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timofeeve [1]
1 year ago
13

Find the area of a circular sector in cm² with angle 15° if the length of the intercepted arc is pi/2 cm

Mathematics
1 answer:
UkoKoshka [18]1 year ago
3 0

Answer:

0.52cm2

Step-by-step explanation:

l of Arc =tita/360×2πr

15/360×2×22/7×2

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Consider the function f(x, y) = x2 + xy + y2 defined on the unit disc, namely, d = {(x, y)| x2 + y2 ≤ 1}. use the method of lagr
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First we note that the partial derivatives vanish simultaneously at one point:

\begin{cases}f_x=2x+y=0\\f_y=x+2y=0\end{cases}\implies(x,y)=(0,0)

so there is one critical point within the region D.

The Lagrangian is

L(x,y,\lambda)=x^2+xy+y^2+\lambda(x^2+y^2-1)

and has partial derivatives

L_x=2x+y+2\lambda x
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Set each partial derivative to 0 to find the possible critical points within the disk D. Then we notice that

yL_x=2xy+y^2+2\lambda xy=0
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Since x^2+y^2=1, we have

x^2-y^2=x^2+y^2-2y^2=0\implies1=2y^2\implies y^2=\dfrac12\implies y=\pm\dfrac1{\sqrt2}

And since x^2-y^2=0, or x^2=y^2, we also have

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So we have four possible additional critical points to consider:

f(0,0)=0
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f\left(-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac12
f\left(\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac12
f\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac32

It should be clear enough which of these correspond to the absolute extrema of f over D.
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Step-by-step explanation:

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