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3 years ago

10

The sum of the areas of two circles is 8pie square meters. Find the length of a radius of each circle ifone of them is twice as

long as the other.what is the radius of the larger circle?

1 answer:

NeX [460]3 years ago

5 0

Nice

area=pir^2
the areas are
pir^2+pi(2r)^2
they add to 8pi

8pi=pir^2+pi(2r)^2
undistribute pi
8pi=pi(r^2+(2r)^2)
divide both sides by pi
8=r^2+(2r)^2
expand
8=r^2+4r^2
add
8=5r^2
divide by 5
8/5=r^2
sqrt both sides
√(8/5)=r
double since it is double
2√(8/5) is radius of bigger aka (4√10)/5

radius=(4√10)/5

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Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

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$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

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Distance is calculated as:

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So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

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Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

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Differentiate using chain rule:

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$u = x^2 - 5x +9$

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$d = \sqrt u$

$d = u^\frac{1}{2}$

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$d' = \frac{du}{dx} * \frac{dd}{du}$

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$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

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3 years ago