Mathematics

Mathematics

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Subject: Differential Equations. Type: Test. Due date: 2 days. Price: $8

Problem 1:  y''' + 2y'' + y'' = x2 + x – 1

Solution:

First we are going to solve the homogenous equation: y'''' + 2y''' + y'' = 0

Let’s think about what we know about this equation:

  1. This is a fourth-order linear differential equation.
  2. If k is a distinct real root, then y = ekt is a solution.
  3. If k is a real root and appears r times then y = ekt, y = tekt, y = t2ekt, …, y = tk-1e-kt are all solutions.
  4. We know that the general solution is of the form: y0 = C1 + C2t + C3ekt + C4ektt, where C1, C2, C3, C4 are all constants.

First we will write the characteristic polynomial:

k4 + 2k3 + k2 = 0
k2(k2 + 2k + 1) = 0

To determine the solution to this equation, either k2 or (k2 + 2K + 1) must equal zero.

k2 = 0 yields k1 = 0 and k2 = 0. (We should expect two solutions since the equation is second order. We can then solve k2 + 2K + 1 = 0 using the quadratic function. This yields: k3 = -1 and k4 = -1.

Thus using facts 2-4 written above, we can plug in for our four k constants into y0 = C1 + C2t + C3ekt + C4ektt:

     k1 = 0: C1

     k2 = 0: C2t

     k3 = -1: C3e-t

     k4 = -1: C4e-tt

Thus the net homogenous solution is: y0 = C1 + C2t + C3e-t + C4e-tt.

Now we can proceed and solve for the nonhomogeneous equation g(t) = x2 + x - 1.

We let Y be a generic fourth order polynomial matching our original equation:

Y = Ax4 + Bx3 + Cx2

g(t)' = 4At3 + 3Bt2 + 2C

g(t)'' = 12At2 + 6Bt + 2C

g(t)''' = 24At + 6B

g(t)'''' = 24A

We plug each derivative of g(t) to the corresponding derivate of y in the original left hand side of the fourth-order linear equation:

24A + 2(24Ax + 6B) + 12Ax2 + 6Bx + 2C = x2 + x - 1

Now we can combine like terms and simplify the left hand side:

12Ax2 + (48A + 6B)x + (2C + 24A + 12B) = x2 + x - 1

We can then setup the system of linear equations by equating the coefficients of equal powers:

$$\left.\begin{matrix} x^{2} : 12A = 1\\ x : 48A + 6B = 1\\ 1: 2C + 24A + 12B = -1\\ \end{matrix}\right.\rightarrow \left.\begin{matrix} A = \frac{1}{12}\\ B = -\frac{1}{2}\\ C = \frac{3}{2}\\ \end{matrix}\right.\rightarrow Y = \frac{x^{4}}{12} - \frac{x^{3}}{2} + \frac{3}{2}x^{2}$$

The complete solution is the sum of the homogenous and nonhomogeneous solutions:

y = y0 + g(t)

$$y = C_{1} + C_{2}t + C_{3}e^{-t} + C_{4}e^{-t}t + \frac{t^{4}}{12} - \frac{t^{3}}{2} + \frac{{3}}{2}t^{2}$$

Problem 2: y''' - 4y'' + 4y' = (x - 1) ex

Solution:

First we are going to solve the homogenous equation: y''- 4y''+ 4y' = 0

Let's think about what we know about this equation:

  1. This is a second-order linear differential equation.
  2. If k is a distinct real root, then y=ekt is a solution.
  3. If k is a real root and appears r times then y=ekt, y=tekt, y=t2ekt, ..., y=tk-1e-kt are all solutions.
  4. We know that the general solution is of the form: y0=C1+C2ekt+C3ekt, where C1, C2, C3 are all constants.

First we will write the characteristic polynomial:

k3 - 4k2 + 4k2 = 0
k(k2 - 4k + 4) = 0

To determine the solution to this equation, either k or  (k2 - 4k + 4) must equal zero.

k=0 yields k1=0. We can then solve k2 - 4k + 4=0 using the quadratic function. This yields: k2=2 and k3=2.

Thus using facts 2-4 written above, we can plug in for our four k constants into y0=C1+C2ekt+C3ektt:

     k1 = 0: C1

     k2 = 2: C2ekt

     k3 = 2: C3ektt

Thus the net homogenous solution is: y0 = C1 + C2e2t + C3e2tt.

Now we can proceed and solve for the nonhomogeneous equation g(t) = (t - 1)et.

