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Differential
calculus - derivatives
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Applications of the derivative
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Angle
between two curves
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| Angle between two curves is the angle subtended by tangent lines at the point where the curves
intersect. |
| If curves
f1(x)
and f2(x)
intercept at P(x0,
y0)
then |
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| as
shows the right figure. |
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Angle
between two curves, examples
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| Example:
Find
the angle between cubic y
= -
x3 + 6x2 -
14x + 14 and quadratic y
= -
x2 + 6x -
6 functions. |
| Solution:
To
find the point where the curves intersect we should solve their
equations as the system of two equations
in two unknowns simultaneously. Therefore, |
| -
x3 + 6x2 -
1
4x + 14
= -
x2 + 6x -
6
or x3
-
7x2 + 20x -
20 = 0
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| the
root of the cubic equation we calculate using the formula |
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| where, |
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Then, we
calculate the slopes of the tangents drown to the given cubic and the
quadratic polynomial by evaluating
their derivatives at x
= 2. Thus, |
| taking
f2(x)
= -
x3 + 6x2 -
14x + 14
so that f
'2(x)
= -
3x2 + 12x -
14
then f '2(2)
= -
2 |
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and f1(x)
= -
x2 + 6x -
6
so that f '1(x)
= -
2x + 6
then f '1(2)
= 2. |
| Finally
we plug the slopes of tangents into the formula to find the angle
between given curves, as shows the
figure below. |
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| We
sketch the graphs of the quadratic and the cubic
by calculating coordinates of translations x0
and
y0,
as they are at the same time the coordinates
of the maximum and the point of inflection
respectively, thus |
| (x0)2
= xmax
= -
a1/(2a2) = 3,
y0
= f1(x0)
= 3, |
| (x0)3
= xinfl
= -
a2/(3a3) = 2,
y0
= f2(x0)
= 2. |
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Differential
of a function
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| If
given y = f
(x)
is differentiable function then, the derivative of y
or f
(x) |
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| as
the instantaneous rate of change of y
with respect to x,
can also be written as |
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where, the increment
Dx
= dx is called differential of
the independent variable and dy is the differential of
y.
Therefore,
the differential of a function
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| dy
= f '(x) dx |
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represent
the main part of the increment Dy
which is linear concerning the increment Dx
= dx, as is shown in
the figure below.
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| Example:
Find
the increment Dy
and the differential dy
of the
function f
(x)
= x2 -
4x,
and
calculate their values
if x
= 3
and Dx
= 0.01. |
| Solution:
Since
Dy
= f(x + Dx)
-
f(x)
then
Dy
= (x + Dx)2
-
4(x
+ Dx)
-
(x2 -
4x)
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| Dy
= (2x
-
4)Dx
+ Dx2
therefore,
dy
= (2x
-
4)Dx
= (2x
-
4)dx |
| or
f '(x)
= 2x
-
4,
dy
= f '(x)dx
= (2x
-
4)dx. |
| Let
calculate the values of Dy
and dy
for
x
= 3
and Dx
= 0.01, |
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Dy
= (2x
-
4)Dx
+ Dx2
= (2
·
3
-
4)
·
0.01
+ 0.012
= 0.0201, |
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and
dy
= f '(x)Dx,
dy
= (2x
-
4)Dx
= (2
·
3
-
4)
·
0.01
= 0.02. |
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Contents
J
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Copyright
© 2004 - 2020, Nabla Ltd. All rights reserved.
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