Mixing
Computing the arc length of the graph $y=sqrt{x-x^2}+arcsin(sqrt{x})$
$begingroup$
Computing the arc length of the graph $y=sqrt{x-x^2}+arcsin(sqrt{x})$
Is this done right?
I found the interval of this function which is $[0,1]$
I know the arc length formula is $$ L=int_a^bsqrt{1+(y')^2}dx$$
The derivative of the function is $$y'=sqrt{frac{1-x}{x}}$$
and the the square of $y'$ is $$(y')^2=frac{1-x}{x}$$
When I plug in these results on the integral I get $$L=int_0^1sqrt{frac{1}{x}d}x$$
with the substitution of $sqrt{x}=t$ we get the result $L=2$
This is the graph of the function
Thanks, @DougM for pointing out my mistake
calculus definite-integrals arc-length
asked Jan 22 at 21:37
J.DaneJ.Dane
368114
$endgroup$
add a comment |
$begingroup$
Computing the arc length of the graph $y=sqrt{x-x^2}+arcsin(sqrt{x})$
Is this done right?
I found the interval of this function which is $[0,1]$
I know the arc length formula is $$ L=int_a^bsqrt{1+(y')^2}dx$$
The derivative of the function is $$y'=sqrt{frac{1-x}{x}}$$
and the the square of $y'$ is $$(y')^2=frac{1-x}{x}$$
When I plug in these results on the integral I get $$L=int_0^1sqrt{frac{1}{x}d}x$$
with the substitution of $sqrt{x}=t$ we get the result $L=2$
This is the graph of the function
Thanks, @DougM for pointing out my mistake
calculus definite-integrals arc-length
asked Jan 22 at 21:37
J.DaneJ.Dane
368114
$endgroup$
1
$begingroup$
Check your differentiation. Get that right and everything that follows will be much simpler.
$endgroup$
– Doug M
Jan 22 at 21:44
$begingroup$
Oh wow, what mistake. Thanks for pointing it though. It's so easy now
$endgroup$
– J.Dane
Jan 22 at 21:52
add a comment |
$begingroup$
Computing the arc length of the graph $y=sqrt{x-x^2}+arcsin(sqrt{x})$
Is this done right?
I found the interval of this function which is $[0,1]$
I know the arc length formula is $$ L=int_a^bsqrt{1+(y')^2}dx$$
The derivative of the function is $$y'=sqrt{frac{1-x}{x}}$$
and the the square of $y'$ is $$(y')^2=frac{1-x}{x}$$
When I plug in these results on the integral I get $$L=int_0^1sqrt{frac{1}{x}d}x$$
with the substitution of $sqrt{x}=t$ we get the result $L=2$
This is the graph of the function
Thanks, @DougM for pointing out my mistake
calculus definite-integrals arc-length
asked Jan 22 at 21:37
J.DaneJ.Dane
368114
$endgroup$
Computing the arc length of the graph $y=sqrt{x-x^2}+arcsin(sqrt{x})$
Is this done right?
I found the interval of this function which is $[0,1]$
I know the arc length formula is $$ L=int_a^bsqrt{1+(y')^2}dx$$
The derivative of the function is $$y'=sqrt{frac{1-x}{x}}$$
and the the square of $y'$ is $$(y')^2=frac{1-x}{x}$$
When I plug in these results on the integral I get $$L=int_0^1sqrt{frac{1}{x}d}x$$
with the substitution of $sqrt{x}=t$ we get the result $L=2$
This is the graph of the function
Thanks, @DougM for pointing out my mistake
calculus definite-integrals arc-length
calculus definite-integrals arc-length
asked Jan 22 at 21:37
J.DaneJ.Dane
368114
asked Jan 22 at 21:37
J.DaneJ.Dane
368114
edited Jan 22 at 22:01
J.Dane
asked Jan 22 at 21:37
J.DaneJ.Dane
368114
asked Jan 22 at 21:37
J.DaneJ.Dane
368114
asked Jan 22 at 21:37
J.DaneJ.Dane
368114
368114
1
$begingroup$
Check your differentiation. Get that right and everything that follows will be much simpler.
$endgroup$
– Doug M
Jan 22 at 21:44
$begingroup$
Oh wow, what mistake. Thanks for pointing it though. It's so easy now
$endgroup$
– J.Dane
Jan 22 at 21:52
add a comment |
1
$begingroup$
Check your differentiation. Get that right and everything that follows will be much simpler.
$endgroup$
– Doug M
Jan 22 at 21:44
$begingroup$
Oh wow, what mistake. Thanks for pointing it though. It's so easy now
$endgroup$
– J.Dane
Jan 22 at 21:52
1
1
$begingroup$
Check your differentiation. Get that right and everything that follows will be much simpler.
$endgroup$
– Doug M
Jan 22 at 21:44
$begingroup$
Check your differentiation. Get that right and everything that follows will be much simpler.
$endgroup$
– Doug M
Jan 22 at 21:44
$begingroup$
Oh wow, what mistake. Thanks for pointing it though. It's so easy now
$endgroup$
– J.Dane
Jan 22 at 21:52
$begingroup$
Oh wow, what mistake. Thanks for pointing it though. It's so easy now
$endgroup$
– J.Dane
Jan 22 at 21:52
add a comment |
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$begingroup$
Check your differentiation. Get that right and everything that follows will be much simpler.
$endgroup$
– Doug M
Jan 22 at 21:44
$begingroup$
Oh wow, what mistake. Thanks for pointing it though. It's so easy now
$endgroup$
– J.Dane
Jan 22 at 21:52