We let Y be a generic second order polynomial matching our original equation:

Y=(Ax+B)ex

g(t)' = Aet + (At + B)et

g(t)'' = Aet + Aet + (At + B)et = 2Aet + (At + B)et

g(t)''' = 2Aet + Aet + (At + B)et = 3Aet + (At + B)et

We plug each derivative of g(t) to the corresponding derivate of y in the original left hand side of the second-order linear equation:

3Aex + (Ax + B)ex - 4(2Aex + (Ax + B)ex) + 4(Aex + (Ax + B)ex) = (x - 1)ex

Now we can combine like terms and simplify the left hand side:

-Aex + (Ax + B)ex = (x - 1)ex

We can then setup the system of linear equations by equating the coefficients of equal powers:

 $$\left.\begin{matrix} e^{x} : -A + B = -1\\ xe^{x}: A = 1\\ \end{matrix}\right.\rightarrow \left.\begin{matrix} B = 0\\ A = 1\\ \end{matrix}\right.\rightarrow Y = xe^{x}$$

The complete solution is the sum of the homogenous and nonhomogeneous solutions:

y = y0 + g(t)

y = C1 + C2e2t + C3e2tt + tet

Problem 3: y'' - 2y' = ex(sin x + cos x)

Solution:

First we are going to solve the homogenous equation: y'' - 2y' = 0

Let's think about what we know about this equation:

  1. This is a second-order linear differential equation.
  2. If k is a distinct real root, then y = ekt is a solution.
  3. If k is a real root and appears r times then y = ekt, y = tekt, y = t2ekt, ..., y = tk-1e-kt are all solutions.
  4. We know that the general solution is of the form: y0 = C1 + C2ekt, where C1, C2 are all constants.

First we will write the characteristic polynomial:

k2 + 2k = 0

k(k + 2) = 0

To determine the solution to this equation, either k or (k+2) must equal zero.

k=0 yields k1=0. We can then solve k + 2 = 0 using the quadratic function. This yields: k2 = -2.

Thus using facts 2-4 written above, we can plug in for our four k constants into y0 = C1 + C2ekt:

k1 = 0: C1

k2 = -2: C2e-2t

Thus the net homogenous solution is: y0 = C1 + C2e-2t:

 Now we can proceed and solve for the nonhomogeneous equation g(t) = et (sin⁡ t+cos ⁡t).

 We let Y be a generic second order polynomial matching our original equation: Y = ex(A sin x + B cos ⁡x)

g(t)' = et(A sin t + B cos⁡t) + et(A sin t - B cos t)

g(t)''= et(A sin t + B cos t) + et(A sin t - B cos t)+ et(A sin t - B cos t) + et(-A sin t - B cos ⁡t)

We plug each derivative of g(t) to the corresponding derivate of y in the original left hand side of the second-order linear equation:

ex(A sin x + B cos ⁡x) + ex(A sin x - B cos x) + ex(A sin x - B cos⁡ x) + ex(A sin x - B cos⁡ x) + 2(ex(A sin x + B cos ⁡x) + ex(-A sin x - B cos ⁡x)) = ex(sin x + cos ⁡x)

Now we can combine like terms and simplify the left hand side:

3ex(A sin x + B cos ⁡x) +3(A sin x - B cos x) + 2ex(-A sin x - B cos ⁡x) = ex(sin x + cos ⁡x)

We can then setup the system of linear equations by equating the coefficients of equal powers:

 $$\left.\begin{matrix} -4B + 2A = 1\\ 4A + 2B = 1\\ \end{matrix}\right.\rightarrow \left.\begin{matrix} A = \frac{3}{10}\\ B = -\frac{1}{10}\\ \end{matrix}\right.\rightarrow Y = e^t(\frac{3}{10}sin x + -\frac{1}{10}cos x)$$

The complete solution is the sum of the homogenous and nonhomogeneous solutions: \[{y=\ y}_0\ +g\left(t\right)\] \[y\ =\ C_1+\ C_2e^{-2t}+e^t(\frac{3}{10}{sin t\ }+\ -\frac{1}{10}{cos t)\ }\]

Problem 4: y'' - 4y' = 16 cosh 4x

Solution:

First we are going to solve the homogenous equation: y'' - 4y' = 0

Let's think about what we know about this equation:

  1. This is a second-order linear differential equation.
  2. If k is a distinct real root, then y = ekt is a solution.
  3. If k is a real root and appears r times then y = ekt, y = tekt, y = t2ekt, ..., y = tk-1e-kt are all solutions.
  4. We know that the general solution is of the form: y0 = C1 + C2ekt, where C1, C2 are all constants.

First we will write the characteristic polynomial:

k2 - 4k = 0

k(k - 4) = 0

To determine the solution to this equation, either k or (k-4) must equal zero.

k = 0 yields k1 = 0. We can then solve k - 4 = 0 using the quadratic function. This yields: k2 = 4.

Thus using facts 2-4 written above, we can plug in for our four k constants into y0 = C1 + C2ekt:

    k1 = 0: C1

    k2 = 4: C2e4t

Thus the net homogenous solution is: y0 = C1 + C2e4t.

 Now we can proceed and solve for the nonhomogeneous equation g(t) = 16 cosh 4t.

 We let Y be a generic second order polynomial matching our original equation: Y =  C1(t) + C2e4t.

