From Förberedande kurs i matematik 2

Revision as of 13:46, 19 August 2008 (edit) Per (Talk | contribs) ← Previous diff		Current revision (10:30, 21 October 2008) (edit) (undo) Tek (Talk | contribs) m
(4 intermediate revisions not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	The channel holds most water when its cross-sectional area is greatest.
-	<center> [[Bild:1_3_5-1(4).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:1_3_5-2(4).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:1_3_5-3(4).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:1_3_5-4(4).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	[[Bild:1_3_5_1_1.gif]]	+	[[Image:1_3_5_1_1.gif|center]]
-	[[Bild:1_3_5_1_2.gif]]	+
		+	By dividing up the channel's cross-section into a rectangle and two triangles we can, with the help of a little trigonometry, determine the lengths of the rectangle's and triangles' sides, and thence the cross-sectional area.
		+
		+	[[Image:1_3_5_1_2.gif|center]]
		+
		+	The area of the cross-section is
		+
		+	{{Displayed math||<math>\begin{align}
		+	A(\alpha) &= 10\cdot 10\cos\alpha + 2\cdot\frac{1}{2}\cdot 10\cos\alpha \cdot 10\sin\alpha\\[5pt]
		+	&= 100\cos \alpha (1+\sin\alpha)\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	If we limit the angle to lie between <math>0</math> and <math>\pi/2</math>, the problem can be formulated as:
		+
		+	::Maximise the function <math>A(\alpha) = 100\cos\alpha (1+\sin\alpha)</math> when
		+	<math>0\le \alpha \le {\pi }/{2}\,</math>.
		+
		+	The area function is a differentiable function and the area is least when <math>\alpha=0</math> or <math>\alpha=\pi/2\,</math>, so the area must assume its maximum at a critical point of the area function.
		+
		+	We differentiate the area function:
		+
		+	{{Displayed math||<math>\begin{align}
		+	A'(\alpha) &= 100\cdot (-\sin\alpha)\cdot (1+\sin\alpha) + 100\cdot\cos\alpha \cdot \cos\alpha\\[5pt]
		+	&= -100\sin\alpha - 100\sin^2\!\alpha + 100\cos^2\!\alpha\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	At a critical point <math>A'(\alpha)=0</math> and this gives us the equation
		+
		+	{{Displayed math||<math>\sin\alpha + \sin^2\!\alpha - \cos^2\!\alpha = 0</math>}}
		+
		+	after eliminating the factor -100. We replace <math>\cos^2\!\alpha</math> with <math>1-\sin^2\!\alpha</math> (according to the Pythagorean identity) to obtain an equation solely in <math>\sin\alpha\,</math>,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\sin\alpha + \sin^2\!\alpha - (1-\sin^2\!\alpha) &= 0\\[5pt]
		+	2\sin^2\!\alpha + \sin\alpha - 1 &= 0\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	This is a second-degree equation in <math>\sin\alpha</math> and completing the square gives that
		+
		+	{{Displayed math||<math>\begin{align}
		+	2\bigl(\sin\alpha + \tfrac{1}{4}\bigr)^{2} - 2\bigl(\tfrac{1}{4}\bigr)^2 - 1 &= 0\\[5pt]
		+	\bigl(\sin\alpha + \tfrac{1}{4}\bigr)^2 &= \frac{9}{16}
		+	\end{align}</math>}}
		+
		+	and we obtain <math>\sin\alpha = -\tfrac{1}{4}\pm \tfrac{3}{4}</math>, i.e. <math>\sin \alpha = -1</math> or <math>\sin \alpha = \tfrac{1}{2}\,</math>.
		+
		+	The case <math>\sin \alpha =-1</math> is not satisfied for <math>0\le \alpha \le \pi/2</math> and <math>\sin \alpha = \tfrac{1}{2}</math> gives <math>\alpha = \pi/6</math>. Thus <math>\alpha = \pi/6\,</math> is a critical point.
		+
		+
		+	If we summarize, we know therefore that the cross-sectional area has local minimum points at <math>\alpha = 0</math> and <math>\alpha = \pi/2</math> and that we have a critical point at <math>\alpha = \pi/6\,</math>. This critical point must be a maximum, which we can also show using the second derivative,
		+
		+	{{Displayed math||<math>\begin{align}
		+	A''(\alpha) &= -100\cos\alpha - 100\cdot 2\sin\alpha\cdot\cos\alpha + 100\cdot 2\cos\alpha \cdot (-\sin\alpha)\\[5pt]
		+	&= -100\cos\alpha (1+4\sin\alpha)\,,
		+	\end{align}</math>}}
		+
		+	which is negative at <math>\alpha = \pi/6</math>,
		+
		+	{{Displayed math||<math>\begin{align}
		+	A''(\pi/6) &= -100\cos\frac{\pi}{6}\cdot \Bigl(1+4\sin\frac{\pi}{6}\Bigr)\\[5pt]
		+	&= -100\cdot\frac{\sqrt{3}}{2}\cdot \Bigl( 1+4\cdot \frac{1}{2} \Bigr)<0\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	There are no local maximum points other than <math>\alpha = \pi/6\,</math>, which must therefore also be a global maximum.

