Higher_Mathematics_Part_1
.pdfWe have
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PN = y , QN = MN tgα = |
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x f (x) = |
f (x)dx = dy. |
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So, the differential of function |
f (x) |
under given values of х and |
x equals |
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the increment QN of secant MQ ordinate, which is drawn to the curve |
y = f (x) at |
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point M , when argument receives increment x. |
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19.2. General properties of the differential |
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Let |
u(x) , |
v(x) be differentiable functions. Then the following equalities |
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are true: |
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1. |
dC = 0 ( C = const ). |
2. d(u +v) = du +dv . |
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3. |
d(uv) = udv +vdu . |
4. d(Cu) = Cdu . |
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5. |
d (u ) = vdu − udv , v ≠ 0. |
6. df (u) = f '(u)du, u = u(x). |
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19.3.The application of differentials in calculus of approximation and in error theory
At small values of x the following formula is true:
y ≈ dy
or
f (x + x) ≈ f (x) + f ' (x) x
Let y = f (x) , and the value х is defined approximately, in other words with
some absolute error x , then a value of function у will have its own absolute error y too. The relative error of computation of the value of function y can be
approximately calculated with the help of differential:
y |
≈ |
dy |
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19.4. Tangent and normal
We remember that tangent equation, constructed to the graph of function y = f (x) at point P0 (x0 , y0 ) is the following:
y − y0 = f ′(x0 )(x − x0 ) , |
(3.10) |
where y0 = f (x0 ) . Equation of normal is
y − y0 = − |
1 |
(x − x0 ) . |
(3.11) |
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f ' (x0 ) |
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Micromodule 19
EXAMPLES OF PROBLEMS SOLUTION
Example 1. Find the differential of function y = x2 +2x at point x = 2 . Solution. The first method. As the differential is the main, linear with respect
to x , part of increment of function at point |
x , let’s find the increment of given |
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function at point |
x = 2 . Therefore |
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y = y(2 + x) − y(2) = (2 + |
x)2 + 2(2 + |
x) − 8 = 6 x + ( x)2 . |
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The linear part of increment is the expression 6 |
x . Finally, dy(2) = 6 |
x . |
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The second method. As dy = f ' (x)dx , then we have |
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y' = 2x + 2 ; |
y' (2) = 6 ; |
dy(2) = 6dx. |
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Example 2. Find the differential of function |
y = |
1− x2 + ln sin x. |
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Solution. We have |
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dy = d( 1−x2 |
+ln sin x) = ( 1− x2 |
+ ln sin x)'dx = |
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= ( |
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− x |
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+ ctg x)dx . |
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1− x2 |
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Example 3. Calculate approximate value ( with the help of differential) of |
15 . |
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Solution. Let’s consider a function f (x) = |
x . Let |
x =16 , x + |
x = 15 , |
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x = −1 . Then |
x + x ≈ |
x + |
1 |
x , or |
15 ≈ 4 + |
1 |
(−1) = 31 = |
3.875 . |
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2 4 |
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Example 4. Calculate approximate value of |
tg 54° . |
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Solution. Let’s consider a function |
f (x) = tg x . Then we receive: |
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tg(x + x) ≈ tg x + (tg x)′ |
x ; |
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tg(x + |
x) ≈ tg x + |
1 |
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x . |
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cos2 |
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x |
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Let’s convert degrees to radians: 54° = 54π |
= 3π . |
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π , x + x = |
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180 |
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10 |
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Let x = |
3π , |
x = |
3π − π |
= |
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π |
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, then |
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10 |
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tg 3π |
≈ tg |
π |
+ |
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1 |
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π |
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=1+ |
π |
≈ 1,314 . |
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cos2 π |
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Finally, |
tg 54° ≈ 1,314. |
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Example 5. Find tangent equation and equation of normal, drawn to the curve y = x ln x +1 at point with ordinate 1.
