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sweet-ann [11.9K]
2 years ago
6

What’s the area of the figure ?

Mathematics
1 answer:
larisa86 [58]2 years ago
8 0

Answer:

8x^3y

Step-by-step explanation:

The area of a rectangle is length * width.

area = 4x^2y * 2x =

= 4 * 2 * x^2 * x * y

area = 8x^3y

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Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).
lutik1710 [3]

Answer:

The arc length is \dfrac{21}{16}

Step-by-step explanation:

Given that,

The given curve between the specified points is

x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}

The points from (\dfrac{9}{16},1) to (\dfrac{9}{8},2)

We need to calculate the value of \dfrac{dx}{dy}

Using given equation

x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}

On differentiating w.r.to y

\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})

\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})

\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})

\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}

We need to calculate the arc length

Using formula of arc length

L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}

Put the value into the formula

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}

L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}

L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}

L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}

Put the limits

L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})

L=\dfrac{21}{16}

Hence, The arc length is \dfrac{21}{16}

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2 years ago
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