Find the exact length of the curve. x=y48+14y2, 1≤y≤2

You need to apply the next formula L=∫cd1+(dxdy)2dy Where 1≤y≤2 First step, to find dxdy x=y48+14y2=y48+y−24 Differentiate both sides with respect to y, you get ddy[x]=ddy[y48+y−24] dxdy=48y3−24y−3 Simplify dxdy=y32−y−32 Now calculate (dxdy)2, so (dxdy)2=(y32−y−32)2=y64−24+y−64 Simplify (dxdy)2=y64−12+y−64 Now calculate (dxdy)2+1, adding 1 to both sides (dxdy)2+1=y64−12+y−64+1 You can see thaty64+12+y−64, is a perfect square trinomial, so (dxdy)2+1=(y32+y−32)2 Now calculate (dxdy)2+1 (dxdy)2+1=(y32+y−32)2 (dxdy)2+1=y32+y−32 Then L=∫cd1+(dxdy)2dy=∫12(y32+y−32)dt Integrate you get: L=[y42(4)+y−22(−2)]12 L=[y48−y−24]12 Apply the fundamental theorem of calculus L=(248−2−24)−(148−1−24(=(2−116)−(18−14) L=3116+18=3316 Answer: L=3316

y48+14y2, 1≤y≤2x=18y4+14y−2dxdy=12y3−12y−3=(y32−y−32)(dxdy)2=(432−4−32)21+(dxdy)2=1+y64−12+y−64=y6+12+y−64=(y32+y−32)23116+18=3316

Sorry, I don't know the answer to your question

You need to apply the next formula L=∫cd1+(dxdy)2dy Where 1≤y≤2 First step, to find dxdy x=y48+14y2=y48+y−24 Differentiate both sides with respect to y, you get ddy[x]=ddy[y48+y−24] dxdy=48y3−24y−3 Simplify dxdy=y32−y−32 Now calculate (dxdy)2, so (dxdy)2=(y32−y−32)2=y64−24+y−64 Simplify (dxdy)2=y64−12+y−64 Now calculate (dxdy)2+1, adding 1 to both sides (dxdy)2+1=y64−12+y−64+1 You can see thaty64+12+y−64, is a perfect square trinomial, so (dxdy)2+1=(y32+y−32)2 Now calculate (dxdy)2+1 (dxdy)2+1=(y32+y−32)2 (dxdy)2+1=y32+y−32 Then L=∫cd1+(dxdy)2dy=∫12(y32+y−32)dt Integrate you get: L=[y42(4)+y−22(−2)]12 L=[y48−y−24]12 Apply the fundamental theorem of calculus L=(248−2−24)−(148−1−24(=(2−116)−(18−14) L=3116+18=3316 Answer: L=3316

y48+14y2, 1≤y≤2x=18y4+14y−2dxdy=12y3−12y−3=(y32−y−32)(dxdy)2=(432−4−32)21+(dxdy)2=1+y64−12+y−64=y6+12+y−64=(y32+y−32)23116+18=3316

Sorry, I don't know the answer to your question

Get the solution to "Find the exact length of the curve. x=y48+14y2,"

Sandra Allison

Answered question

2021-12-30

Find the exact length of the curve.

