Mixing
Let C be the graph of hypocycloid $x^{2/3} + y^{2/3} = 1$ oriented clockwise. Parametrize the curve and find...
$begingroup$
Let C be the graph of hypocycloid $x^{2/3} + y^{2/3} = 1$ oriented clockwise. Parametrize the curve and find its arc length.
Attempt:
let $x = sin^{3}t, y = cos^3({t})$
$r(t) = (sin^{3}(t), cos^3(t)), 0 le t le 2pi$
$r'(t) = (3 sin^2(t) cos(t), -3 cos^2(t) sin(t))$
begin{align*}
||r'(t)||
& = sqrt{3 sin^2(t) + cos(t))^2 + (3 cos^2(t) + sin(t)^2} \
& = sqrt{9 sin^4 + cos^2(t)+ (sin^2(t) + cos^2(t)) } \
& = sqrt{9 sin^2(t) + cos^2(t)}
= sqrt{(3 sin(t) cos(t))^2} \
& = 3 sin(t) cos(t)
= frac{3}{2} (2 sin(t) cos(t))
= frac{3}{2} sin(2t).
end{align*}
Now arc length:
$$
L = int_{0}^{2pi} frac{3}{2} sin(2t)dt
= 4 cdot frac{3}{2} cdot frac{1}{2}bigg[-cos(2t) bigg]_{0}^{2pi}
= 3(1+1)
= 6.$$
Is this correct?
multivariable-calculus parametrization
edited Jan 21 at 0:10
Viktor Glombik
1,0031527
asked Jan 20 at 23:11
first n lastnfirst n lastn
61
$endgroup$
add a comment |
$begingroup$
Let C be the graph of hypocycloid $x^{2/3} + y^{2/3} = 1$ oriented clockwise. Parametrize the curve and find its arc length.
Attempt:
let $x = sin^{3}t, y = cos^3({t})$
$r(t) = (sin^{3}(t), cos^3(t)), 0 le t le 2pi$
$r'(t) = (3 sin^2(t) cos(t), -3 cos^2(t) sin(t))$
begin{align*}
||r'(t)||
& = sqrt{3 sin^2(t) + cos(t))^2 + (3 cos^2(t) + sin(t)^2} \
& = sqrt{9 sin^4 + cos^2(t)+ (sin^2(t) + cos^2(t)) } \
& = sqrt{9 sin^2(t) + cos^2(t)}
= sqrt{(3 sin(t) cos(t))^2} \
& = 3 sin(t) cos(t)
= frac{3}{2} (2 sin(t) cos(t))
= frac{3}{2} sin(2t).
end{align*}
Now arc length:
$$
L = int_{0}^{2pi} frac{3}{2} sin(2t)dt
= 4 cdot frac{3}{2} cdot frac{1}{2}bigg[-cos(2t) bigg]_{0}^{2pi}
= 3(1+1)
= 6.$$
Is this correct?
multivariable-calculus parametrization
edited Jan 21 at 0:10
Viktor Glombik
1,0031527
asked Jan 20 at 23:11
first n lastnfirst n lastn
61
$endgroup$
$begingroup$
I'm not getting how you are calculating $||r'(t)||$
$endgroup$
– user289143
Jan 20 at 23:17
$begingroup$
Starting with $||r'(t)||$ there are typos or mistakes: fix your $+$, some of them should be multiplications. Mismatched parenthesis ...
$endgroup$
– user376343
Jan 20 at 23:18
add a comment |
$begingroup$
Let C be the graph of hypocycloid $x^{2/3} + y^{2/3} = 1$ oriented clockwise. Parametrize the curve and find its arc length.
