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3 years ago

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PLEASE HELP WILL GIVE BRAINLIEST

1 answer:

OlgaM077 [116]3 years ago

3 0

Answer:

Step-by-step explanation:

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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 $\dfrac{21}{16}$

6 0

3 years ago

1 bag of grape tomatoes coast $3 how many packages can you buy for $15

hoa [83]

Answer:

5 package

Step-by-step explanation:

1 bag = $3

X bag = $15

1 * 15/3 = 5 bags

hence : 5 bag can be bought out of $15

6 0

3 years ago

Answer the question below from I ready

san4es73 [151]

Your answer is going to be the first one: 3^2

6 0

3 years ago

Read 2 more answers

What is HCF and LCM ? # confused

GaryK [48]

HCF= highest common factor
LCM= least common multiple

7 0

3 years ago

Read 2 more answers

Find the length of diagonal HJ. Round to the nearest hundredth.

Alisiya [41]

Answer:

The length of the diagonal HJ is 10.82 units

Step-by-step explanation:

* Lets revise the rule of the distance between two points

- $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ , where

$(x_{1},y_{1})$ and $(x_{2},y_{2})$ are the two points

* Lets use this rule to find the length of the diagonal HJ

∵ The coordinates of point H are (-4 , 3)

∵ The coordinates of point J are (5 , -3)

∴ $x_{1}=-4$ and $x_{2}=5$

∴ $y_{1}=3$ and $y_{2}=-3$

- Lets find the length of the diagonal HJ by using the rule above

∴ HJ = $\sqrt{(5-(-4))^{2}+(-3-3)^{2}}=\sqrt{(5+4)^{2}+(-6)^{2}}$

∴ HJ = $\sqrt{(9)^{2}+36}=\sqrt{81+36}=\sqrt{117}=10.81665$

∴ HJ = 10.82

* The length of the diagonal HJ is 10.82 units

7 0

3 years ago