In mathematics, an equation is a mathematical formula that expresses the equality of two expressions, by connecting them with the equals sign =. The word equation and its cognates in other languages may have subtly different meanings; for example, in French an équation is defined as containing one or more variables, while in English, any well-formed formula consisting of two expressions related with an equals sign is an equation.Solving an equation containing variables consists of determining which values of the variables make the equality true. The variables for which the equation has to be solved are also called unknowns, and the values of the unknowns that satisfy the equality are called solutions of the equation. There are two kinds of equations: identities and conditional equations. An identity is true for all values of the variables. A conditional equation is only true for particular values of the variables.The "=" symbol, which appears in every equation, was invented in 1557 by Robert Recorde, who considered that nothing could be more equal than parallel straight lines with the same length.
I've tried a few things. I did one method to try to accomplish the removal of the -70 in the derivative boundary condition. It came out as below. When plotting it however it gave a solution that didn't make sense.
(Yes, I'm guilty of passing along something incomplete and of utterly unknown provenance that someone saw somewhere and didn't understand. I feel dirty.)
∀t(T(x,t)⟹(¬H(x,t)∨H(x,t+Δt)))
Certainly, without specifying any of the variables, it's got to be useless. At least we can assume t and Δt...
Was the approach used by the referenced tutor correct?
In my understanding, the set up equation ought to be,
##s=-h, u=16## as the stone is being thrown upwards and ##g=-9.8##
giving me,
##-h = 64 - \dfrac{1}{2}×9.8 ×16##
##-h=-14.4##m
##h=14.4##m , or it does not really matter.
The paper follows (free):
https://arxiv.org/abs/2111.03203v2
I was trying to read it, but couldn't understand where equation 2 comes from (page 3). Is this just a standard diffraction or interference equation? Could anyone provide a reference that explains that equation?
My attempt and solution :
$$2^x=3^y+509\Longrightarrow 2^x-512=3^y+509-512\Longrightarrow 2^x-2^9=3^y-3$$
$$\Longrightarrow 2^9(2^{x-9}-1)=3(3^{y-1}-1)$$
$$\Longrightarrow (x,~y)=\boxed{(9,~1)}$$
İs there any solution?
Would we also use Newton's method, or are there more powerful methods? Are there general methods for solving equations with fractional exponents?
The solution to this problem is;
https://www.wolframalpha.com/input?i=solve+x%5E%2817%2F6%29+%2B+x%5E%2821%2F25%29+%3D+15
Is there a better way of solving this,
My steps,
##5(x+1)^{1.5}-4x-28=0##
##(x+1)^{1.5}=\dfrac{4x+28}{5}##
##(x+1)^3=(\dfrac{4x+28}{5})^2##
##x^3+3x^2+3x+1=\dfrac{16(x+7)^2}{25}##
##25x^3+75x^2+75x+25=16(x^2+14x+49)##
##25x^3+59x^2-149x-759=0##
Letting
##f(x)=25x^3+59x^2-149x-759=0##
and using...
I figured since ## dr/dt ## is simply the velocity of the target mass, the velocity ##u_x## would simply have to be changed by the Lorentz transformation. Since the rest mass doesn't change, I think this should be as simple as taking the Lorentz transformation for velocity, and substituting the...
Proof of kinetic energy
work done equals a change in kinetic energy in a mechanical system.
δW = F.ds
W = ∫F.ds
W = m∫a.ds
W = m∫(dv/dt).ds
W = m∫v.dv
here if v and dv are in the same direction the change in kinetic energy will be the usual equation. what happens if both are in different...
I'm currently taking Math Methods in Physics, and we're working on infinite series right now. In one of the examples, in order to find the limit, they do what you can see below in "Example 3"
As you can see, they turn the original equation into a ln equation, so that they can bring the...
I have this expression, $$T=2 P r-\frac{q^2}{4 \pi r^3}+\frac{1}{4 \pi r}$$. Now I want to solve this equation for ##r## perturbatively. This will give the expression $$r=\frac{T}{2 P}-\frac{1}{4 \pi T}+\frac{P \left(8 \pi P q^2-1\right)}{8 \left(\pi ^2 T^3\right)}+.......$$. I was reading...