Solution. By conditions y0 =1. Let’s find an abscissa of touching point. We have
1= x ln x +1, x ln x = 0 ; x ≠ 0 . Therefore ln x = 0 , then |
x0 =1 . |
Let’s find derivative |
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y′ = (x ln x + 1)′ = ln x + (ln x)′ x = ln x + 1. |
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Then y′(1) = ln1+ 1 = 1. Substituting values of x0 , y0 and |
y′(x0 ) in |
formula of tangent (3.10) and normal (3.11), we will obtain tangent equation
y − 1 = 1 (x − 1) , then y = x , and equation of normal |
y − 1 = − 1 |
(x −1) |
, or |
y = 2 − x. |
1 |
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Example 6. Find an angle between the tangent, constructed to the curve
y = x3 − 2x2 − 5x at point x =1 , and positive direction of |
Ox . |
Solution. Let’s find derivative y′ = 3x2 − 4x − 5 . Then |
y′(2) = 12 − 8 − 5 = −1 . |
According to the geometrical interpretation of derivative, solve the equation tg α =−1 . We obtain α = 34π (let’s point out that 0 ≤ α < π ).
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Example 7. Write the tangent equation and equation of normal to the curve x3 +2 y3 = 5xy at point M (2; 1) .
Solution. Substituting coordinates of point M in equation of the curve you will obtain right equality 10=10. Then point М will belong to this curve and
x0 = 2, y0 =1. |
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Let’s find derivative of function |
y at point x0 . We have: |
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(x3 + 2y3 )′ = (5xy)′ , 3x2 + 6y2 y ' = 5( y + xy ') , (6y2 − 5x) y ' = 5y − 3x2 , |
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y' = |
5y − 3x3 |
y' (2) = |
5 1− 3 8 |
= |
19 |
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Then: |
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6y 2 − |
5x |
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6 1− 5 2 |
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y − 1 = 19 (x − 2) , or 19x − 4y − 34 = 0 is equation of tangent, |
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y − 1 = − |
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(x − 2) , |
or 4x + 19 y − 27 = 0 is equation of normal. |
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19 |
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Example 8. Compose tangent equation for an ellipse, given by equations
x = 2 sin t , |
y = 3cost, |
at point M0 (1; y0 ) , where |
y0 > 0. |
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Solution. Equation |
of |
tangent is |
y − y0 |
= f ' (x0 )(x − x0 ) . By conditions |
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x =1 . Let’s find an ordinate of touching point |
y . t = π |
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and |
t = |
5π |
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are |
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6 |
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solutions |
of |
the |
equation |
1= 2 sin t |
if |
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t [0; 2π) . Taking |
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account |
the |
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condition |
y |
> 0, |
that is, cost > 0, we obtain |
t = |
π . Then |
y |
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= cos π = |
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3 |
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Next we shall find the derivative f ′(x0 ) of the parametrically given function: |
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y' (t) = −3sin t , |
y '(t) = 2cost. |
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Therefore |
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y x' = |
−3sin t |
= − |
3 |
tgt ; |
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f ' (1) = − 3 tg π = − |
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3 . |
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2 cos t |
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2 |
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Tangent equation is |
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y − |
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3 |
= − |
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(x − 1) , |
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or |
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y = − |
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x + |
3. |
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Micromodule 19
CLASS AND HOME ASSIGNMENT
1. Find the increment and the differential of the function y = x2 −4x +3 , if:
а) |
x =1 , x = 0,1 ; b) |
x = 3 , |
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x = 0,05 . |
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Find the differential of a function: |
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2. y = (4−x2 )2x . 3. |
y = |
sin x + tg2 x . |
4. y = ln arcsin x . |
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5. |
y = x2x+1 . 6. y = |
arccos x |
+arctg |
1 |
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7. y = 4 x + x log2 tg x . |
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x2 |
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Find the differential of an implicit function at the point M0 (x0 ; y0 ) :
8.x3 + y3 +3xy −15 = 0 , M0 (1; 2) .
9.ln x2 + y2 = arctg xy , M0 (0; 1) .
Calculate an approximate value with the help of the differential:
10. 3 131 . 11. (0, 95)6 . 12. sin 9° . 13. cos155° .
14. arctg1, 05 . 15. arccos 0, 02 . 16. ctg 85° .
