My calculus method of finding the circumference of a circle gives $16r-8rsqrt{2}$. Where's my mistake?
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To find the circumference of a circle using calculus, I've seen the approach of starting with $(ds)^2 = (dx)^2 + (dy)^2$, but my approach should work, too.
I take a small slice of the circle $dθ$. I take the radius on a point $θ$ and $θ + dθ$. I suppose that $ds$ is the base of an isosceles triangle with the other two sides being equal to $r$. Ι take the height and find that $ds = 2rsin(θ/2)$. I integrate up to $pi/2$, and I multiply by $4$. The result is equal to $16r-8rsqrt{2}$. It's very wrong, but I can't see the mistake in my thought process.
EDIT: After getting some help from the comments I think I found the solution.
I, once again, take a small slice of the circle $dθ$. I take the radius on a point $θ$ and $θ + Δθ$. I throw out the isosceles triangle idea because while Δθ tends to 0 the angle that r and Δs form tends to $πover2$.
So:
$$S(Δθ + θ) - S(θ)=4r*sin(Δθ)$$
$$frac{S(Δθ + θ) - S(θ)}{Δθ}=4frac{r*sin(Δθ)}{Δθ}$$
$$lim_{Δθto 0}frac{S(Δθ + θ) - S(θ)}{Δθ}=4lim_{Δθto 0}frac{r*sin(Δθ)}{Δθ}$$
$$frac{dS}{dθ}=4rlim_{Δθto 0}frac{sin(Δθ)}{Δθ}$$
$$frac{dS}{dθ}=4r$$
$$int_0^{frac{pi}{2}}frac{dS}{dθ}dθ=4int_0^{frac{pi}{2}}rdθ $$
$$S = 4rfrac{pi}{2}$$
$$S = 2pi r$$
calculus geometry
asked Dec 10 '18 at 17:06
Mario DrougaMario Drouga
235
$endgroup$
add a comment |
$begingroup$
To find the circumference of a circle using calculus, I've seen the approach of starting with $(ds)^2 = (dx)^2 + (dy)^2$, but my approach should work, too.
I take a small slice of the circle $dθ$. I take the radius on a point $θ$ and $θ + dθ$. I suppose that $ds$ is the base of an isosceles triangle with the other two sides being equal to $r$. Ι take the height and find that $ds = 2rsin(θ/2)$. I integrate up to $pi/2$, and I multiply by $4$. The result is equal to $16r-8rsqrt{2}$. It's very wrong, but I can't see the mistake in my thought process.
EDIT: After getting some help from the comments I think I found the solution.
I, once again, take a small slice of the circle $dθ$. I take the radius on a point $θ$ and $θ + Δθ$. I throw out the isosceles triangle idea because while Δθ tends to 0 the angle that r and Δs form tends to $πover2$.
So:
$$S(Δθ + θ) - S(θ)=4r*sin(Δθ)$$
$$frac{S(Δθ + θ) - S(θ)}{Δθ}=4frac{r*sin(Δθ)}{Δθ}$$
$$lim_{Δθto 0}frac{S(Δθ + θ) - S(θ)}{Δθ}=4lim_{Δθto 0}frac{r*sin(Δθ)}{Δθ}$$
$$frac{dS}{dθ}=4rlim_{Δθto 0}frac{sin(Δθ)}{Δθ}$$
$$frac{dS}{dθ}=4r$$
$$int_0^{frac{pi}{2}}frac{dS}{dθ}dθ=4int_0^{frac{pi}{2}}rdθ $$
$$S = 4rfrac{pi}{2}$$
$$S = 2pi r$$
calculus geometry
asked Dec 10 '18 at 17:06
Mario DrougaMario Drouga
235
$endgroup$
1
$begingroup$
The angle is $rm{d}theta$ not $theta.$
$endgroup$
– saulspatz
Dec 10 '18 at 17:16
1
$begingroup$
Informally, $sin ({1 over 2} d theta) = {1 over 2} d theta$.
$endgroup$
– copper.hat
Dec 10 '18 at 17:20
$begingroup$
How did you get to that conclusion? dθ is close to 0 shouldn't $sin({1over 2 }dθ) = 0$??
$endgroup$
– Mario Drouga
Dec 10 '18 at 17:59
$begingroup$
"$dtheta$ is close to $0$ shouldn't $sin(frac12dtheta)=0$?" That would be true if you believe that "$dtheta$ is close to $0$" means "$dtheta=0.$" But as long as $dthetaneq0,$ if $dtheta$ is small then $sin(frac12dtheta)$ will be much closer to $frac12dtheta$ than to $0.$
$endgroup$
– David K
Dec 10 '18 at 18:26
$begingroup$
@Mario $lim_{x to 0}frac{sin x}{x}=1$ answers your question in comment.