We can now create a system of equations:

C1(t) + C2(t)e4t = 0

C1(t)*0 + 4C2(t)e4t = 16 cosh 4t

Now we can solve the system of equations:

\[\mathit{\Delta}\ =\ \left| \begin{array}{cc} 1 & e^{4t} \\ 0 & 4e^{4t} \end{array} \right| = 4e^{4t}\]

\[{\mathit{\Delta}}_1\ =\ \left| \begin{array}{cc} 0 & e^{4t} \\ 16\ {cosh 4t\ } & 4e^{4t} \end{array} \right| = -16\ {e^{4t}cosh 4t\ }\] \[{\mathit{\Delta}}_2\ =\ \left| \begin{array}{cc} 1 & 0 \\ 0 & 16\ {cosh 4t\ } \end{array} \right| = 16\ {cosh 4t\ }\]

We take the derivate of C1 and C2 and then integrate:

\[C_1'=\ \frac{{\mathit{\Delta}}_1}{\mathit{\Delta}}=\ \frac{-16\ {e^{4t}cosh 4t\ }}{4e^{4t}}= -4{cosh 4t\ }\] \[C_1\left(t\right)=\ \int{-4{cosh 4t\ }\ dt}=\ -\int{{cosh 4t\ }\ d(4t})=-{sinh 4t+e_1\ }\] \[C_2'=\ \frac{{\mathit{\Delta}}_2}{\mathit{\Delta}}=\ \frac{16\ {cosh 4t\ }}{4e^{4t}}\ = \frac{4{cosh 4t\ }}{e^{4t}}\ = \frac{4}{e^{4t}} * \frac{e^{4t}+\ e^{-4t}}{2} = 2 (1 + e^{-8t}\ )\] \[C_2\left(t\right)=\ 2\int{\left(1\ +\ e^{-8t}\right)dt}=\ 2t-\ \frac{1}{4}e^{-8t}\ +\ C_2\ \]

Now we substitute in C1(t) and C2(t):

$$Y=\ {-sinh 4t\ }+\ C_1+ (2t-\ \frac{1}{4}e^{-8t}\ +\ C_2)\ e^{4t},$$ where C1 = C2 = 0

The complete solution is the sum of the homogenous and nonhomogeneous solutions:

\[{y=\ y}_0\ +Y\] \[y\ =\ C_1+\ C_2e^{4t}\ {-sinh 4t\ }+(2te^{4t}-\ \frac{1}{4}e^{-8t})\]


Subject: Mathematical Analysis. Multivariable functions. Type: Test. Due date: 3 days. Price: $12

Problem 1.

For the given function, find the directional derivate for the vector a at the point A.,
f(x,y) = xy - x2 cos y2, a = (2,4) A(3;1)

Solution.

Let's first find the first-derivative. \[\begin{array}{l} {f'_{x} (x,y)=y-2x\cos y^{2} } \\ {f'_{x} (3,1)=1-2\cdot 3\cos 1=1-6\cos 1\approx -2.24} \end{array}\] \[f'_{y} (x,y)=x-x^{2} \cdot (-\sin y^{2} )\cdot 2y=x+2yx^{2} \sin y^{2} \] \[f'_{y} (3,1)=3+2\cdot 9\cdot \sin 1\approx 18.15\] Let's find the the cosine directional of the vector $$\bar{a}=(2,4)$$ \[\cos \alpha =\frac{2}{\sqrt{2^{2} +4^{2} } } =\frac{2}{\sqrt{20} } =\frac{1}{\sqrt{5} } \approx 0.4472\] \[\cos \beta =\frac{4}{\sqrt{2^{2} +4^{2} } } =\frac{4}{\sqrt{20} } =\frac{2}{\sqrt{5} } \approx 0.8944\] The directional derivative is then: \[\begin{array}{l} {\left. f'_{v} \right|_{A} =f'_{x} (3,1)\cos \alpha +f'_{y} (3,1)\cos \beta =-2.24\cdot 0.4472+18.15\cdot 0.8944\approx -{\rm 1,001728}+{\rm 16,23336}\approx 15.23} \end{array}\]

Problem 2.

For the given plane and curve, find the angles between them at their points of intersection.
Curve: x = t; y = 2t3; z = t2; plane: 6x2 - z2 = 1.

Solution.

Let's find the coordinates of the points of intersection, then we'll plug in the parametric expressions for each curve in the function of the plane.

\[6t^{2} -(t^{2} )^{2} =1\] \[\begin{array}{l} {t^{4} -6t^{2} +1=0} \\ {t^{2} =\frac{6\pm \sqrt{36-4} }{2} =\frac{6\pm 4\sqrt{2} }{2} =3\pm 2\sqrt{2} } \end{array}\] \[\begin{array}{l} {t_{1} =+\sqrt{3-2\sqrt{2} } =\sqrt{2} -1;} \\ {t_{2} =-\sqrt{3-2\sqrt{2} } =1-\sqrt{2} } \\ {t_{3} =\sqrt{3+2\sqrt{2} } =\sqrt{2} +1} \\ {t_{4} =-\sqrt{3+2\sqrt{2} } =-1-\sqrt{2} } \end{array}\]

We get 4 points of intersection:

\[\begin{array}{l} {M_{1} (\sqrt{2} -1;10\sqrt{2} -14;3-2\sqrt{2} )} \\ {M_{2} (1-\sqrt{2} ;14-10\sqrt{2} ;3-2\sqrt{2} )} \\ {M_{3} (\sqrt{2} +1;14+10\sqrt{2} ;3+2\sqrt{2} )} \\ {M_{4} (-\sqrt{2} -1;-14-10\sqrt{2} ;3+2\sqrt{2} )} \end{array}\]