Compose an equation of a tangent line and a normal to a curve, which is given by a parametric equation:
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x = t −sin t , y =1−cos t , at the point t = |
π . |
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π . |
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x = cos3 t , y = sin3 t , at the point t = |
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4 |
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π . |
19. |
x = 2 cos t −cos 2t , y = 2 sin t −sin 2t , at the point t = |
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x = tet , y = te−t , at the point x = e . |
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Compose an equation of a tangent line and a normal to a curve at point М:
21.4x4 +6xy − y4 = 0 , M (1; 2) .
22.x2 (2x −y) = 2x −y3 , M (1;1) .
23.x2 / 3 + y2 / 3 = 8 , M (8; 8) .
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Answers |
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1. a) y = −0,19 |
, dy = −0,2 ; b) dy = 0,1, |
y = 0,1025. 2. |
2x[(4 − x2 )ln 2 − 2x]dx . |
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5. x2x (2x ln x + 2x + |
1)dx . 8. −3dx / 5 . 9. dx. |
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5,08. 11. 0,7. |
12. 0,157. 13. – 0,908. |
14. 0,81. 15. 1,55. 16. 0,087. 17. y = x + 2 − π / 2 ; y = − x + π / 2 . 18. y + x = 2 / 2 ;
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y = x . 19. y + x = 3 ; |
y = x + 1 . 20. y = 1/ e ; x = e . 21. 14x − 13y + 12 = 0 ; |
13x + 14y − 41 = 0 . 22. |
y = x ; y = − x + 2 . 23. y + x =16; y = x . |
Micromodule 19
SELF-TEST ASSIGNMENT
19.1. Calculate an approximate value with the help of the first differential:
19.1.1. cos 61° . |
19.1.2. e0,2 . |
19.1.3. sin 33° . |
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19.1.4. arctg1, 05 . |
19.1.5. |
120 . |
19.1.6. |
3 340 . |
19.1.7. 3 66 . |
19.1.8. |
5 33 . |
19.1.9. |
6 70 . |
19.1.10. cos 85° . |
19.1.11. sin 8° . |
9.1.12. sin 28° . |
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19.1.13. arctg 0, 95 . |
19.1.14. arctg 0, 9 . |
19.1.15. e0,3 . |
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19.1.16. ln1, 05 . |
19.1.17. ln 0, 97 . |
19.1.18. ln1, 08 . |
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19.1.19. tg 47° . |
19.1.20. ctg 50° . |
19.1.21. (1, 02)5 . |
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19.1.22. arccos 0, 45 . |
19.1.23. arcsin 0, 52 . |
19.1.24. (1, 97)6 . |
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19.1.25. (2, 04)4 . |
19.1.26. ln tg 48° . |
19.1.27. ln tg 43° . |
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19.1.28. cos 86° . |
19.1.29. sin 26° . |
19.1.30. tg 40° . |
19.2. Solve the tasks for composition of an equation of the normal and the tangent to a curve.
19.2.1. Write the equation of a normal and a tangent to the curve y = 13 x3 −12 x2 + 56 at the points, where the slope of the tangent is equal to 2.
19.2.2. Write the equation of a normal and a tangent to the ellipse |
x2 |
+ |
y2 |
= 1 |
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at the points, where the slope of the normal line is equal to 12 .
19.2.3. Write |
the |
equation of |
a normal |
and a tangent to the ellipse |
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x = 3cost, y = 2sin t |
at the points, |
where the |
tangent |
is parallel |
to the line |
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y =− |
2 |
x +4 . |
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19.2.4. Write |
the |
equation of |
a normal |
and a |
tangent to |
the curve |
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y = 3x3 + 5x2 + 5x + 4 |
at the points, where the tangents are parallel to the line |
y = 4x − 2 .
