unfortunately, because of the way the replacement works, Mathematica will print a warning that m/s and m are incompatible units, then the substitution happens and it realizes the units are fine.

vectorDirection(vec_) := ArcTan(vec((2))/vec((1)))*180/(Pi)

r0 = Quantity({0, 0}, "Meters");
s0 = Quantity(37, "Meters"/"Seconds");
v0 = AngleVector(s0, 53.1 Degree)
a0 = Quantity({0, -9.8}, "Meters"/"Seconds"^2);

r0 + v0 Integrate(1, t) + a0 Integrate(1, t, t) /. 
 t -> Quantity(2, "Seconds")

  v0 + a0 Integrate(1, t) /. t -> Quantity(2, "Seconds")), 
  v0 + a0 Integrate(1, t) /. t -> Quantity(2, "Seconds"))}

r1 = {x, y};
v1 = {v1x, Quantity(0, ("Meters")/("Seconds"))};
Solve({r1 == r0 + v0 Integrate(1, t) + a0 Integrate(1, t, t), 
  v1 == v0 + a0 Integrate(1, t)}, {x, y, t, v1x})

r1 = {x, Quantity(0, "Meters")};
v1 = {v1x, v1y};
Last@Solve({r1 == r0 + v0 Integrate(1, t) + a0 Integrate(1, t, t), 
   v1 == v0 + a0 Integrate(1, t)}, {x, t, v1x, v1y})

Of course all the quantities and integrals can be entered in the more traditional format when using Mathematica, but it doesn’t translate well to StackExchange. I do think it’s a bit strange to carry around all these integrals and repeatedly solve the same equations over and over. My usual process would be to solve the integral once, if possible, and then be done with it. Perhaps you have a reason you want to keep the integrals, but I think it’s cleaner without. With the initial values defined as above:

pos = Integrate(a, t, t, GeneratedParameters -> C) /. {C(1) -> r0, C(2) -> v0,  a -> a0}
pos /. t -> Quantity(2, "Seconds")
{Norm(#), vectorDirection(#)} &@(D(pos, t) /. 
   t -> Quantity(2, "Seconds"))
Solve({{x, y} == pos, {v1x, Quantity(0, ("Meters")/("Seconds"))} == 
   D(pos, t)}, {x, y, t, v1x})
Last@Solve({{x, Quantity(0, "Meters")} == pos, {v1x, v1y} == 
    D(pos, t)}, {x, t, v1x, v1y})

If we were using the derivative of the position a lot, we could also just define a new equation like vel = D(pos, t). I also switched the definition of position here. From a physics perspective, I don’t think the original definition is quite right. It gets the same answer, but a double indefinite integral should have two unknowns (initial velocity and initial position, here). When using the double integral, we’re really solving the differential equation r''(t) == a and integrating both sides twice to get r(t) = 1/2 a t^2 + v0 t + r0, so we don’t need the definition to contain 3 separate parts.

This is a calculus 3question referring to finding the total mass of the solid in kilograms

These two questions are put together and the first one will be needed to be done to be worked upon on the second one .

calculus – Picard-Lindelöf theorem variations

There are so many slight variations of this theorem around. I wrote the following in my own words from different sources online. Is my version correct? I want to make sure I have understood the theorem correctly.

Let $U$ be an open subset of $mathbb{R}times mathbb{R}^m$ containing $(t_0,y_0)$ and let $f:Uto mathbb{R}^m$. If $f$ is continuous in $t$ and satisfies for some $Kgeqslant 0$, $|f(t,x)-f(t,y)|leqslant K,|x-y|$ for every $x,y in U$. Then for some $tau >0$ there exists a unique solution $y(t)$ on the interval $(t_0 – tau, t_0 +tau)$ for the initial value problem
$$y'(t)=f(t,y(t)), ,,,,, y(t_0)=y_0$$

calculus and analysis – Why the integration command for this function gives no answer?