Now let's find the tangential vector to the curve at these points

\[x'=1;\quad y'=6t^{2} ;\quad z'=2t\] \[\begin{array}{l} {\bar{l}_{1} =(1;18-12\sqrt{2} ;2\sqrt{2} -2)} \\ {\bar{l}_{2} =(1;18-12\sqrt{2} ;2-2\sqrt{2} )} \\ {\bar{l}_{3} =(1;18+12\sqrt{2} ;2\sqrt{2} +2)} \\ {\bar{l}_{4} =(1;18+12\sqrt{2} ;-2-2\sqrt{2} )} \end{array}\]

Let's build the function for the tangential vector to the plane at the determined points

$$\begin{array}{l} {F(x,y,z)=6x^{2} -z^{2} -1=0} \\ {F'_{x} =12x;\quad F'_{y} =0;\quad F'_{z} =-2z} \end{array}$$

At this point

$$M_{1} (\sqrt{2} -1;10\sqrt{2} -14;3-2\sqrt{2} )$$ the tangent to the plane is: $$\begin{array}{l} {(12\sqrt{2} -12)(x-\sqrt{2} +1)+(4\sqrt{2} -6)(z+2\sqrt{2} -3)=0} \\ {(12\sqrt{2} -12)x+(4\sqrt{2} -6)z-2=0} \\ {(6\sqrt{2} -6)x+(2\sqrt{2} -3)z-1=0} \\ {\bar{n}_{1} =(6\sqrt{2} -6;0;2\sqrt{2} -3)} \end{array}$$

At this point

$$M_{2} (1-\sqrt{2} ;14-10\sqrt{2} ;3-2\sqrt{2} )$$ the tangent to the plane is: \[\begin{array}{l} {(12-12\sqrt{2} )(x+\sqrt{2} -1)+(4\sqrt{2} -6)(z+2\sqrt{2} -3)=0} \\ {(6-6\sqrt{2} )x+(2\sqrt{2} -3)z-1=0} \end{array}\] \[\bar{n}_{2} =(6-6\sqrt{2} ;0;2\sqrt{2} -3)\]

At this point

$$M_{3} (\sqrt{2} +1;14+10\sqrt{2} ;3+2\sqrt{2} )$$

the tangent to the plane is:

\[\begin{array}{l} {(12\sqrt{2} +12)(x-\sqrt{2} -1)+(-6-12\sqrt{2} )(z-3-2\sqrt{2} )=0} \\ {(12\sqrt{2} +12)x-(6+12\sqrt{2} )z+30+24\sqrt{2} =0} \\ {(2\sqrt{2} +2)x-(1+2\sqrt{2} )z+5+4\sqrt{2} =0} \end{array}\] \[\bar{n}_{3} =(2\sqrt{2} +2;0;-2\sqrt{2} -1)\]

At this point

$$M_{4} (-\sqrt{2} -1;-14-10\sqrt{2} ;3+2\sqrt{2} )$$

the tangent to the plane is:

\[\begin{array}{l} {-(12\sqrt{2} +12)(x+\sqrt{2} +1)+(-6-12\sqrt{2} )(z-3-2\sqrt{2} )=0} \\ {-(12\sqrt{2} +12)x-(6+12\sqrt{2} )z+30+24\sqrt{2} =0} \\ {-(2\sqrt{2} +2)x-(1+2\sqrt{2} )z+5+4\sqrt{2} =0} \end{array}\] \[\bar{n}_{4} =(-2\sqrt{2} -2;0;-2\sqrt{2} -1)\]

The angle between the curve and the plane is considered to be the angle between the tangent to the line and the tangent to the plane. It can be calculated by the scalar of the vector derivate perpendicular to the plane and the vector tangent to the line.

\[\begin{array}{l} {\bar{l}_{1} =(1;18-12\sqrt{2} ;2\sqrt{2} -2)} \\ {\bar{l}_{2} =(1;18-12\sqrt{2} ;2-2\sqrt{2} )} \\ {\bar{l}_{3} =(1;18+12\sqrt{2} ;2\sqrt{2} +2)} \\ {\bar{l}_{4} =(1;18+12\sqrt{2} ;-2-2\sqrt{2} )} \end{array}\]

\[\begin{array}{l} {\alpha _{1} =\arcsin \frac{\bar{l}_{1} \bar{n}_{1} }{\left|\bar{l}_{1} \right|\left|\bar{n}_{1} \right|} =\arcsin \frac{6\sqrt{2} -6+0+(2\sqrt{2} -2)(2\sqrt{2} -3)}{\sqrt{1+(18-12\sqrt{2} )^{2} +(2\sqrt{2}-2)^{2} } \sqrt{(6\sqrt{2}-6)^{2} +0+(2\sqrt{2} -3)^{2} } } = \arcsin \frac{8-4\sqrt{2} }{\sqrt{625-440\sqrt{2} } \sqrt{125-84\sqrt{2} } } \approx \arcsin (0.4049)\approx 0.42(rad)} \end{array}\]