Attempt:
let $x = sin^{3}t, y = cos^3({t})$
$r(t) = (sin^{3}(t), cos^3(t)), 0 le t le 2pi$
$r'(t) = (3 sin^2(t) cos(t), -3 cos^2(t) sin(t))$
begin{align*}
||r'(t)||
& = sqrt{3 sin^2(t) + cos(t))^2 + (3 cos^2(t) + sin(t)^2} \
& = sqrt{9 sin^4 + cos^2(t)+ (sin^2(t) + cos^2(t)) } \
& = sqrt{9 sin^2(t) + cos^2(t)}
= sqrt{(3 sin(t) cos(t))^2} \
& = 3 sin(t) cos(t)
= frac{3}{2} (2 sin(t) cos(t))
= frac{3}{2} sin(2t).
end{align*}
Now arc length:
$$
L = int_{0}^{2pi} frac{3}{2} sin(2t)dt
= 4 cdot frac{3}{2} cdot frac{1}{2}bigg[-cos(2t) bigg]_{0}^{2pi}
= 3(1+1)
= 6.$$
Is this correct?
multivariable-calculus parametrization
edited Jan 21 at 0:10
Viktor Glombik
1,0031527
asked Jan 20 at 23:11
first n lastnfirst n lastn
61
$endgroup$
Let C be the graph of hypocycloid $x^{2/3} + y^{2/3} = 1$ oriented clockwise. Parametrize the curve and find its arc length.
Attempt:
let $x = sin^{3}t, y = cos^3({t})$
$r(t) = (sin^{3}(t), cos^3(t)), 0 le t le 2pi$
$r'(t) = (3 sin^2(t) cos(t), -3 cos^2(t) sin(t))$
begin{align*}
||r'(t)||
& = sqrt{3 sin^2(t) + cos(t))^2 + (3 cos^2(t) + sin(t)^2} \
& = sqrt{9 sin^4 + cos^2(t)+ (sin^2(t) + cos^2(t)) } \
& = sqrt{9 sin^2(t) + cos^2(t)}
= sqrt{(3 sin(t) cos(t))^2} \
& = 3 sin(t) cos(t)
= frac{3}{2} (2 sin(t) cos(t))
= frac{3}{2} sin(2t).
end{align*}
Now arc length:
$$
L = int_{0}^{2pi} frac{3}{2} sin(2t)dt
= 4 cdot frac{3}{2} cdot frac{1}{2}bigg[-cos(2t) bigg]_{0}^{2pi}
= 3(1+1)
= 6.$$
Is this correct?
multivariable-calculus parametrization
multivariable-calculus parametrization
edited Jan 21 at 0:10
Viktor Glombik
1,0031527
asked Jan 20 at 23:11
first n lastnfirst n lastn
61
edited Jan 21 at 0:10
Viktor Glombik
1,0031527
asked Jan 20 at 23:11
first n lastnfirst n lastn
61
edited Jan 21 at 0:10
Viktor Glombik
1,0031527
edited Jan 21 at 0:10
Viktor Glombik
1,0031527
edited Jan 21 at 0:10
Viktor Glombik
1,0031527
1,0031527
asked Jan 20 at 23:11
first n lastnfirst n lastn
61
asked Jan 20 at 23:11
first n lastnfirst n lastn
61
asked Jan 20 at 23:11
first n lastnfirst n lastn
61
61
$begingroup$
I'm not getting how you are calculating $||r'(t)||$
$endgroup$
– user289143
Jan 20 at 23:17
$begingroup$
Starting with $||r'(t)||$ there are typos or mistakes: fix your $+$, some of them should be multiplications. Mismatched parenthesis ...
$endgroup$
– user376343
Jan 20 at 23:18
add a comment |
$begingroup$
I'm not getting how you are calculating $||r'(t)||$
$endgroup$
– user289143
Jan 20 at 23:17
$begingroup$
Starting with $||r'(t)||$ there are typos or mistakes: fix your $+$, some of them should be multiplications. Mismatched parenthesis ...
$endgroup$
– user376343
Jan 20 at 23:18
$begingroup$
I'm not getting how you are calculating $||r'(t)||$
$endgroup$
– user289143
Jan 20 at 23:17
$begingroup$
I'm not getting how you are calculating $||r'(t)||$
$endgroup$
– user289143
Jan 20 at 23:17
$begingroup$
Starting with $||r'(t)||$ there are typos or mistakes: fix your $+$, some of them should be multiplications. Mismatched parenthesis ...