I have a function wfVAR2(r1, r2) which has the following form:

wfVAR2(r1_, r2_):=-(1/(r1 r2))
 2 E^(-3.99999997*10^-8 r1 - 1.24 r1^2 - 0.2379728 r1 r2 - 
   1.24 r2^2) π (-4.202160919 E^(
     0.1189864 r1^2 + 
      r1 (1.999999985*10^-8 + 0.4759456 r2) + (-1.999999985*10^-8 + 
         0.1189864 r2) r2) + 
    4.202160919 E^(
     0.1189864 r1^2 + 0.1189864 r2^2 + 1.999999985*10^-8 (r1 + r2)) - 
    2.159230221*10^-7 E^(r1 (3.99999997*10^-8 + 0.2379728 r2))
      Erfi(2.899020819*10^-8 - 0.3449440534 r1 + 0.3449440534 r2) + 
    2.159230221*10^-7 E^(r1 (3.99999997*10^-8 + 0.2379728 r2))
      Erfi(2.899020819*10^-8 - 0.3449440534 (r1 + r2)) - 
    2.64697796*10^-23 E^(r1 (3.99999997*10^-8 + 0.2379728 r2))
      Erfi(2.899020819*10^-8 - 0.3449440534 Sqrt(r1^2 + r2^2)))

when I try this

gVAR = Integrate(wfVAR2(r1, r2), {r2, r-r1, r+r1})

it’s really amazing that I get no answer! I really don’t know what is the problem?! Any idea?

calculus – Prooving that the transversal intersection of regular surfaces is a regular curve.

On his book on differential geometry, Do Carmo has an excercise on section 2-4, essentially excercise 17.

That states:

Show that the transversal intersection of two regular surfaces $S_1, S_2$ is a regular curve.
A transversal intersection is one such that $p in S_1 cap S_2 implies T_p(S1) neq T_p(S_2)$.

Slader has one possible solution: But although I am able to follow the math per se. Something isn’t clicking. Moreover the author used a theorem that isn’t in the prior sections of the book, and it is quite involved (I understand the result but I have never seen the proof and so this won’t help me get the intuition right).

The way I tried to do it was, the intersection must be a curve a point or the empty set, otherwise if it is a surface the tangent spaces at that point cannot disagree.

Assume the intersection is a curve $C$. $C$ contains $p$ by construction.

Then since $S_1$ is a regular surface $forall p in S_1 exists f: Usubset R^2 rightarrow V subset S_1$ Such that $p in f(U)$ $f$ is differentiable, a homeomorphism, and the rank of the jacobian of any point in $f(U)$ is 2.

Consider $C_p = C cap V$ A subsection of $C$ that is contained in $V$ and contains $p$.

$forall p in C_p; exists x in U$ such that $f(x) = p$, since $f$ is a homeomorphism.

Here I get stuck, I want to justify that either $C_p$ is continuous or that $f^{-1}(C_p)$ is continuous. If I get either one of those conditions, I think I can continue since the homeomorphism condition implies the other must also be continuous and then:

Since $f^{-1}(C_p) = I$ is continuous then since $f$ is differentiable we get that $f'(x_p) = v in T_p(S_1)$ where $x_p in I$.

An analogous argument on $S_2$ will give us $w in T_p(S_2)$.

And then I want to show that it is not possible for both of these vectors to be 0 simultaneously (somehow) which implies the curve is regular since it has a non zero tangent vector everywhere.

calculus – Theorem for convergence of a series

I was studying Infinite Series. The book I use is Thomas and Finney – Calculus and Analytical Geometry 9th Edition. On pg 635, the following theorem 6 was provided :

enter image description here

Theorem 6 says that when the sum , say

S = a1 + a2 + a3 + a4 + ….. + a(n)

is continued for infinite terms and converges resulting into a unique and finite value, then the limit of n tending to infinity of a(n) must be zero. This means that as we go further and further with our sum S, we reach to a term that would become, well, very close to zero.

I have tried thinking about it, but I could not really justify this theorem in any manner. The proof of the theorem is not given in the book. Could anyone please help ?