Introduction to tough math problems:
In this article we are going to see about the tough math problems. Mathematics is the learning of quantity, arrangement, space, and change. Math seeks out patterns that originate the new conjecture, and ascertain truth by precise deduction from properly selected axioms and definitions. Math is used throughout the whole world that has fundamental tool in various fields that include natural science, engineering, medicine, and the social sciences Understanding Business Math Problems is always challenging for me but thanks to all math help websites to help me out.
Example Problems for Tough Math Problems:
Tough math problems – Example: 1
Find the externals of the functional `J(y)=\int_0^\pi\[y'(x)\]^2dx` subject to the constraint `int_0^\pi \[ y(x)\]^2dx=1, y(0)=y(\pi)=0`
Solution:
Let `H(y,\lambda) = \int_0^\pi ([y'(x)]^2 + \lambda [y(x)]^2)dx` where \lambda is a Lagrange multiplier. The Euler equation is
`\frac{d}{dx}\frac{\partial (y'^2 + \lambda y^2)}{\partial y'} - \frac{\partial(y'^2 + \lambda y^2)}{\partial y} = 0`
`\implies y'' - \lambda y = 0`
(i) Try `\lambda > 0;` let `\lambda=\alpha^2.` Then
`y'' - \alpha^2 y = 0`
`\implies y = a e^{\alpha x} + b e^{-\alpha x}`
`y(0) = 0 \implies a + b = 0;\quad y(\pi) = 0 \implies a e^{\alpha\pi} + b e^{-\alpha \pi} = 0`
But the solution for this is a = b = 0, i.e., `y\equiv 0` which does not satisfy the constraint `\int_0^\pi y^2dx = 1`
(ii) Try \lambda = 0 Then
`y'' = 0`
`\implies y = a x + b`
`y(0) = 0 \implies b = 0 so y = a x;` then `y(\pi) = 0 \implies a = 0 \implies y \equiv 0` which, again, does not satisfy the constraint.
`(iii) \lambda < 0; let \lambda=-\alpha^2.` Then
`y'' + \alpha^2 y = 0`
`\implies y = a \cos(\alpha x) + b \sin(\alpha x)`
`y(0) = 0 \implies a = 0; y(\pi) = 0 \implies b = 0 or \sin(\alpha \pi) = 0`
`b = 0 \implies y\equiv 0,` so we reject this and take `\sin(\alpha \pi) = 0` , which `implies \alpha = \pm 1, \pm 2, \dots`
Thus `y = b \sin(nx)` where n is any nonzero integer.
b is found by imposing the constraint:
`\int_0^\pi b^2 \sin^2(nx)dx = 1 \implies \int_0^\pi b^2 \frac{1-\cos(2nx)}{2}dx = 1 \implies\frac{b^2}{2} \[x - \frac{\sin(2nx)}{2n}\]_0^\pi = 1`
`\implies b = \pm\sqrt{2/\pi}`
Hence, the extremes of J subject to the given constraint are `y = \pm\sqrt{2/\pi}\sin(nx), n=\pm1, \pm2, \dots`
Tough math problems – Example: 2
Constraint problem: Minimize `T(y)=\int_0^1\(y'^2+x^2\)dx s.t. K(y)=\int_0^1y^2dx=2.`
Solution:
Using the method of Lagrange multipliers,
`\int_0^1\(y'^2+x^2+\lambda y^2\)dx = J(y)`
If the Euler-Lagrange equations are satisfied then this integral has an extremum.
`\frac{\partial}{\partial y}\[y'^2+x^2+\lambda y^2\]=\frac{d}{dx}\frac{\partial}{\partial y'}\[y'^2+x^2+\lambda y^2\]`
`2\lambda y = 2y''`
`y''=\lambda y so y(x) = c_1\cos\sqrt{\lambda}x + c_2\sin\sqrt{\lambda}x`
`y(0)=0` gives `c_1=0`
`y(x)=c_2\sin\sqrt{\lambda}x`
`y(1)=0 gives 0=c_2\sin\sqrt{\lambda} so \sqrt{\lambda}=n\pi, n=1,2,...`
The solution is
`y(x)=c_2 \sin n\pi x`
`2=\int_0^1y^2dx = c_2^2\int_0^1\sin^2(n\pi x)dx`
`2=\frac{c_2^2}{2}\int_0^1(1-\cos 2n\pi x)dx = \frac{c_2^2}{2}`
`c_2=\pm 2`
Tough math problems – Example: 3
Find the extreme of `x^2+y^2+z^2` subject to the constraint `x^2+2y^2-z^2-1=0.`
Solution:
Using Lagrange multipliers, make a new function:
`f = x^2 + y^2 + z^2 + \lambda x^2 + 2 \lambda y^2 - \lambda z^2 - \lambda = 0`
Take the partial derivatives of the function with the respect to x, y, and z.