$endgroup$
– user376343
Jan 20 at 23:18
$begingroup$
Starting with $||r'(t)||$ there are typos or mistakes: fix your $+$, some of them should be multiplications. Mismatched parenthesis ...
$endgroup$
– user376343
Jan 20 at 23:18
add a comment |
1 Answer
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oldest
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$begingroup$
Now $||r'(t)||=sqrt{(3 sin^2(t) cos (t))^2+(-3 cos^2(t) sin (t))^2}=sqrt{9sin^4(t)cos^2(t)+9cos^4(t)sin^2(t)}=sqrt{9sin^2(t)cos^2(t)(sin^2(t)+cos^2(t)}=sqrt{9sin^2(t)cos^2(t)}=|3sin(t)cos(t)|=|frac{3}{2}sin(2t)|$
$$L=int_0^{2pi}frac{3}{2} |sin(2t)| dt=frac{3}{2}int_0^{2pi}|sin(2t)| dt=frac{3}{2}big(int_0^{frac{pi}{2}}sin(2t) dt-int_{frac{pi}{2}}^{pi}sin(2t) dt+int_pi^{frac{3pi}{2}}sin(2t) dt-int_{frac{3pi}{2}}^{2pi}sin(2t) dtbig)=frac{3}{2}cdot frac{1}{2}[(-cos(2t))_0^{ frac{pi}{2}}+(cos(2t))_{frac{pi}{2}}^{ pi}+(-cos(2t))_{pi}^{frac{ 3pi}{2}}+(cos(2t))_{frac{3pi}{2}}^{2 pi}]=frac{3}{4}(1+1+1+1+1+1+1+1)=6$$
answered Jan 20 at 23:41
user289143user289143
1,002313
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Now $||r'(t)||=sqrt{(3 sin^2(t) cos (t))^2+(-3 cos^2(t) sin (t))^2}=sqrt{9sin^4(t)cos^2(t)+9cos^4(t)sin^2(t)}=sqrt{9sin^2(t)cos^2(t)(sin^2(t)+cos^2(t)}=sqrt{9sin^2(t)cos^2(t)}=|3sin(t)cos(t)|=|frac{3}{2}sin(2t)|$
$$L=int_0^{2pi}frac{3}{2} |sin(2t)| dt=frac{3}{2}int_0^{2pi}|sin(2t)| dt=frac{3}{2}big(int_0^{frac{pi}{2}}sin(2t) dt-int_{frac{pi}{2}}^{pi}sin(2t) dt+int_pi^{frac{3pi}{2}}sin(2t) dt-int_{frac{3pi}{2}}^{2pi}sin(2t) dtbig)=frac{3}{2}cdot frac{1}{2}[(-cos(2t))_0^{ frac{pi}{2}}+(cos(2t))_{frac{pi}{2}}^{ pi}+(-cos(2t))_{pi}^{frac{ 3pi}{2}}+(cos(2t))_{frac{3pi}{2}}^{2 pi}]=frac{3}{4}(1+1+1+1+1+1+1+1)=6$$
answered Jan 20 at 23:41
user289143user289143
1,002313
$endgroup$
add a comment |
$begingroup$
Now $||r'(t)||=sqrt{(3 sin^2(t) cos (t))^2+(-3 cos^2(t) sin (t))^2}=sqrt{9sin^4(t)cos^2(t)+9cos^4(t)sin^2(t)}=sqrt{9sin^2(t)cos^2(t)(sin^2(t)+cos^2(t)}=sqrt{9sin^2(t)cos^2(t)}=|3sin(t)cos(t)|=|frac{3}{2}sin(2t)|$