$endgroup$
– user376343
Dec 10 '18 at 18:39
add a comment |
$begingroup$
To find the circumference of a circle using calculus, I've seen the approach of starting with $(ds)^2 = (dx)^2 + (dy)^2$, but my approach should work, too.
I take a small slice of the circle $dθ$. I take the radius on a point $θ$ and $θ + dθ$. I suppose that $ds$ is the base of an isosceles triangle with the other two sides being equal to $r$. Ι take the height and find that $ds = 2rsin(θ/2)$. I integrate up to $pi/2$, and I multiply by $4$. The result is equal to $16r-8rsqrt{2}$. It's very wrong, but I can't see the mistake in my thought process.
EDIT: After getting some help from the comments I think I found the solution.
I, once again, take a small slice of the circle $dθ$. I take the radius on a point $θ$ and $θ + Δθ$. I throw out the isosceles triangle idea because while Δθ tends to 0 the angle that r and Δs form tends to $πover2$.
So:
$$S(Δθ + θ) - S(θ)=4r*sin(Δθ)$$
$$frac{S(Δθ + θ) - S(θ)}{Δθ}=4frac{r*sin(Δθ)}{Δθ}$$
$$lim_{Δθto 0}frac{S(Δθ + θ) - S(θ)}{Δθ}=4lim_{Δθto 0}frac{r*sin(Δθ)}{Δθ}$$
$$frac{dS}{dθ}=4rlim_{Δθto 0}frac{sin(Δθ)}{Δθ}$$
$$frac{dS}{dθ}=4r$$
$$int_0^{frac{pi}{2}}frac{dS}{dθ}dθ=4int_0^{frac{pi}{2}}rdθ $$
$$S = 4rfrac{pi}{2}$$
$$S = 2pi r$$
calculus geometry
asked Dec 10 '18 at 17:06
Mario DrougaMario Drouga
235
$endgroup$
To find the circumference of a circle using calculus, I've seen the approach of starting with $(ds)^2 = (dx)^2 + (dy)^2$, but my approach should work, too.
I take a small slice of the circle $dθ$. I take the radius on a point $θ$ and $θ + dθ$. I suppose that $ds$ is the base of an isosceles triangle with the other two sides being equal to $r$. Ι take the height and find that $ds = 2rsin(θ/2)$. I integrate up to $pi/2$, and I multiply by $4$. The result is equal to $16r-8rsqrt{2}$. It's very wrong, but I can't see the mistake in my thought process.
EDIT: After getting some help from the comments I think I found the solution.
I, once again, take a small slice of the circle $dθ$. I take the radius on a point $θ$ and $θ + Δθ$. I throw out the isosceles triangle idea because while Δθ tends to 0 the angle that r and Δs form tends to $πover2$.
So:
$$S(Δθ + θ) - S(θ)=4r*sin(Δθ)$$
$$frac{S(Δθ + θ) - S(θ)}{Δθ}=4frac{r*sin(Δθ)}{Δθ}$$
$$lim_{Δθto 0}frac{S(Δθ + θ) - S(θ)}{Δθ}=4lim_{Δθto 0}frac{r*sin(Δθ)}{Δθ}$$
$$frac{dS}{dθ}=4rlim_{Δθto 0}frac{sin(Δθ)}{Δθ}$$
$$frac{dS}{dθ}=4r$$
$$int_0^{frac{pi}{2}}frac{dS}{dθ}dθ=4int_0^{frac{pi}{2}}rdθ $$
$$S = 4rfrac{pi}{2}$$
$$S = 2pi r$$
calculus geometry
calculus geometry
asked Dec 10 '18 at 17:06
Mario DrougaMario Drouga
235
asked Dec 10 '18 at 17:06
Mario DrougaMario Drouga
235
edited Dec 10 '18 at 19:09
Mario Drouga
asked Dec 10 '18 at 17:06
Mario DrougaMario Drouga
235
asked Dec 10 '18 at 17:06
Mario DrougaMario Drouga
235
asked Dec 10 '18 at 17:06
Mario DrougaMario Drouga
235
235
1
$begingroup$
The angle is $rm{d}theta$ not $theta.$
$endgroup$
– saulspatz
Dec 10 '18 at 17:16
1
$begingroup$
Informally, $sin ({1 over 2} d theta) = {1 over 2} d theta$.
$endgroup$
– copper.hat
Dec 10 '18 at 17:20
$begingroup$
How did you get to that conclusion? dθ is close to 0 shouldn't $sin({1over 2 }dθ) = 0$??
$endgroup$
– Mario Drouga
Dec 10 '18 at 17:59
$begingroup$
"$dtheta$ is close to $0$ shouldn't $sin(frac12dtheta)=0$?" That would be true if you believe that "$dtheta$ is close to $0$" means "$dtheta=0.$" But as long as $dthetaneq0,$ if $dtheta$ is small then $sin(frac12dtheta)$ will be much closer to $frac12dtheta$ than to $0.$
$endgroup$
– David K
Dec 10 '18 at 18:26
$begingroup$
@Mario $lim_{x to 0}frac{sin x}{x}=1$ answers your question in comment.