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19.2.5. Write the |
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equation |
of a |
normal and a tangent to the |
curve |
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y = 2 x3 |
+ 5x2 + 9x + |
2 |
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at |
the |
points, |
where the tangents are parallel |
to the |
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3 |
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line y = x −2 . |
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19.2.6. Write |
the |
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equation of a |
normal and a tangent to the |
curve |
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y = 1 x3 |
+ |
7x2 + 15x + 1 |
at |
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points, |
where the slope of the tangent |
line is |
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3 |
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3 |
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equal to 2. |
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19.2.7. |
Write |
the |
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equation of a normal and a tangent to the curve |
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y = x3 + 7x2 + 11x + |
8 |
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where the tangent is parallel to the line y = 3x −4 . |
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19.2.8. |
Write |
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equation of a |
normal and a tangent to the |
curve |
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the |
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y = 5 x3 |
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11x2 + 13x + |
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8 |
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at the points, where the slope of the normal line is |
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75 |
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3 |
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equal to |
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19.2.9. |
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the |
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equation of a |
normal and a tangent to the |
curve |
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Write |
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y = 4 x3 |
+ |
5x2 + 3x + |
4 |
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at the points, where tangents and the x-axis form an |
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3 |
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3 |
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angle of1350 . |
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19.2.10. |
Write |
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the |
equation of a |
normal and a tangent to the |
curve |
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= x3 + 6x2 + |
1 x + 5 at the points, where the slope of the normal line is equal |
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to 1 . |
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equation of a |
normal and a tangent to the |
curve |
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19.2.11. |
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y = 1 x3 |
+ 7x2 + 10x + 130 |
at the points, where the tangents are parallel to the |
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line y = −3x +2 . |
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19.2.12. |
Write |
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the |
equation of a |
normal and a tangent to the |
curve |
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y = x3 + 6x2 + 2 x + |
8 |
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at the points, where the tangents are parallel to the line |
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y = −3x + 1 . |
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19.2.13. |
Write |
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the |
equation of a |
normal and a tangent to the |
curve |
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y = 3x3 + 6x2 − x + 1 |
at the points, where the slope of the tangent lines is equal |
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to −4 . |
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the |
equation of a |
normal and a tangent to the |
curve |
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19.2.14. |
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y = 3x2 + 6x2 − 6x + 25 |
at the points, where the tangents and the x- axis form |
9
the angle 1350 .
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19.2.15. |
Write |
the |
equation of a normal and |
a |
tangent |
to |
the |
line |
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y = |
x2 −3x +6 |
at the point with the abscissa x |
= 3 . |
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19.2.16. |
Write |
the |
equation of |
a |
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normal |
and |
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a |
tangent |
to the curve |
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y = (x +1)3 |
3−x |
at the point with the abscissa |
x |
= 2 . |
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19.2.17. Write the equation of |
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a |
normal |
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tangent |
to |
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astroid x = 2 |
2 cos3 t, |
y = 2 2 sin3 t , if t |
= π . |
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19.2.18 Write |
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the |
equation |
of |
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normal |
and |
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tangent |
to the |
cycloid |
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x = 2(t −sin t), |
y = 2(1−cos t) , if |
t = |
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π . |
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19.2.19. Write |
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the |
equation |
of |
a |
normal |
and |
a |
tangent |
to |
the |
parabola |
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y2 −y +2x −4 = 0 at the point with the abscissa x |
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= −4 . |
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19.2.20. |
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Write |
the |
equation of |
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normal |
and |
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a |
tangent |
to the curve |
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x2 +2xy +2y4 = 5 at the point |
M0 (1; 1) . |
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19.2.21. Write |
the |
equation |
of |
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a |
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normal |
and |
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a |
tangent to the |
cycloid |
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x = 3(t −sin t), |
y = 3(1−cos t) , if |
t |
= |
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3π |
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Write |
the |
equation of |
a |
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normal |
and |
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tangent |
to the curve |
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y = 3x3 −6x2 −6x + |
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at the points, where the tangents and the x- axis form |
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19.2.23. |
Write |
the |
equation of |
a |
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normal |
and |
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a |
tangent |
to the curve |
y = 3x3 −6x2 −x +1 at the points, where the tangents are parallel to the line
y = −4x +5 . |
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19.2.24. |
Write |
the |
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equation of |
a |
normal |
and |
a |
tangent |
to |
the |
curve |
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x4 +3xy2 +3y4 =1 at the point M0 (−1;1) . |
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19.2.25. |
Write |
the |
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equation of |
a |
normal |
and |
a |
tangent |
to |
the |
curve |
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y = |
2 |
x |
3 |
−5x |
2 |
+9x + |
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at the points, where the tangents are parallel to the |
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bisectrix of the first coordinate quarter (angle). |
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19.2.26. |
Write |
the |
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equation of |
a |
normal |
and |
a |
tangent |
to |
the |
curve |
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y = |
1 |
x |
3 |
−3x |
2 |
+6x − |
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at the points, where the tangents are parallel |
to the |
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line y = −2x +3 . |
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19.2.27. |
Write |
the |
equation |
of |
a normal |
and a tangent to the |
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curve x = t2 , |
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y = t3 , if t = 2 . |
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19.2.28. Write the equation of a normal and a tangent to the curve
y = |
4 |
x |
3 |
−5x |
2 |
+3x + |
1 |
at the points, where the tangents are parallel to the |
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bisectrix of the second coordinate quarter (angle).