\[\begin{array}{l} {\alpha _{2} =\arcsin \frac{\bar{l}_{2} \bar{n}_{2} }{\left|\bar{l}_{2} \right|\left|\bar{n}_{2} \right|} =\arcsin \frac{6-6\sqrt{2} +0-(2\sqrt{2} -2)(2\sqrt{2} -3)}{\sqrt{1+(18-12\sqrt{2} )^{2} +(2-2\sqrt{2})^{2} } \sqrt{(6-6\sqrt{2})^{2} +0+(2\sqrt{2} -3)^{2} } } = \arcsin \frac{8-4\sqrt{2} }{\sqrt{625-440\sqrt{2} } \sqrt{125-84\sqrt{2} } } \approx \arcsin (-0.4049)\approx -0.42(rad)} \end{array}\]

\[\begin{array}{l} {\alpha _{3} =\arcsin \frac{\bar{l}_{3} \bar{n}_{3} }{\left|\bar{l}_{3} \right|\left|\bar{n}_{3} \right|} =\arcsin \frac{2\sqrt{2}+2+0-(2\sqrt{2} +1)(2\sqrt{2} +2)}{\sqrt{1+(18+12\sqrt{2} )^{2} +(2+2\sqrt{2})^{2} } \sqrt{(2\sqrt{2} +2)^{2} +0+(2\sqrt{2} +1)^{2} } } = \arcsin \frac{-8-4\sqrt{2} }{\sqrt{625+440\sqrt{2} } \sqrt{21+12\sqrt{2} } } \approx \arcsin (-0.0628)\approx -0.06(rad)} \end{array}\]

\[\begin{array}{l} {\alpha _{4} =\arcsin \frac{\bar{l}_{4} \bar{n}_{4} }{\left|\bar{l}_{4} \right|\left|\bar{n}_{4} \right|} =\arcsin \frac{-2\sqrt{2}-2+0+(2\sqrt{2} +1)(2\sqrt{2} +2)}{\sqrt{1+(18+12\sqrt{2} )^{2} +(2+2\sqrt{2})^{2} } \sqrt{(2\sqrt{2} +2)^{2} +0+(2\sqrt{2} +1)^{2} } } = \arcsin \frac{8+4\sqrt{2} }{\sqrt{625+440\sqrt{2} } \sqrt{21+12\sqrt{2} } } \approx \arcsin (0.0628)\approx 0.06(rad)} \end{array}\]

Problem 3.

For the linear one-dimensional figure, bounded by the given curves, find the value and the position of the center of mass.

y = 0; y = 2x; y = 3 - x2

Solution.

Let's build the figure D in the Cartesian coordinate system.

 math2 3

The bounds are:

\[0\le y\le 2;\frac{y}{2} \le x\le \sqrt{3-y} \]

The mass of the shaded cross-section:

\[\begin{array}{l} {M=\mathop{\int\!\!\!\!\int}\nolimits _{D}\mu (x,y)dxdy =\int _{0}^{2}dy \int _{y/2}^{\sqrt{3-y} }1\cdot dx =\int _{0}^{2}dy \int _{y/2}^{\sqrt{3-y} }dx =\int _{0}^{2}(\sqrt{3-y} -\frac{y}{2} )dy =-\frac{2}{3} \left. \sqrt{(3-y)^{3} } \right|_{0}^{2} -\left. \frac{y^{2} }{4} \right|_{0}^{2} =} \\ {=-\frac{2}{3} +\frac{2}{3} \sqrt{27} -1+0=2\sqrt{3} -\frac{2}{3} \approx 2.80(mass)} \end{array}\]

The coordinates of the center of gravity:

\[\begin{array}{l} {x_{C} =\frac{M_{x} }{M} =\frac{\mathop{\int\!\!\!\!\int}\nolimits _{D}xdxdy }{M} =\frac{\int _{0}^{2}dy \int _{y/2}^{\sqrt{3-y} }x\cdot dx }{M} =\frac{\int _{0}^{2}(3y-\frac{y^{2} }{4} )dy }{2M} =\frac{1}{2M} \left. (\frac{3y^{2} }{2} -\frac{y^{3} }{12} )\right|_{0}^{2} =} \\ {=\frac{1}{2M} (6-0-\frac{2}{3} +0)=\frac{5.3333}{5.6} \approx 0.95} \end{array}\] \[\begin{array}{l} {y_{C} =\frac{M_{y} }{M} =\frac{\mathop{\int\!\!\!\!\int}\nolimits _{D}ydxdy }{M} =\frac{\int _{0}^{2}ydy \int _{y/2}^{\sqrt{3-y} }dx }{M} =\frac{\int _{0}^{2}y(\sqrt{3-y} -\frac{y}{2} )dy }{M} =\frac{1}{M} ((-\left. \frac{y^{3} }{6} )\right|_{0}^{2} +\int _{0}^{2}y\sqrt{3-y} dy )=} \\ {=\frac{1}{M} (-\frac{4}{3} +0+\int _{0}^{2}y\sqrt{3-y} dy )=\left\{\begin{array}{c} {3-y=t^{2} } \\ {dy=-2tdt} \end{array}\right\}=-\frac{4}{3M} +\frac{1}{M} \int _{\sqrt{3} }^{1}(2t^{4} -6t^{2} )dt =} \\ {=-\frac{4}{3M} +\frac{1}{M} \left. (\frac{2t^{5} }{5} -2t^{3} )\right|_{\sqrt{3} }^{1} =-\frac{4}{3M} +\frac{1}{M} (\frac{2}{5} -2-\frac{18\sqrt{3} }{5} +6\sqrt{3} )=-\frac{4}{3M} -\frac{8}{5M} +\frac{12\sqrt{3} }{5M} =} \\ {=\frac{-44+36\sqrt{3} }{15M} \approx \frac{112.99}{42} \approx 0.44} \end{array}\]

Problem 4.