$endgroup$
– user376343
Dec 10 '18 at 18:39
add a comment |
1
$begingroup$
The angle is $rm{d}theta$ not $theta.$
$endgroup$
– saulspatz
Dec 10 '18 at 17:16
1
$begingroup$
Informally, $sin ({1 over 2} d theta) = {1 over 2} d theta$.
$endgroup$
– copper.hat
Dec 10 '18 at 17:20
$begingroup$
How did you get to that conclusion? dθ is close to 0 shouldn't $sin({1over 2 }dθ) = 0$??
$endgroup$
– Mario Drouga
Dec 10 '18 at 17:59
$begingroup$
"$dtheta$ is close to $0$ shouldn't $sin(frac12dtheta)=0$?" That would be true if you believe that "$dtheta$ is close to $0$" means "$dtheta=0.$" But as long as $dthetaneq0,$ if $dtheta$ is small then $sin(frac12dtheta)$ will be much closer to $frac12dtheta$ than to $0.$
$endgroup$
– David K
Dec 10 '18 at 18:26
$begingroup$
@Mario $lim_{x to 0}frac{sin x}{x}=1$ answers your question in comment.
$endgroup$
– user376343
Dec 10 '18 at 18:39
1
1
$begingroup$
The angle is $rm{d}theta$ not $theta.$
$endgroup$
– saulspatz
Dec 10 '18 at 17:16
$begingroup$
The angle is $rm{d}theta$ not $theta.$
$endgroup$
– saulspatz
Dec 10 '18 at 17:16
1
1
$begingroup$
Informally, $sin ({1 over 2} d theta) = {1 over 2} d theta$.
$endgroup$
– copper.hat
Dec 10 '18 at 17:20
$begingroup$
Informally, $sin ({1 over 2} d theta) = {1 over 2} d theta$.
$endgroup$
– copper.hat
Dec 10 '18 at 17:20
$begingroup$
How did you get to that conclusion? dθ is close to 0 shouldn't $sin({1over 2 }dθ) = 0$??
$endgroup$
– Mario Drouga
Dec 10 '18 at 17:59
$begingroup$
How did you get to that conclusion? dθ is close to 0 shouldn't $sin({1over 2 }dθ) = 0$??
$endgroup$
– Mario Drouga
Dec 10 '18 at 17:59
$begingroup$
"$dtheta$ is close to $0$ shouldn't $sin(frac12dtheta)=0$?" That would be true if you believe that "$dtheta$ is close to $0$" means "$dtheta=0.$" But as long as $dthetaneq0,$ if $dtheta$ is small then $sin(frac12dtheta)$ will be much closer to $frac12dtheta$ than to $0.$
$endgroup$
– David K
Dec 10 '18 at 18:26
$begingroup$
"$dtheta$ is close to $0$ shouldn't $sin(frac12dtheta)=0$?" That would be true if you believe that "$dtheta$ is close to $0$" means "$dtheta=0.$" But as long as $dthetaneq0,$ if $dtheta$ is small then $sin(frac12dtheta)$ will be much closer to $frac12dtheta$ than to $0.$
$endgroup$
– David K
Dec 10 '18 at 18:26
$begingroup$
@Mario $lim_{x to 0}frac{sin x}{x}=1$ answers your question in comment.
$endgroup$
– user376343
Dec 10 '18 at 18:39
$begingroup$
@Mario $lim_{x to 0}frac{sin x}{x}=1$ answers your question in comment.
$endgroup$
– user376343
Dec 10 '18 at 18:39
add a comment |
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1
$begingroup$
The angle is $rm{d}theta$ not $theta.$
$endgroup$
– saulspatz
Dec 10 '18 at 17:16
1
$begingroup$
Informally, $sin ({1 over 2} d theta) = {1 over 2} d theta$.
$endgroup$
– copper.hat
Dec 10 '18 at 17:20
$begingroup$
How did you get to that conclusion? dθ is close to 0 shouldn't $sin({1over 2 }dθ) = 0$??
$endgroup$
– Mario Drouga
Dec 10 '18 at 17:59
$begingroup$
"$dtheta$ is close to $0$ shouldn't $sin(frac12dtheta)=0$?" That would be true if you believe that "$dtheta$ is close to $0$" means "$dtheta=0.$" But as long as $dthetaneq0,$ if $dtheta$ is small then $sin(frac12dtheta)$ will be much closer to $frac12dtheta$ than to $0.$
$endgroup$
– David K
Dec 10 '18 at 18:26
$begingroup$
@Mario $lim_{x to 0}frac{sin x}{x}=1$ answers your question in comment.
$endgroup$
– user376343
Dec 10 '18 at 18:39