I initially thought about the different forms of energy present at each of the points:
Total energy at starting point: PEA+ KEA= mgH
at point D:
KE_D = 1/2mv2f PED= mgD
Energy at point D: PED+ KED
D = mgD + 1/2 mv2f
because EA= ED
mgH = mgD = 1/2 mv2f
mg(H-D) = 1/2 mv2f
g(H-D) = 1/2...
Proof:
Let ## x=\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}} ##.
Then ## x^3=(\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}})^3 ##.
Note that ## (a+b)^3=a^3+3ab(a+b)+b^3 ## where ## a=\sqrt[3]{18+\sqrt{325}} ## and ## b=\sqrt[3]{18-\sqrt{325}} ##.
This gives ##...
I've seen a lot of people use implication and equivalence logic incorrectly. For example, when solving equations (i.e. ##x - 2 = 3 \implies x = 5##). Implication is not reversible, thus it only works in one way. By saying, ##x - 2 = 3 \implies x = 5##, you are essentially saying that it is...
Balancing the complete combustion of acetic acid equation to carbon dioxide and water is straightforward if you remember hydrogen is always +1, elemental oxygen is zero and combined oxygen is always -2. just balance the exchange of electrons between oxygen and carbon. But how do you balance...
Who and when first time introduced below equations(dont have to be in same notation, content is important)?
If this formula is always the same, what is contribution of Navier, what of Stokes, what changes all these years?
For this problem,
The solution is,
However, I have a question about the solution. Does someone please know why they write out ##\frac{dF}{dx} = \frac{\partial F}{\partial y}y' + \frac{\partial F}{\partial y'}y''## since we already know that ##\frac{dF}{dx} = 0##?
Thanks!
$$(b-a)^2-4(b-c)(c-a)=0$$
$$b^2-2ab+a^2=4(bc-ab-c^2+ac)$$
$$b^2-2ab+a^2+4ab=4bc-4c^2+4ac$$
$$(b+a)^2-4ac=4c(b-c)$$
$$b-c=\frac{(b+a)^2-4ac}{4c}$$
I don't know how to continue and not even sure what I did is useful.
Thanks
I have having trouble understanding Maxwell's Equations. Can anyone recommend some good book or website that can help me to understand these Equations? How can electric and magnetic fields travel perpendicular to each other? What causes electromagnetic waves to first radiate from its source? I...
This is a question from the MIT Open courseware website.
(1). d = vt + ut let t = time it takes
d = (u + v)t
t = d / (u + v)
(2). d = vt + ut
d - vt = ut. Substitute t with d / (u + v)
d - v*(d/(u+v)) = u*(d/(u+v))
d - v*(d/(u+v)) = “distance...
This question is from the MIT Courseware. I’m having difficulty finding the general equation to solve the problem
(1). d = vt + ut
d = (u + v)t
t = d/(u + v)
(2). d = vt + ut
d - vt = ut sub t with d/(u+v)
d - (v*d)/(u+v) = (u*d)/(u+v)
I’m done with the...
I was able to solve with a rather longer way; there could be a more straightforward approach;
My steps are along these lines;
##\sinh^{-1} x = 2 \ln (2+ \sqrt{3})##
##\sinh^{-1} x = \ln (7+ 4\sqrt{3})##
##x = \sinh[ \ln (7+ 4\sqrt{3})]##
##x = \dfrac {e^{\ln (7+ 4 \sqrt{3})} - e^{-[\ln 7+ 4...
Hello there, I am trying to solve the Following equation for r,
$$2 a Q^4+5 r^4 \left(3 c (\omega +1) r^{1-3 \omega }-2 r (r-3 M)-4 Q^2\right)=0$$
Clearly this is unsolvable. But if we substitute a=0 and c=0 we get one of the solution, ##r=\frac{1}{2} \left(\sqrt{9 M^2-8 Q^2}+3 M\right)##. Can I...
Solve the given PDE for ##u(x,t)##;
##\dfrac{∂u}{∂t} +8 \dfrac{∂u}{∂x} = 0##
##u(x,0)= \sin x##
##-∞ <x<∞ , t>0##
In my working (using the method of characteristics) i have,
##x_t =8##
##x(t) = 8t + a##
##a = x(t) - 8t## being the first characteristic.
For the second...
My confusion refers to this question above.
If I were to ask you, what is the equation of the radial line, what would you say? I know that the general equation the radial line with cartesian gradient of m has an equation of θ = arctan(m). Clearly here the angle between the radial line and...