For the one-dimensional figure bounded by the given planes, find the mass and the moment of inertia relative to the z-axis.

z = 0, 2z = y2, x - y = 0, x + y = 9

Solution.

math2 4

First we'll build the figure in the Cartesian coordinate system and will determine its bounds.

\[V:\left\{\begin{array}{c} {0\le y\le 9/2} \\ {y\le x\le 9-y} \\ {0\le z\le y^{2} /2} \end{array}\right. \]

Since the figure is uniform, we can determine the momentum of intertia Iz using the formula

\[\begin{array}{l} {I_{z} =\mathop{\int\!\!\!\!\int\!\!\!\!\int}\nolimits _{V}(x^{2} +y^{2} )dxdydz =\int _{0}^{9/2}dy \int _{y}^{9-y}dx \int _{0}^{y^{2} /2}(x^{2} +y^{2} )dz =\int _{0}^{9/2}dy \int _{y}^{9-y}dx\cdot \left. (x^{2} +y^{2} )z\right| _{0}^{y^{2} /2} =} \\ {=\int _{0}^{9/2}dy \int _{y}^{9-y}(x^{2} \frac{y^{2} }{2} +\frac{y^{4} }{2} )dx=\int _{0}^{9/2}dy \left. \left(\frac{y^{2} x^{3} }{6} +\frac{y^{5} x^{2} }{2} \right)\right|_{y}^{9-y} =} \\ {=\int _{0}^{9/2}(\frac{y^{2} (9-y)^{3} }{6} +\frac{y^{5} (9-y)^{2} }{2} -\frac{y^{5} }{6} -\frac{y^{7} }{2} )dy =} \\ {=\int _{0}^{9/2}(\frac{y^{2} (729-y^{3} +27y^{2} -243y)}{6} +\frac{y^{5} (81+y^{2} -18y)}{2} -\frac{y^{5} }{6} -\frac{y^{7} }{2} )dy =} \\ {=\int _{0}^{9/2}(\frac{729y^{2} -y^{5} +27y^{4} -243y^{3} }{6} +\frac{81y^{5} +y^{7} -18y^{6} }{2} -\frac{y^{5} }{6} -\frac{y^{7} }{2} )dy =} \\ {=\int _{0}^{9/2}(\frac{243y^{2} }{2} -\frac{121y^{5} }{3} +\frac{9y^{4} }{2} -\frac{81y^{3} }{2} +\frac{y^{7} }{2} -9y^{6} -\frac{y^{5} }{6} -\frac{y^{7} }{2} )dy =} \\ {=\int _{0}^{9/2}(\frac{243y^{2} }{2} -\frac{81y^{5} }{2} +\frac{9y^{4} }{2} -\frac{81y^{3} }{2} -9y^{6} )dy } \\ {=\left. \left(\frac{243\cdot y^{3} }{6} -\frac{81\cdot y^{6} }{6} +\frac{9\cdot y^{5} }{10} -\frac{81\cdot y^{4} }{8} -\frac{9y^{7} }{7} \right)\right|_{0}^{9/2} =} \\ {=\frac{81\cdot 91.125}{2} -\frac{27\cdot {\rm 8303,765625}}{2} +{\rm 1660,753125}-\frac{81\cdot {\rm 410,0625}}{8} -\frac{9\cdot {\rm 37366,9453125}}{7} =} \\ {={\rm 3690,5625-112100,8359375}+{\rm 1660,753125-4151,8828125-48043,215402}=} \\ {={\rm -158944,6185}} \end{array}\]

Problem 5.

Given the vector field F and the plane p, which with the coordinate planes creates a pyramid. Additionally σ, the boundary of the pyramid, is part of the mentioned plane , and λ, the contour, that bounds σ. Find the flux F through the surface σ, the flux F through the surface of the pyramid itself and using the formula of Ostrogradskii, and circulation of F along the contour λ itself and directly using Stokes' formula. The normal to the surface of the pyramid is external, the direction of the contour corresponding to the right hand rule.

Field: F = (y - x + z)i, plane -x + 2y + z + 4 = 0

Solution.