$$L=int_0^{2pi}frac{3}{2} |sin(2t)| dt=frac{3}{2}int_0^{2pi}|sin(2t)| dt=frac{3}{2}big(int_0^{frac{pi}{2}}sin(2t) dt-int_{frac{pi}{2}}^{pi}sin(2t) dt+int_pi^{frac{3pi}{2}}sin(2t) dt-int_{frac{3pi}{2}}^{2pi}sin(2t) dtbig)=frac{3}{2}cdot frac{1}{2}[(-cos(2t))_0^{ frac{pi}{2}}+(cos(2t))_{frac{pi}{2}}^{ pi}+(-cos(2t))_{pi}^{frac{ 3pi}{2}}+(cos(2t))_{frac{3pi}{2}}^{2 pi}]=frac{3}{4}(1+1+1+1+1+1+1+1)=6$$
answered Jan 20 at 23:41
user289143user289143
1,002313
$endgroup$
add a comment |
$begingroup$
Now $||r'(t)||=sqrt{(3 sin^2(t) cos (t))^2+(-3 cos^2(t) sin (t))^2}=sqrt{9sin^4(t)cos^2(t)+9cos^4(t)sin^2(t)}=sqrt{9sin^2(t)cos^2(t)(sin^2(t)+cos^2(t)}=sqrt{9sin^2(t)cos^2(t)}=|3sin(t)cos(t)|=|frac{3}{2}sin(2t)|$
$$L=int_0^{2pi}frac{3}{2} |sin(2t)| dt=frac{3}{2}int_0^{2pi}|sin(2t)| dt=frac{3}{2}big(int_0^{frac{pi}{2}}sin(2t) dt-int_{frac{pi}{2}}^{pi}sin(2t) dt+int_pi^{frac{3pi}{2}}sin(2t) dt-int_{frac{3pi}{2}}^{2pi}sin(2t) dtbig)=frac{3}{2}cdot frac{1}{2}[(-cos(2t))_0^{ frac{pi}{2}}+(cos(2t))_{frac{pi}{2}}^{ pi}+(-cos(2t))_{pi}^{frac{ 3pi}{2}}+(cos(2t))_{frac{3pi}{2}}^{2 pi}]=frac{3}{4}(1+1+1+1+1+1+1+1)=6$$
answered Jan 20 at 23:41
user289143user289143
1,002313
$endgroup$
Now $||r'(t)||=sqrt{(3 sin^2(t) cos (t))^2+(-3 cos^2(t) sin (t))^2}=sqrt{9sin^4(t)cos^2(t)+9cos^4(t)sin^2(t)}=sqrt{9sin^2(t)cos^2(t)(sin^2(t)+cos^2(t)}=sqrt{9sin^2(t)cos^2(t)}=|3sin(t)cos(t)|=|frac{3}{2}sin(2t)|$
$$L=int_0^{2pi}frac{3}{2} |sin(2t)| dt=frac{3}{2}int_0^{2pi}|sin(2t)| dt=frac{3}{2}big(int_0^{frac{pi}{2}}sin(2t) dt-int_{frac{pi}{2}}^{pi}sin(2t) dt+int_pi^{frac{3pi}{2}}sin(2t) dt-int_{frac{3pi}{2}}^{2pi}sin(2t) dtbig)=frac{3}{2}cdot frac{1}{2}[(-cos(2t))_0^{ frac{pi}{2}}+(cos(2t))_{frac{pi}{2}}^{ pi}+(-cos(2t))_{pi}^{frac{ 3pi}{2}}+(cos(2t))_{frac{3pi}{2}}^{2 pi}]=frac{3}{4}(1+1+1+1+1+1+1+1)=6$$
answered Jan 20 at 23:41
user289143user289143
1,002313
answered Jan 20 at 23:41
user289143user289143
1,002313
answered Jan 20 at 23:41
user289143user289143
1,002313
answered Jan 20 at 23:41
user289143user289143
1,002313
1,002313
add a comment |
add a comment |
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$begingroup$
I'm not getting how you are calculating $||r'(t)||$
$endgroup$
– user289143
Jan 20 at 23:17
$begingroup$
Starting with $||r'(t)||$ there are typos or mistakes: fix your $+$, some of them should be multiplications. Mismatched parenthesis ...
$endgroup$
– user376343
Jan 20 at 23:18