`1) \frac{ \partial f}{\partial x} = 2x+2 \lambda x = 0`
`2) \frac{ \partial f}{\partial y} = 2y+4 \lambda y = 0`
`3) \frac{ \partial f}{\partial z} = 2z-2 \lambda z = 0`
For each equation above, find a value of lambda that works and force that value into the other two equations. Choose the other two variables appropriately to satisfy the equalities.
From 1), λ = − 1, and in that case `y = z = 0.` Plugging into our constraint equation yields` x = 1, - 1.`
From 2), λ = − 1 / 2, and in that case `x = z = 0,` `y = 1/\sqrt{2}, -1/\sqrt{2}.`
From 3), λ = 1, and in that case `x = y = 0,z = i, - i.`
Combining these results, the extreme are
`(1,0,0), (-1,0,0), (0,1/\sqrt{2},0), (0,-1/\sqrt{2},0), (0,0,i), (0,0,-i)`
The last two solutions may be disregarded if we are restricting to real variables.Having problem with Continuous Quantitative Variable keep reading my upcoming posts, i will try to help you.
Practice Problems for Tough Math Problems:
1. Find the Euler equation for the functional `J(u)=\int int_G\[u_x^2+u_y^2+2f(x,y)u(x,y)\]dxdy`
`Answer: f(x, y) = u_{xx} + u_{yy}`
2. Find the externals of `J(y) = \int_0^1(yy'+y''^2)dx, y(0)=0, y'(0)=1, y(1)=2, y'(1)=4`
`Answer: y = x^3 + x`
In this article we are going to see about the tough math problems. Mathematics is the learning of quantity, arrangement, space, and change. Math seeks out patterns that originate the new conjecture, and ascertain truth by precise deduction from properly selected axioms and definitions. Math is used throughout the whole world that has fundamental tool in various fields that include natural science, engineering, medicine, and the social sciences Understanding Business Math Problems is always challenging for me but thanks to all math help websites to help me out.
Example Problems for Tough Math Problems:
Tough math problems – Example: 1
Find the externals of the functional `J(y)=\int_0^\pi\[y'(x)\]^2dx` subject to the constraint `int_0^\pi \[ y(x)\]^2dx=1, y(0)=y(\pi)=0`
Solution:
Let `H(y,\lambda) = \int_0^\pi ([y'(x)]^2 + \lambda [y(x)]^2)dx` where \lambda is a Lagrange multiplier. The Euler equation is
`\frac{d}{dx}\frac{\partial (y'^2 + \lambda y^2)}{\partial y'} - \frac{\partial(y'^2 + \lambda y^2)}{\partial y} = 0`
`\implies y'' - \lambda y = 0`
(i) Try `\lambda > 0;` let `\lambda=\alpha^2.` Then
`y'' - \alpha^2 y = 0`
`\implies y = a e^{\alpha x} + b e^{-\alpha x}`
`y(0) = 0 \implies a + b = 0;\quad y(\pi) = 0 \implies a e^{\alpha\pi} + b e^{-\alpha \pi} = 0`
But the solution for this is a = b = 0, i.e., `y\equiv 0` which does not satisfy the constraint `\int_0^\pi y^2dx = 1`
(ii) Try \lambda = 0 Then
`y'' = 0`
`\implies y = a x + b`
`y(0) = 0 \implies b = 0 so y = a x;` then `y(\pi) = 0 \implies a = 0 \implies y \equiv 0` which, again, does not satisfy the constraint.
