Question

Find the arc length of y=x^/16+1/2x^2 x=2 x=3

Answer 1

Use this site I'll help https://www.wolframalpha.com/

Answer 2

Thanks but I kinda want a more in depth explanation do how to do it

Answer 3

find the derivative,
and plug it in arclength formula

Answer 4

What would the derivative be? @ganeshie8

Answer 5

dy/dx = ?

Answer 6

you need help in finding the derivative ?

Answer 7

Well when I took the derivative I got x^3/64-1/2x^2

Answer 8

I don't think that's right

Answer 9

whats ur function ?

Answer 10

can u type it agian here ?

Answer 11

there seem to be some formatting error in ur original question

Answer 12

y=x^/16+1/2x^2

Answer 13

??

Answer 14

Oh sorry it's y=x^4/16+1/2x^2

Answer 15

ty :)

\(\large y = \frac{x^4}{16} + \frac{1}{2x^2}\)

Answer 16

like this ?

Answer 17

Yep

Answer 18

\(\large y = \frac{x^4}{16} + \frac{1}{2x^2}\)


\(\large \frac{dy}{dx} = \frac{d}{dx} \left(\frac{x^4}{16} + \frac{1}{2x^2}\right)\)


\(\large ~~~~~= \frac{d}{dx} \left(\frac{x^4}{16}\right) + \frac{d}{dx}\left(\frac{1}{2x^2}\right)\)


\(\large ~~~~~= \frac{1}{16}\frac{d}{dx} \left(x^4\right) + \frac{1}{2}\frac{d}{dx}\left(x^{-2}\right)\)

Answer 19

whats the derivative of x^4 ?
whats the derivative of x^(-2) ?

Answer 20

The derivative of x^4 is 4x^3

Answer 21

And x^-2 is -2x^-3

Answer 22

yes.

Answer 23

now find dy/dx

Answer 24

Uhh would it be 3/4x^2+3x^-4?

Answer 25

it's
\[ \frac{1}{16}\frac{d}{dx} \left(x^4\right) + \frac{1}{2}\frac{d}{dx}\left(x^{-2}\right) \]

replace d/dx(x^4) with your result. same for the other

Answer 26

Yeah I did that .-.

Answer 27

you should get this
\[ \frac{1}{16}\left(4 x^3\right) + \frac{-2}{2} \left(x^{-3}\right) \]
which can be simplified

Answer 28

1/4x^3+ (-1x^-3)

Answer 29

yes.

\[ y' = \frac{x^3}{4} - x^{-3} \]
you now need (y')^2

Answer 30

we are going to use the arc length formula
see http://tutorial.math.lamar.edu/Classes/CalcII/ArcLength.aspx
for info.

Answer 31

Can I just distribute the 2 or do I have to foil it?

Answer 32

you have to foil.
(y')^2 means y' * y' or in this case
\[ \left( \frac{x^3}{4} - x^{-3} \right)\left( \frac{x^3}{4} - x^{-3} \right)\]

Answer 33

It looks a bit ugly, but we can deal with it.

Answer 34

Okay let me see what I get

Answer 35

X^6/16-1/2+x^-6?

Answer 36

yes.
the arc length formula uses
\[ \sqrt{ 1 + (y')^2 } \]
so let's add 1 to what you got:
\[ \frac{x^6}{16} + x^{-6} - \frac{1}{2} +1 \]
and simplify to
\[ \frac{x^6}{16} + x^{-6} + \frac{1}{2} \]

to make this easier to work with, multiply and divide by x^6
\[\frac{1}{x^6} \left( x^6 \left(\frac{x^6}{16} + x^{-6} + \frac{1}{2}\right) \right) \]

Answer 37

distribute the x^6

Answer 38

this will get rid of the x^-6
we will also factor out 1/16 from the inner expression, to get rid of the fractions

Answer 39

you should get
\[ \frac{1}{x^6} \left( \frac{x^{12}}{16} + 1 + \frac{x^6}{2} \right) \]

if you also factor out 1/16 you get
\[ \frac{1}{16x^6} \left( x^{12}+ 16 + 8x^6 \right) \]
or
\[ \frac{1}{16x^6} \left( x^{12}+ 8x^6+ 16 \right) \]

Answer 40

notice that you have a perfect square. can you factor it ?

Answer 41

Factor the x^12+8x^6+16?

Answer 42

Is it (x^6+4)?

Answer 43

yes (x^6+4)^2

Answer 44

that means
\[ \sqrt{1 + (y')^2 } = \sqrt{ \frac{(x^6+4)^2}{16x^6} }\]
can you simplify that ?

Answer 45

Um x^2+4/4x^6?

Answer 46

Oops x^6+4/4x^6

Answer 47

almost. the bottom x^6 should be x^3

Answer 48

Because of the square root ?

Answer 49

yes. \[ \sqrt{x^6} = x^3 \]
and \( x^3 \cdot x^3 = x^6 \)

Answer 50

or if you use
\[ \sqrt{x^6} = \left(x^6\right)^{\frac{1}{2}} = x^{6\cdot\frac{1}{2}}= x^3 \]

Answer 51

now you are ready to find the arc length
\[ \int \sqrt{1 + (y')^2 } dx =\int_2^{3} \frac{(x^6+4)}{4x^3} \ dx\]
we can simplify that to

\[ \int_2^{3} \frac{x^3}{4}+x^{-3} \ dx\]

Answer 52

So the integral is x^4/16-x^-4/4

Answer 53

ok on x^4 / 16
but when you integrate you add +1 to the exponent. -3+1 is not -4

Answer 54

Oh sorry x^-2

Answer 55

so the integral is
\[ \frac{x^4}{16} - \frac{1}{2} x^{-2} \]

Answer 56

now evaluate it from x=2 to x= 3 to find the arc length.