Higher_Mathematics_Part_2
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8.3.12. |
x = 3 cos t, y = 8 sin t, y = 4 ( y ≥ 4 ). |
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8.3.13. |
x = 6(t − sin t), y = 6(1 − cost), |
y = 0 , y = 6 ( 0 ≤ y ≤ 6, 0 ≤ x ≤ 2π ). |
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8.3.14. |
x = 8 cos3 t, y = 4 sin3 t, x = 3 |
3 ( x ≥ 3 3 ). |
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8.3.15. |
x = 6 cos t, y = 4 sin t, y = 2 |
3 ( y ≥ 2 3 ). |
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8.3.16. |
x = 10(t − sin t), y = 10(1− cos t), y = 15 (15 ≤ y , 0 ≤ x ≤ 6π ) . |
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8.3.17. |
x = 2 2 cos3 t, y = |
2 sin3 t, |
x = −1 ( x ≤ −1). |
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8.3.18. |
x = |
2 cos t, y = 4 |
2 sin t, y = −4 ( y ≤ −4 ). |
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8.3.19. |
y = t − sin t, x = 1− cos t, x = 1 (1 ≤ x , 0 ≤ y ≤ 2π ). |
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8.3.20. |
x = 9 cos t, y = 4 sin t, y = 2 ( y ≥ 2 ). |
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8.3.21. |
x = 8(t − sin t), y = 8(1− cos t), |
y = 12 (12 ≤ y , 0 ≤ x ≤ 6π ). |
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8.3.22. |
x = 24 sin3 t, y = 2 cos3 t, x = 9 |
3 ( x ≥ 9 3 ). |
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8.3.23. |
x = 2(t − sin t), y = 2(1 − cost), |
y = 2 , y = 3 ( 2 ≤ y ≤ 3 , 0 ≤ x ≤ 4π ). |
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8.3.24. |
x = 4 2 cos3 t, y = |
2 sin3 t, |
x = −2 ( x ≥ −2 ). |
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8.3.25. |
x = 2 |
2 cos t, y = 5 |
2 sin t, |
y = −5 ( y ≥ −5 ). |
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8.3.26. |
x = 4(t − sin t), y = 4(1− cos t), |
y = 6 ( 6 ≤ y , 0 ≤ x ≤ 4π ). |
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8.3.27. |
x = 8 cos3 t, y = 2 sin3 t, x = −1 ( x ≤ −1). |
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8.3.28. |
x = 2 cos t, y = 3sin t, x = −1 , x = 1 ( −1 ≤ x ≤ 1). |
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8.3.29. x = 8(t − sin t), y = 8(1 − cost), y = 4 , y = 12 ( 4 ≤ y ≤ 12 , 0 ≤ x ≤ 2π ).
8.3.30. x = 4 cos3 t, y = 8 sin3 t, y = 1 , y = 3
3 (1 ≤ y ≤ 3
3 ).
8.4. Find the arc length.
8.4.1. y = 
1 − x2 + arcsin x , x [0;1] .
8.4.2. y = ln |
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, x [ 3; 8] . |
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8.4.3. y = (x2 |
− 2 ln x) / 4 , x [1; 2] . |
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8.4.4. y = ln x , x [
3; 
15] .
8.4.5. y = 1 − x2 |
+ arccos x , x [0; 8 / 9] . |
8.4.6. y = 2 + ch x , |
x [0;1] . |
8.4.7. y = e x − 1 , |
x [ln 8; ln 15] . |
8.4.8. y = e2x + e−2x , x [0; 2] .
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8.4.9.y = ln sin x , x [π / 3; π / 2] .
8.4.10.y = 3 − ch x , x [0;1] .
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8.4.11. y = 1 + ln cos x , x [0; π / 3] .
8.4.12. y = arcsin e− x , x [0;1] .
8.4.13. y = ln sin 2x , x [π / 6; π / 3] . 2
8.4.14.y = ln 7x , x [
15; 
24] .