Let's rewrite the equation of the plane in segments and build the pyramid. \[\frac{x}{4} +\frac{y}{-2} +\frac{z}{-4} =1\]

math2 5

The normal to the plane (unit vector)

\[\bar{n}=(\frac{1}{4} \bar{i}-\frac{1}{2} \bar{j}-\frac{1}{4} \bar{k})/(\sqrt{3} /2\sqrt{2} )=\frac{\sqrt{2} }{2\sqrt{3} } \bar{i}-\frac{\sqrt{2} }{\sqrt{3} } \bar{j}-\frac{\sqrt{2} }{2\sqrt{3} } \bar{k}\]

a) Let's find the immediate flux of the vector field through the surfuce.

\[\frac{x}{4} +\frac{y}{-2} +\frac{z}{-4} =1\] \[\bar{n}=(\frac{1}{4} \bar{i}-\frac{1}{2} \bar{j}-\frac{1}{4} \bar{k})/(\sqrt{3} /2\sqrt{2} )=\frac{\sqrt{2} }{2\sqrt{3} } \bar{i}-\frac{\sqrt{2} }{\sqrt{3} } \bar{j}-\frac{\sqrt{2} }{2\sqrt{3} } \bar{k}\] \[z=-4+x-2y;\quad z'_{x} =1;z'_{y} =-2\] \[\begin{array}{l} {{\rm \; }\Phi _{p} {\rm (}\bar{{\rm F}}{\rm )=}\mathop{\int\!\!\!\!\int}\nolimits _{\sigma }\bar{{\rm F}}\cdot \bar{n}ds =\mathop{\int\!\!\!\!\int}\nolimits _{p}\frac{\sqrt{2} }{2\sqrt{3} } (y-x+z)ds =\frac{\sqrt{2} }{2\sqrt{3} } \mathop{\int\!\!\!\!\int}\nolimits _{p}(y-x+z)ds =} \\ {=\frac{\sqrt{2} }{2\sqrt{3} } \int _{-2}^{0}dx \int _{\frac{1}{2} x-2}^{0}(y-x+z)\cdot \sqrt{1+(z'_{x} )^{2} +(z'_{y} )^{2} } dy=\frac{\sqrt{2} }{2\sqrt{3} } \int _{-2}^{0}dx \int _{\frac{1}{2} x-2}^{0}(y-x-4+x-2y)\cdot \sqrt{6} dy=} \\ {=\int _{-2}^{0}dx \int _{\frac{1}{2} x-2}^{0}(-y-4) dy=\int _{-2}^{0}dx \cdot \left. (-\frac{y^{2} }{2} -4y)\right|_{\frac{1}{2} x-2}^{0} =\int _{-2}^{0}dx \cdot (0+\frac{(\frac{1}{2} x-2)^{2} }{2} +4(\frac{1}{2} x-2))=} \\ {=\int _{-2}^{0}(\frac{x^{2} }{4} -x+2+2x-8)dx =\int _{-2}^{0}(\frac{x^{2} }{4} +x-6)dx =\left. \left(\frac{x^{3} }{12} -6x+\frac{x^{2} }{2} \right)\right|_{-2}^{0} =0-(\frac{-8}{12} +12+2)=-\frac{40}{3} } \end{array}\]

b) Let's find the immediate flux of the vector field

$$\bar{F}=(y-x+z)\bar{i}$$ through the surface $$S1:z=0$$ Normal (orth) $$\bar{n}=(0,0,-1)$$ \[\Phi _{s1} (\bar{F})=\mathop{\int\!\!\!\!\int}\nolimits _{S1}\bar{F}\cdot \bar{n}ds =-\mathop{\int\!\!\!\!\int}\nolimits _{s1}0ds =0\]

c) Let's find the immediate flux of the vector field through the surfuce $$S2:x=0$$

Normal (orth) $$\bar{n}=(-1,0,0)$$ \[\begin{array}{l} {{\rm \; }\Phi _{S2} {\rm (}\bar{{\rm F}}{\rm )=}\mathop{\int\!\!\!\!\int}\nolimits _{S2}\bar{{\rm F}}\cdot \bar{n}ds =-\mathop{\int\!\!\!\!\int}\nolimits _{S2}(y-x+z)ds =-\int _{-2}^{0}dy \int _{-4-2y}^{0}(y+z)dz =-\int _{-2}^{0}dy \left. (yz+\frac{z^{2} }{2} )\right|_{-4-2y}^{0} =} \\ {=-\int _{-2}^{0}\left(0-y(-4-2y)-\frac{1}{2} (-4-2y)^{2} \right)dy =-\int _{-2}^{0}\left(+4y+2y^{2} -2(4+4y+y^{2} )\right)dy =} \\ {=-\int _{-2}^{0}\left(4y+2y^{2} -8-8y-2y^{2} \right)dy =-\int _{-2}^{0}\left(-4y-8\right)dy =\int _{-2}^{0}\left(4y+8\right)dy =\left. (2y^{2} +8y)\right|_{-2}^{0} =} \\ {=0-(8-16)=8} \end{array}\]

