# integrals and length of a curve

• Nov 12th 2009, 04:57 PM
calcbeg
integrals and length of a curve
HI

So let curve C have equation y = 4x^(3/2)
--------
3
Let a > 2. The portion of C btwn the points (2, (4(sqrt 2)^2)/3) and (a,(4a^(3/2))/3) has length 49/3

Find the value of a

So if y = (4x^(3/2))/3 then dy/dx = 2x^(1/2)

Correct me anywhere you see a mistake!!

Then L = integral sign (S) a= 2 and b= a sq rt( 1 + (dy/dx)^2) dx
= S a=2 b=a sq rt (1 + 4x) dx

SO then I let u = 1 + 4x so du = 4 dx

so when x = 2 then u = 9 and when x = a then u = 1 + 4a

L = 4 S a= 9 and b=1+4a sq rt u du which is 4(2/3)u^(3/2) a= 9 b = 1+4a

but when I do the substitution I get 265/8 = sq rt ((1+4a)^3) and since this is not a easy equation to solve I don't know what a is

Any idea where I went wrong???

Thanks

Struggling calculus beginner
• Nov 12th 2009, 05:16 PM
skeeter
$\displaystyle y = \frac{4}{3}x^{\frac{3}{2}}$

$\displaystyle y' = 2x^{\frac{1}{2}}$

$\displaystyle (y')^2 = 4x$

$\displaystyle \int_2^a \sqrt{1+4x} \, dx = \frac{49}{3}$

$\displaystyle \frac{1}{6}\left[(1+4x)^{\frac{3}{2}}\right]_2^a = \frac{49}{3}$

$\displaystyle \left[(1+4x)^{\frac{3}{2}}\right]_2^a = 98$

$\displaystyle (1+4a)^{\frac{3}{2}} - 27 = 98$

$\displaystyle (1+4a)^{\frac{3}{2}} = 125$

$\displaystyle 1+4a = 125^{\frac{2}{3}}$

$\displaystyle 1+4a = 25$

$\displaystyle a = 6$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}