8.4.15.y = ln(1 − x2 ) , x [0;1/ 4] .
8.4.16. y = |
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2x |
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8.4.17. y = x − |
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x3 |
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8.4.18.y = ln(x2 − 1) , x [3; 4] .
8.4.19.y = − ln cos x, x [0; π / 6] .
8.4.20.y = x4 + 1 , x [1; 2] . 32 x2
8.4.21.y = arcsin x − 
1 − x2 , x [0;15 /16] .
8.4.22.y = − arccos 
x + 
x − x2 , x [0;1/ 4] .
8.4.23. y = 1 + arcsin x − 
1 − x2 , x [0; 3 / 4] .
=x + sin 2x − π
8.4.24.y tg x , x [0; / 4] . 8 16
8.4.25.y = − arccos 
x + 
x − x2 + 2 , x [1/ 9;1] .
8.4.26.y = 1 − arccos x + 
1 − x2 , x [0; 9 /16] .
8.4.27.y = 1− x2 , x [0; 8 / 9] .
8.4.28. y = |
x2 |
ln x − |
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ln ln x |
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x [e; e2 ] . |
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8.4.29. y = 4 arccos x − |
x − x2 |
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8.4.30. y = 3 − e x , |
x [ln 15; ln |
24] . |
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8.5. Find the arc length. |
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8.5.1. x = et (cos t + sin t), y = et (cos t − sin t), |
0 ≤ t ≤ 2π . |
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8.5.2. x = 2(cos t + t sin t), y = 2(sin t − t cos t), |
0 ≤ t ≤ π / 2 . |
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8.5.3. x = 4(2 cos t − cos 2t), y = 4(2 sin t − sin 2t), 0 ≤ t ≤ π .
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8.5.4. |
x = (t2 − 2) sin t + 2t cos t, y = (2 − t2 ) cos t + 2t sin t, 0 ≤ t ≤ π . |
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8.5.5. |
x = |
1+ t2 , y = |
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t |
1+ t2 − |
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ln(t + |
1+ t2 ), 0 ≤ t ≤ 1 . |
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8.5.6. |
x = |
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t3 ln t − |
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t3 , y = t ln t − t, |
1 ≤ t ≤ 3 . |
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8.5.7. x = 2 cos t − cos 2t, y = 2 sin t − sin 2t, 0 ≤ t ≤ 2π . |
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8.5.8. x = ln(1+ sin t) − ln cos t, y = ln(1− cos t) − ln sin t, π / 6 ≤ t ≤ π / 3 .
8.5.9. x = tet (sin t − cost) + et cost, y = tet (cost + sin t) − et sin t, |
0 ≤ t ≤ 2π . |
8.5.10. x = (1 − 2t2 )cos 2t + 2t sin 2t, y = (2t2 − 1)sin 2t + 2t cos 2t, |
0 ≤ t ≤ 2π . |
8.5.11. x = 5(t − sin t), y = 5(1− cos t), 0 ≤ t ≤ π . |
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8.5.12. x = 2t − sin 2t, y = 2 sin2 t, 0 ≤ t ≤ π . |
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8.5.13. x = 2 arctg t, y = ln(1+ t2 ), 0 ≤ t ≤ 1 .
8.5.14.x = cos t + t sin t, y = sin t − t cos t, 0 ≤ t ≤ 2π .
8.5.15.x = et sin(t + π / 4), y = et cos(t + π / 4), 0 ≤ t ≤ π .
8.5.16.x = 3(t − sin t), y = 3(1− cos t), π ≤ t ≤ 3π .
8.5.17.x = 2 cos3 t, y = 2 sin3 t, 0 ≤ t ≤ π / 3 .
8.5.18.x = 6 cos3 t, y = 6 sin3 t, π / 2 ≤ t ≤ π .
8.5.19.x = et cos(t − π / 4), y = et sin(π / 4 − t), 0 ≤ t ≤ π .