19.2.29. Write the equation of a normal and a tangent to the curve
x3 −3xy2 + y3 = −3 at the point |
M0 (1; 2) . |
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19.2.30. Write |
the |
equation |
of a normal and a tangent to the curve |
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y = x3 −5x2 + x − |
1 |
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at the points, where the normal is parallel to the line |
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27 |
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y = 12 x +1 .
Micromodule 20
BASIC THEORETICAL INFORMATION.
HIGHER ORDER DERIVATIVES AND DIFFERENTIALS
Higher order derivatives of functions given explicitly, implicitly, or parametrically. Leibniz's formula. Higher order differentials.
Literature: [3, chapter 4, items 4.1—4.2], [4, part 5], [6, chapter 5, § 4], [7, chapter 6, § 18], [9], [10, chapter 4, § 12, 20], [12, chapter 3, §§ 22—24].
20.1. Higher order derivatives
Suppose differentiable function y = f (x) is defined on an interval (a; b), then its derivative f ′(x) , called also the first-order derivative or the first derivative, is a function of x too. The function f ′(x) may have derivative at the interval (a; b) too.
Definition 3.22. The later derivative is called the second derivative (or the second-order derivative).
It is denoted with one of the following symbols:
y'', f '' |
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d 2 y |
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d 2 |
f |
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dy |
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dx dx |
The second derivative has the following mechanical matter. If a mass point occurs in accordance with the law S = f (t) , then derivative S′ is a velocity at
the given time point and S′′ is an acceleration in the same time point t.
Definition 3.23. Derivative of the second-order derivative, if it is existed, is called the third-order derivative.
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y′′′ = |
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d 2 y |
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So, according to definition we write |
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2 |
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dx |
dx |
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The first derivative of the (n – 1)-th derivative, if it is existed, is called the n- th derivative:
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( y ( n − 1) ) ′ , or y |
( n ) |
= |
d |
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d n − 1 y |
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y ( n ) |
= |
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d x |
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Derivatives having the order higher than one are called the higher order derivatives.
20.2 Leibniz’s formula
Suppose y = uv , where u(x) and v(x) are the n times differentiated functions. Then
k |
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n! |
where Cn |
= |
(n−k)!k!, |
n
(uv)(n) = ∑ Cnk u(n−k )v(k ) , k =0
n ! = 1 2 3 k ( n − 1) n , 0 ! = 1 .
Particularly,
(u v ) ' |
= u ' v + u v ' (n = 1) , |
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(u v ) ′′ = u ′′v + 2 u ′v ′ + u v ′′ |
(n = 2), |
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(u v ) ′′ = u ′′′v + |
3u ′′v ′ + 3u ′v ′′ + |
u v ′′′ (n = 3). |
20.3. Calculation of the higher order derivatives of parametrically given functions
If some function is given parametrically by equations x = x(t) , y = y(t) , then derivatives are calculated by the following formulas:
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dy ′ |
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dy |
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y′ |
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d 2 y |
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dx |
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dx |
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xt′ |
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