---

Current revision

The channel holds most water when its cross-sectional area is greatest.

By dividing up the channel's cross-section into a rectangle and two triangles we can, with the help of a little trigonometry, determine the lengths of the rectangle's and triangles' sides, and thence the cross-sectional area.

The area of the cross-section is

\displaystyle \begin{align} A(\alpha) &= 10\cdot 10\cos\alpha + 2\cdot\frac{1}{2}\cdot 10\cos\alpha \cdot 10\sin\alpha\\[5pt] &= 100\cos \alpha (1+\sin\alpha)\,\textrm{.} \end{align}

If we limit the angle to lie between \displaystyle 0 and \displaystyle \pi/2, the problem can be formulated as:

Maximise the function \displaystyle A(\alpha) = 100\cos\alpha (1+\sin\alpha) when

\displaystyle 0\le \alpha \le {\pi }/{2}\,.

The area function is a differentiable function and the area is least when \displaystyle \alpha=0 or \displaystyle \alpha=\pi/2\,, so the area must assume its maximum at a critical point of the area function.

We differentiate the area function:

\displaystyle \begin{align} A'(\alpha) &= 100\cdot (-\sin\alpha)\cdot (1+\sin\alpha) + 100\cdot\cos\alpha \cdot \cos\alpha\\[5pt] &= -100\sin\alpha - 100\sin^2\!\alpha + 100\cos^2\!\alpha\,\textrm{.} \end{align}

At a critical point \displaystyle A'(\alpha)=0 and this gives us the equation

\displaystyle \sin\alpha + \sin^2\!\alpha - \cos^2\!\alpha = 0

after eliminating the factor -100. We replace \displaystyle \cos^2\!\alpha with \displaystyle 1-\sin^2\!\alpha (according to the Pythagorean identity) to obtain an equation solely in \displaystyle \sin\alpha\,,

\displaystyle \begin{align} \sin\alpha + \sin^2\!\alpha - (1-\sin^2\!\alpha) &= 0\\[5pt] 2\sin^2\!\alpha + \sin\alpha - 1 &= 0\,\textrm{.} \end{align}

This is a second-degree equation in \displaystyle \sin\alpha and completing the square gives that

\displaystyle \begin{align} 2\bigl(\sin\alpha + \tfrac{1}{4}\bigr)^{2} - 2\bigl(\tfrac{1}{4}\bigr)^2 - 1 &= 0\\[5pt] \bigl(\sin\alpha + \tfrac{1}{4}\bigr)^2 &= \frac{9}{16} \end{align}

and we obtain \displaystyle \sin\alpha = -\tfrac{1}{4}\pm \tfrac{3}{4}, i.e. \displaystyle \sin \alpha = -1 or \displaystyle \sin \alpha = \tfrac{1}{2}\,.

The case \displaystyle \sin \alpha =-1 is not satisfied for \displaystyle 0\le \alpha \le \pi/2 and \displaystyle \sin \alpha = \tfrac{1}{2} gives \displaystyle \alpha = \pi/6. Thus \displaystyle \alpha = \pi/6\, is a critical point.

If we summarize, we know therefore that the cross-sectional area has local minimum points at \displaystyle \alpha = 0 and \displaystyle \alpha = \pi/2 and that we have a critical point at \displaystyle \alpha = \pi/6\,. This critical point must be a maximum, which we can also show using the second derivative,

\displaystyle \begin{align} A''(\alpha) &= -100\cos\alpha - 100\cdot 2\sin\alpha\cdot\cos\alpha + 100\cdot 2\cos\alpha \cdot (-\sin\alpha)\\[5pt] &= -100\cos\alpha (1+4\sin\alpha)\,, \end{align}

which is negative at \displaystyle \alpha = \pi/6,

\displaystyle \begin{align} A''(\pi/6) &= -100\cos\frac{\pi}{6}\cdot \Bigl(1+4\sin\frac{\pi}{6}\Bigr)\\[5pt] &= -100\cdot\frac{\sqrt{3}}{2}\cdot \Bigl( 1+4\cdot \frac{1}{2} \Bigr)<0\,\textrm{.} \end{align}

There are no local maximum points other than \displaystyle \alpha = \pi/6\,, which must therefore also be a global maximum.