8.5.20.x = 4 cos3 t, y = 4 sin3 t, π ≤ t ≤ 2π .
8.5.21.x = 2 cos t − cos 2t, y = 2 sin t − sin 2t, 0 ≤ t ≤ π / 3 .
8.5.22. |
x = (t2 − 2) sin t + 2t cos t, y = (2 − t2 ) cos t + 2t sin t, 0 ≤ t ≤ π . |
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8.5.23. |
x = 2(cos t + t sin t), y = 2(sin t − t cos t), |
0 ≤ t ≤ π . |
8.5.24. |
x = t cos t, y = t sin t, 0 ≤ t ≤ π . |
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8.5.25. |
x = (t2 − 2)sin t + 2t cost, y = (2 − t2 )cost + 2t sin t, π / 2 ≤ t ≤ 2π . |
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8.5.26. |
x = ln(1+ sin 2t) − ln cos 2t, y = ln tg t, |
π / 8 ≤ t ≤ π / 6 . |
8.5.27.x = t cos t, y = t sin t, π / 2 ≤ t ≤ π .
8.5.28.x = 7(t − sin t), y = 7(1− cos t), 0 ≤ t ≤ 4π .
8.5.29.x = 2t − sin 2t, y = 2 cos2 t, 0 ≤ t ≤ 2π .
8.5.30. x = 8 sin3 t, y = 8 cos3 t, 0 ≤ t ≤ π .
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Main concepts |
Т.1 |
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1.1. Definitions and elementary problems
An equation involving differentials or derivatives in its expression is a differential equation. If the equation contains total differentials, total derivatives, or both, but does not contain partial derivatives, it is called an ordinary differential equation; if it contains partial derivatives, it is called a partial differential equation.
The order of a differential equation is the order of the highest-ordered derivative involved in its expression. For instance, the differential equation
y′′ + xy′ + 2y 4 = 0 is of the first order.
The degree of a differential equation is the degree that its highest-ordered derivative would have if the equation were rationalized and cleared of fractions with regard to all derivatives involved in it.
A solution, or integral, of an ordinary differential equation in two variables is a relation between the variables which satisfies the equation.
If, in particular, the solution has the form y = f(x), then replacement of y and the derivatives of y with respect to x by f(x) and its derivatives produces an identity.
The general solution of the first-order differential equation y′ = f (x, y) or F(x, y, y′) = 0 is a function y = ϕ(x, C) , where C represents an arbitrary
constant.
A particular solution to the differential equation is a function y= ϕ(x, C0) such that y(x0 ) = y0
1.2. Variables separable
Only certain types of differential equations can be solved in terms of ordinary functions, and generally each type calls for a special method of integration.
We shall first consider a differential equation of the first order and first degree, which can easily be reduced to the form
M (x)dx + N ( y)dy = 0 , |
(3.1) |
Where M(x) is a function of x alone and N(y) is a function of y alone. This type is referred to as variables separable. Direct integration of equation (3.1) gives the general solution
∫ M (x)dx + ∫ N ( y)dy = C , |
(3.2) |
Where C is an arbitrary constant. It is evident that equation (3.1) would result from differentiating equation (3.2), even if we should write any function of C instead of C.
A differential equation
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Either of the substitutions, y = xu(x), where u(x) is the new unknown function, will reduce any homogeneous differential equation of the first order and first degree to the type of variables separable
For the equations of the form
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a x |
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y′ = |
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a2 x |
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t and v are |
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we could have used the substitutions x = t + α , |
y = v + β , where |
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the new variables, α and β are solutions of the system |
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a1 α+ b1 β+ c1 = |
0, |
(3.4) |
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α+ b2 β+ c2 |
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a2 |
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such circumstances |
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substitution |
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z = a1x + b1 y + c1 transforms the given expression to homogeneous differential equation.