Hi there,
I am going through a book on multi-storey steel structures and I have come to a chapter that gives approximate methods to calculate rotations at the joints (The intersecting members) of a rigid frame. There is a recurrence equation that computes the rotations and this is given below...
Can someone please explain me the steps of calculation of X1:X2 after putting in the lower value of W^2 in equation 9.9 in "Riley, Hobson, Bence - Mathematical Methods for Physics and Engineering 2006 - pg 319"? I have attached the page as a PDF file.
Thank you.
For this problem,
I am confused by the term below. I get all their terms, expect replacing the highlighted term by ##e^{3t}##, does someone please know whether this is yet another typo?
Thanks!
In my line i have,
##\dfrac{∂r}{du} = \vec{i} +\dfrac{1}{2}u \vec{k} = \vec{i} +1.5 \vec{k}##
##\dfrac{∂r}{dv} = \vec{j} -\dfrac{1}{2}v \vec{k} = \vec{j} -0.5\vec{k}##
The normal to plane is given by,
##\dfrac{∂r}{du}× \dfrac{∂r}{dv} = -\dfrac{3}{2} \vec{ i} + \dfrac{1}{2}\vec{j}+\vec{k}##...
So basically i am vary green when it comes to equations/formula
Well to the point i don't even know if proper name for what i'm looking for is called equation or formula but let's stick to equation
So going as simple as i can
I want to throw boomerang creating boomerang like ellipse trajectory...
This is the general suggested approach given in a textbook.
My question is why can I not directly write it in vector form?
E1 vector + E2 vector =0 should be valid no?
Why are they choosing to write E1 mag + E2 mag=0
Then find a vector form
Then convert the magnitude equation into a vector...
Need help solving this question. Can't seem to get the right answer using PV/T=constant
P1V1/T1 = P2V2/T2
Patm = 75.23cmHg T1+20+273=293K
STP: P=1.01 x 10^5 N/m^2 Pabs=41cmOil
P1 = density x g x h = (810 kg/m^3)(9.8 m/s^2)(75.23-41)x10^-2 mOil=2717.18 N/m^2...
I attempted the problem by first finding the radial, theta, and phi equation for the ground state of a hydrogen atom. I multiplied the three equations to get the wave equation. From there, I took each derivative in the Schrodinger Spherical equation and found that ## \frac {\partial^2 \psi}...
Hello! I'm currently working with a problem which allows modelling ball motion
$$\begin{aligned} m \ddot{x} & =-k_x \dot{x} \sqrt{\dot{x}^2+\dot{y}^2} \\ m \ddot{y} & =-k_y \dot{y} \sqrt{\dot{x}^2+\dot{y}^2}-m g \end{aligned}$$
Given that ##k_x, k_y=0.005##, ##m=0.01## and ##g=9.81## and when...
I need insight on the highlighted in Red on how ##\left[\dfrac{dz}{dx} - 1 = \dfrac{dy}{dx}\right]## otherwise the rest of the steps are clear. I just read that ##\dfrac{dx}{dy} \dfrac{dy}{dz} \dfrac{dz}{dx} =-1##
c
Parts (a) and (b) are okay ... though the challenge was on part (a)
My graph had a plot of r on the y-axis vs θ on the x-axis). The sketch of my graph looks like is shown below;
I suspect the ms had θ on the x-axis vs r on the y-axis.
I used the equation ##r=\sqrt{\dfrac {1}{θ^2+1}}##...
My approach - i think similar to ms approach.
The required Equation will be in the form ##y=mx##
##\begin{pmatrix}
a & b^2 \\
c^2 & a
\end{pmatrix} ⋅
\begin{pmatrix}
k \\
mk
\end{pmatrix} =
\begin{pmatrix}
x \\
y
\end{pmatrix}
##
##ak+b^2mk=x##
##kc^2+amk=y##
##x=k(a+b^2m)##...
For this,
I tried solving the differential equation using an alternative method. My alternative method starts at
##tv^{''} + v^{'} = 0##
I substitute ##v(t) = e^{rt}## into the equation getting,
##tr^2e^{rt} + re^{rt} = 0##
##e^{rt}[tr^2 + r] = 0##
##e^{rt} = 0## or ##tr^2 + r = 0##
Note that...