x = \frac{y^{4}}{8} + \frac{1}{4 y^{2}}, 1 \leq y \leq 2

Answer & Explanation

Vasquez

Expert2022-01-08Added 669 answers

You need to apply the next formula
$L = \int_{c}^{d} \sqrt{1 + (\frac{d x}{d y})^{2}} d y$ Where $1 \leq y \leq 2$
First step, to find $\frac{d x}{d y}$
$x = \frac{y^{4}}{8} + \frac{1}{4 y^{2}} = \frac{y^{4}}{8} + \frac{y^{- 2}}{4}$
Differentiate both sides with respect to y, you get
$\frac{d}{d y} [x] = \frac{d}{d y} [\frac{y^{4}}{8} + \frac{y^{- 2}}{4}]$
$\frac{d x}{d y} = \frac{4}{8} y^{3} - \frac{2}{4} y^{- 3}$
Simplify
$\frac{d x}{d y} = \frac{y^{3}}{2} - \frac{y^{- 3}}{2}$
Now calculate $(\frac{d x}{d y})^{2}$ , so
$(\frac{d x}{d y})^{2} = (\frac{y^{3}}{2} - \frac{y^{- 3}}{2})^{2} = \frac{y^{6}}{4} - \frac{2}{4} + \frac{y^{- 6}}{4}$
Simplify
$(\frac{d x}{d y})^{2} = \frac{y^{6}}{4} - \frac{1}{2} + \frac{y^{- 6}}{4}$
Now calculate $(\frac{d x}{d y})^{2} + 1$ , adding 1 to both sides
$(\frac{d x}{d y})^{2} + 1 = \frac{y^{6}}{4} - \frac{1}{2} + \frac{y^{- 6}}{4} + 1$
You can see that $\frac{y^{6}}{4} + \frac{1}{2} + \frac{y^{- 6}}{4}$ , is a perfect square trinomial, so
$(\frac{d x}{d y})^{2} + 1 = (\frac{y^{3}}{2} + \frac{y^{- 3}}{2})^{2}$
Now calculate $\sqrt{(\frac{d x}{d y})^{2} + 1}$
$\sqrt{(\frac{d x}{d y})^{2} + 1} = \sqrt{(\frac{y^{3}}{2} + \frac{y^{- 3}}{2})^{2}}$
$\sqrt{(\frac{d x}{d y})^{2} + 1} = \frac{y^{3}}{2} + \frac{y^{- 3}}{2}$
Then
$L = \int_{c}^{d} \sqrt{1 + (\frac{d x}{d y})^{2}} d y = \int_{1}^{2} (\frac{y^{3}}{2} + \frac{y^{- 3}}{2}) d t$
Integrate you get:
$L = [\frac{y^{4}}{2 (4)} + \frac{y^{- 2}}{2 (- 2)}]_{1}^{2}$
$L = [\frac{y^{4}}{8} - \frac{y^{- 2}}{4}]_{1}^{2}$
Apply the fundamental theorem of calculus
$L = (\frac{2^{4}}{8} - \frac{2^{- 2}}{4}) - (\frac{1^{4}}{8} - \frac{1^{- 2}}{4} (= (2 - \frac{1}{16}) - (\frac{1}{8} - \frac{1}{4})$
$L = \frac{31}{16} + \frac{1}{8} = \frac{33}{16}$
Answer: $L = \frac{33}{16}$

user_27qwe

Skilled2022-01-08Added 375 answers

$\begin{matrix} \frac{y^{4}}{8} + \frac{1}{4 y^{2}}, 1 \leq y \leq 2 \\ x = \frac{1}{8} y^{4} + \frac{1}{4} y^{- 2} \\ \frac{d x}{d y} = \frac{1}{2} y^{3} - \frac{1}{2} y^{- 3} = (\frac{y^{3}}{2} - \frac{y^{- 3}}{2}) \\ (\frac{d x}{d y})^{2} = (\frac{4^{3}}{2} - \frac{4^{- 3}}{2})^{2} \\ 1 + (\frac{d x}{d y})^{2} = 1 + \frac{y^{6}}{4} - \frac{1}{2} + \frac{y^{- 6}}{4} \\ = \frac{y^{6}}{} + \frac{1}{2} + \frac{y^{- 6}}{4} \\ = (\frac{y^{3}}{2} + \frac{y^{- 3}}{2})^{2} \\ \frac{31}{16} + \frac{1}{8} = \frac{33}{16} \end{matrix}$

karton

Expert2022-01-08Added 613 answers

Sorry, I don't know the answer to your question

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Algebra II

Didn't find what you were looking for?