d) Let's find the immediate flux of the vector field through the surfuce $$S3:y=0$$ \[\bar{n}=(0,-1,0)\] \[\Phi _{s3} (\bar{F})=\mathop{\int\!\!\!\!\int}\nolimits _{S3}\bar{F}\cdot \bar{n}ds =\mathop{\int\!\!\!\!\int}\nolimits _{s3}0ds =0\] In order to find the calculated flux through the enclosed surface σ, let's use the ratio \[\Phi _{\sigma } (\bar{F})=\Phi _{s1} (\bar{F})+\Phi _{s2} (\bar{F})+\Phi _{s3} (\bar{F})+\Phi _{p} (\bar{F})=0+8+0-\frac{40}{3} =\frac{24-40}{3} =-\frac{16}{3} \] Field components: \[\begin{array}{l} {P=y-x+z;\quad \frac{\partial P}{\partial x} =-1;\frac{\partial P}{\partial y} =1;\frac{\partial P}{\partial z} =1} \\ {Q=0;\; \frac{\partial Q}{\partial y} =0;\quad R=0;\quad \frac{\partial R}{\partial z} =0} \end{array}\] Divergence $$div\bar{F}=\frac{\partial P}{\partial x} +\frac{\partial Q}{\partial y} +\frac{\partial R}{\partial z} =-1$$ \[\frac{\partial R}{\partial y} -\frac{\partial Q}{\partial z} =0;\frac{\partial P}{\partial z} -\frac{\partial R}{\partial x} =1;\frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y} =-1\] Curl $$curl\bar{F}=\bar{j}-\bar{k}$$

The flux of the vector field over the entire surface using Ostrogradskii's formula

$$\mathop{\int\!\!\!\!\int\!\!\!\!\int}\nolimits _{V}div\bar{F}dxdydz =-\mathop{\int\!\!\!\!\int\!\!\!\!\int}\nolimits _{V}1\cdot dxdydz =-\mathop{\int\!\!\!\!\int\!\!\!\!\int}\nolimits _{V}dxdydz =-V_{OABC} =-\frac{1}{3} \cdot \frac{1}{2} \cdot 2\cdot 4\cdot 4=-\frac{16}{3} $$

Circulation of the vector field F on the contour λ equals the linear integral

\[\oint _{\lambda }\bar{F}d\bar{r} =\oint _{\lambda }(F_{x} dx+F_{y} dy+F_{z} dz)= \oint _{\lambda }(y-x+z)dx \] \[\begin{array}{l} {\lambda _{1} :y=\frac{1}{2} x-2;z=0\quad ;dy=\frac{dx}{2} ;dz=0;\quad x_{1} =4;x_{2} =0} \\ {\lambda _{2} :z=x-4;dz=dx;y=0;dy=0;0\le x\le 4} \\ {\lambda _{3} :z=-2y-4;dz=-2dy;x=0;dx=0;-2\le y\le 0} \end{array}\] \[\begin{array}{l} {\oint _{\lambda }(y-x+z)dx =\int _{\lambda _{1} }(y-x+z)dx +\int _{\lambda _{2} }(y-x+z)dx +\int _{\lambda _{3} }(y-x+z)dx =} \\ {=\int _{4}^{0}(\frac{1}{2} x-2-x+0)dx +\int _{0}^{4}(0-x+x-4)dx +0=} \\ {=\int _{4}^{0}(-\frac{1}{2} x-2)dx -4\int _{0}^{4}dx =\frac{1}{2} \int _{0}^{4}(x+4)dx -4\int _{0}^{4}dx =\frac{1}{2} \left. \left(\frac{x^{2} }{2} +4x\right)\right|_{0}^{4} -16=} \\ {=\frac{1}{2} (8+16)-16=12-16=-4} \end{array}\]

Normal to the plane (unit vector)

\[\bar{n}=(\frac{1}{4} \bar{i}-\frac{1}{2} \bar{j}-\frac{1}{4} \bar{k})/(\sqrt{3} /2\sqrt{2} )=\frac{\sqrt{2} }{2\sqrt{3} } \bar{i}-\frac{\sqrt{2} }{\sqrt{3} } \bar{j}-\frac{\sqrt{2} }{2\sqrt{3} } \bar{k}\] \[curl\bar{F}=\bar{j}-\bar{k}\]

Using Stokes' theorem:

\[\mathop{\int\!\!\!\!\int}\nolimits _{\sigma }\bar{n}curl\bar{F}d\sigma = \mathop{\int\!\!\!\!\int}\nolimits _{\sigma }(\bar{j}-\bar{k})(\frac{\sqrt{2} }{2\sqrt{3} } \bar{i}-\frac{\sqrt{2} }{\sqrt{3} } \bar{j}-\frac{\sqrt{2} }{2\sqrt{3} } \bar{k})d\sigma = \mathop{\int\!\!\!\!\int}\nolimits _{\sigma }(-\frac{\sqrt{2} }{\sqrt{3} } +\frac{\sqrt{2} }{2\sqrt{3} } )d\sigma =-\frac{\sqrt{2} }{2\sqrt{3} } S_{\sigma } \]

The flat figure σ is an isoscles triangle ABC with sides:

\[AB=\sqrt{2^{2} +4^{2} } =\sqrt{20} =2\sqrt{5} ;\quad BC=4\sqrt{2} ;\quad AC=\sqrt{2^{2} +4^{2} } =\sqrt{20} =2\sqrt{5} . \] Then the height $$h=\sqrt{20-8} =\sqrt{12} =2\sqrt{3} $$ $$S_{\sigma } =\frac{1}{2} BC\cdot h=\frac{1}{2} 4\sqrt{2} \cdot 2\sqrt{3} =4\sqrt{6} $$ \[-\frac{\sqrt{2} }{2\sqrt{3} } S_{\sigma } =-\frac{\sqrt{2} }{2\sqrt{3} } \cdot 4\sqrt{6} =-4\]