1.4. Linear differential equation. Equations reducible to linear form
A differential equation M (x, y)dx + N(x, y)dy = 0 of any order is said to be
linear when it is of the first degree in the dependent variable and its derivatives. It follows that the general type of a linear differential equation of the first order is
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y′ + P(x) y = Q(x), |
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(3.5) |
where P and Q are functions of x only. |
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To find an integrating factor of (3.5) let us solve |
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y′ + P(x) y = 0 . |
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(3.6) |
Here the variables are separated and the solution is |
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dy |
dy |
∫ |
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= −∫ P(x)dx ; |
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dx = − P(x) y ; |
y = − P(x)dx ; |
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ln y = −∫ P(x)dx + ln C .
Finally,
y = Ce− ∫ P(x)dx .
Thereafter, to solve an equation having the form (3.5) two following methods by Bernoulli and by Lagrange are used.
Bernoulli’s method. Solution of (3.5) is defined as the product of the two unknown functions u(x) and v(x) , namely
y(x) = u(x)v(x).
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Substituting in (3.5), we get
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v + u |
dv |
+ P(x)uv = Q(x) , |
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or |
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du |
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+ P(x)u v + u |
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Define u(x) such as the expression in a parenthesis be equal to zero, then
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+ P(x)u = 0, |
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The first equation has the following particular solution (C = 0):
u(x) = e− ∫P(x)dx .
The second equation of (3.7) has the form (3.6), hence
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∫ |
P(x)dx |
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= Q(x)e∫ |
P(x)dx |
, v = ∫Q(x)e∫ |
P(x)dx |
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dx + C. |
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−∫ |
P(x)dx |
−∫ P(x)dx |
∫ Q(x)e |
∫ P(x)dx |
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y =uv =Ce |
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Lagrange’s method. Solution of (3.5) is defined in the form |
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y = C(x)e− ∫P(x)dx . |
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(3.9) |
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This has the form (3.6) if C is a constant. If C(x) is a new function substituting (3.9) in (3.5), we get C′(x) = Q(x)e∫P(x)dx . Therefore,
C(x) = ∫ Q(x)e |
∫ P(x)dx |
dx +C1 |
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where C1 is any constant. Then, the general solution is |
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y = e−∫ P(x)dx |
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C1 + |
∫ Q(x)e∫ P(x)dx dx , |
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which is identity with (3.8).
Remark. Sometimes the differential equation is easily computed if x is a function of y, namely x′ + P( y)x = Q( y) .
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u = ∫x M dx + ϕ(y), |
(3.12) |
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where ϕ(y) does not contain x and the superscript x means that y is to be held constant during the integration. Substituting the value of u from (3.12) in
N= ∂∂uy , we obtain
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∂u |
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∂ |
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∂ϕ |
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= |
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∫M dx + |
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∂y |
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or |
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x0 |
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∂ϕ |
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∂ |
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= N − |
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(3.13) |
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∂y |
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derivative with |
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∂N |
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find, by integrating (3.13), that ϕ is |
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ϕ = |
N − |
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dy = |
∫ |
N(x , y)dy + C. |
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Substituting ϕ in (3.12) we get |
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u(x, y) = ∫M (x, y)dx + ∫ N(x0 , y)dy + C = 0. |
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1.6. Simple applications |
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Any differential equation of the first degree may be written in the form |
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M(x, y)dx + N(x, y)dy = 0, or |
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(3.14) |
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N(x, y) |
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As y′ evaluated for a curve represents its slope, it appears that equation (3.14) assigns to every point in the plane of a slope and that any curve represented by a solution of (3.14) must have a slope given by (3.14). If, then, we can find the expression for the slope of a curve, we can find the curve as a solution of a differential equation. If the equation of a particular curve passing through a point is wanted, the constant of integration must be replaced by such a number that the coordinates of the point will satisfy the equation.
If a particle of mass m moves in a line with acceleration a, under the influence of several applied forces whose resultant is F, then, in accordance with
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