`(iii) \lambda < 0; let \lambda=-\alpha^2.` Then
`y'' + \alpha^2 y = 0`
`\implies y = a \cos(\alpha x) + b \sin(\alpha x)`
`y(0) = 0 \implies a = 0; y(\pi) = 0 \implies b = 0 or \sin(\alpha \pi) = 0`
`b = 0 \implies y\equiv 0,` so we reject this and take `\sin(\alpha \pi) = 0` , which `implies \alpha = \pm 1, \pm 2, \dots`
Thus `y = b \sin(nx)` where n is any nonzero integer.
b is found by imposing the constraint:
`\int_0^\pi b^2 \sin^2(nx)dx = 1 \implies \int_0^\pi b^2 \frac{1-\cos(2nx)}{2}dx = 1 \implies\frac{b^2}{2} \[x - \frac{\sin(2nx)}{2n}\]_0^\pi = 1`
`\implies b = \pm\sqrt{2/\pi}`
Hence, the extremes of J subject to the given constraint are `y = \pm\sqrt{2/\pi}\sin(nx), n=\pm1, \pm2, \dots`
Tough math problems – Example: 2
Constraint problem: Minimize `T(y)=\int_0^1\(y'^2+x^2\)dx s.t. K(y)=\int_0^1y^2dx=2.`
Solution:
Using the method of Lagrange multipliers,
`\int_0^1\(y'^2+x^2+\lambda y^2\)dx = J(y)`
If the Euler-Lagrange equations are satisfied then this integral has an extremum.
`\frac{\partial}{\partial y}\[y'^2+x^2+\lambda y^2\]=\frac{d}{dx}\frac{\partial}{\partial y'}\[y'^2+x^2+\lambda y^2\]`
`2\lambda y = 2y''`
`y''=\lambda y so y(x) = c_1\cos\sqrt{\lambda}x + c_2\sin\sqrt{\lambda}x`
`y(0)=0` gives `c_1=0`
`y(x)=c_2\sin\sqrt{\lambda}x`
`y(1)=0 gives 0=c_2\sin\sqrt{\lambda} so \sqrt{\lambda}=n\pi, n=1,2,...`
The solution is
`y(x)=c_2 \sin n\pi x`
`2=\int_0^1y^2dx = c_2^2\int_0^1\sin^2(n\pi x)dx`
`2=\frac{c_2^2}{2}\int_0^1(1-\cos 2n\pi x)dx = \frac{c_2^2}{2}`
`c_2=\pm 2`
Tough math problems – Example: 3
Find the extreme of `x^2+y^2+z^2` subject to the constraint `x^2+2y^2-z^2-1=0.`
Solution:
Using Lagrange multipliers, make a new function:
`f = x^2 + y^2 + z^2 + \lambda x^2 + 2 \lambda y^2 - \lambda z^2 - \lambda = 0`
Take the partial derivatives of the function with the respect to x, y, and z.
`1) \frac{ \partial f}{\partial x} = 2x+2 \lambda x = 0`
`2) \frac{ \partial f}{\partial y} = 2y+4 \lambda y = 0`
`3) \frac{ \partial f}{\partial z} = 2z-2 \lambda z = 0`
For each equation above, find a value of lambda that works and force that value into the other two equations. Choose the other two variables appropriately to satisfy the equalities.
From 1), λ = − 1, and in that case `y = z = 0.` Plugging into our constraint equation yields` x = 1, - 1.`
From 2), λ = − 1 / 2, and in that case `x = z = 0,` `y = 1/\sqrt{2}, -1/\sqrt{2}.`
From 3), λ = 1, and in that case `x = y = 0,z = i, - i.`
Combining these results, the extreme are
`(1,0,0), (-1,0,0), (0,1/\sqrt{2},0), (0,-1/\sqrt{2},0), (0,0,i), (0,0,-i)`
The last two solutions may be disregarded if we are restricting to real variables.Having problem with Continuous Quantitative Variable keep reading my upcoming posts, i will try to help you.
Practice Problems for Tough Math Problems:
1. Find the Euler equation for the functional `J(u)=\int int_G\[u_x^2+u_y^2+2f(x,y)u(x,y)\]dxdy`
`Answer: f(x, y) = u_{xx} + u_{yy}`
2. Find the externals of `J(y) = \int_0^1(yy'+y''^2)dx, y(0)=0, y'(0)=1, y(1)=2, y'(1)=4`
`Answer: y = x^3 + x`
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