Determining the length of a curve using integrals [closed]
$begingroup$
Im having problems solving the following problem.
Determine the distance of the following curve:
begin{cases}
x(t)= cos^3(t),\
y(t)= sin^3(t). \
end{cases}
In this case $0 le t le 2pi$
The answer should be $6$
I would be really grateful if someone could of help me understand it.
integration
edited Dec 26 '18 at 7:28
Dylan
14.3k31127
asked Dec 25 '18 at 22:03
FosorfFosorf
63
$endgroup$
closed as off-topic by Namaste, Eevee Trainer, Paul Frost, KReiser, Lord_Farin Dec 26 '18 at 9:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Eevee Trainer, Paul Frost, KReiser, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Im having problems solving the following problem.
Determine the distance of the following curve:
begin{cases}
x(t)= cos^3(t),\
y(t)= sin^3(t). \
end{cases}
In this case $0 le t le 2pi$
The answer should be $6$
I would be really grateful if someone could of help me understand it.
integration
edited Dec 26 '18 at 7:28
Dylan
14.3k31127
asked Dec 25 '18 at 22:03
FosorfFosorf
63
$endgroup$
closed as off-topic by Namaste, Eevee Trainer, Paul Frost, KReiser, Lord_Farin Dec 26 '18 at 9:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Eevee Trainer, Paul Frost, KReiser, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Im having problems solving the following problem.
Determine the distance of the following curve:
begin{cases}
x(t)= cos^3(t),\
y(t)= sin^3(t). \
end{cases}
In this case $0 le t le 2pi$
The answer should be $6$
I would be really grateful if someone could of help me understand it.
integration
edited Dec 26 '18 at 7:28
Dylan
14.3k31127
asked Dec 25 '18 at 22:03
FosorfFosorf
63
$endgroup$
Im having problems solving the following problem.
Determine the distance of the following curve:
begin{cases}
x(t)= cos^3(t),\
y(t)= sin^3(t). \
end{cases}
In this case $0 le t le 2pi$
The answer should be $6$
I would be really grateful if someone could of help me understand it.
integration
integration
edited Dec 26 '18 at 7:28
Dylan
14.3k31127
asked Dec 25 '18 at 22:03
FosorfFosorf
63
edited Dec 26 '18 at 7:28
Dylan
14.3k31127
asked Dec 25 '18 at 22:03
FosorfFosorf
63
edited Dec 26 '18 at 7:28
Dylan
14.3k31127
edited Dec 26 '18 at 7:28
Dylan
14.3k31127
edited Dec 26 '18 at 7:28
Dylan
14.3k31127
14.3k31127
asked Dec 25 '18 at 22:03
FosorfFosorf
63
asked Dec 25 '18 at 22:03
FosorfFosorf
63
asked Dec 25 '18 at 22:03
FosorfFosorf
63
63
closed as off-topic by Namaste, Eevee Trainer, Paul Frost, KReiser, Lord_Farin Dec 26 '18 at 9:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Eevee Trainer, Paul Frost, KReiser, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, Eevee Trainer, Paul Frost, KReiser, Lord_Farin Dec 26 '18 at 9:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Eevee Trainer, Paul Frost, KReiser, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The length of the curve, known as arc length, is given by
$$L = int_{0}^{2pi}sqrt{x'(t)^{2}+y'(t)^{2}} dt$$
In this case, this gives
begin{align}
L&=int_{0}^{2pi}sqrt{(3cos^{2}{t}(-sin{t}))^{2}+(3sin^{2}{t}(cos{t}))^{2}} dt \
&= int_{0}^{2pi}sqrt{9cos^{4}{t}sin^{2}{t}+9sin^{4}{t}cos^{2}{t}} dt\
&= int_{0}^{2pi}sqrt{9cos^{2}{t}sin^{2}{t}(sin^{2}{t}+cos^{2}{t})} dt\
&= int_{0}^{2pi}3|cos{t}sin{t}| dt\
&= int_{0}^{2pi}frac{3}{2}|sin{2t}| dt
end{align}
You can easily verify that $sin{2t}
geq 0$ when $tin[0,frac{pi}{2}]cup[pi,frac{3pi}{2}]$ and that $sin{2t}leq0$ when $tin[frac{pi}{2},pi]cup[frac{3pi}{2},2pi].$ Therefore
$$L=frac{3}{2}left[int_{0}^{pi/2}sin{2t} dt + int_{pi}^{3pi/2}sin{2t} dt - int_{pi/2}^{pi}sin{2t} dt - int_{3pi/2}^{2pi}sin{2t} dt right]$$
You can also check that the first two integrals are $1$ and the last two are $-1$. Therefore
$$L=frac{3}{2}(4)=6$$
answered Dec 25 '18 at 22:22
pwerthpwerth
3,340417
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If you don't know the formula imagine a little particle moving along this path. Then at each small slice of time, the particle moves its total speed times time. That is, if the particle's path is described a vector-valued function of time $$overrightarrow{r}(t) = left langle begin{matrix} cos ^3 (t) \ sin^3 (t)end{matrix} right rangle $$ then we have that $$ text{d}s = | r'(t) | text{d}t $$ so that $$s = int_{t_1}^{t_2} sqrt{(x'(t))^2 + (y'(t))^2} text{d}t$$ From here on, the solution is identical to pwerth's excellent derivation, so in your case the arc length is $$ s= int_0^{2pi}sqrt{left (frac{text{d}}{text{d}x} cos^3 (t) right)^2 + left (frac{text{d}}{text{d}x} sin^3 (t) right)^2 } text{d}t $$ $$= int_0^{2pi} sqrt{left(3 cos^2(t) cdot (- sin(t)) right)^2 + left(3 sin^2(t) cos(t) right)^2} text{d}t $$ $$= int_0^{2pi} sqrt{9 cos^4(t)sin^2(t) + 9 sin^4(t) cos^2(t)} text{d}t $$ $$=int_0^{2pi} |3sin(t) cos(t)| sqrt{sin^2(t) + cos^2(t)} text{d}t $$ $$ = 3 int_0^{2pi} frac{1}{2} |sin(2t)| text{d}t = frac{3}{2} cdot 4 = 6 $$
answered Dec 25 '18 at 22:27
Cade ReinbergerCade Reinberger
38317
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The length of the curve, known as arc length, is given by
$$L = int_{0}^{2pi}sqrt{x'(t)^{2}+y'(t)^{2}} dt$$
In this case, this gives
begin{align}
L&=int_{0}^{2pi}sqrt{(3cos^{2}{t}(-sin{t}))^{2}+(3sin^{2}{t}(cos{t}))^{2}} dt \
&= int_{0}^{2pi}sqrt{9cos^{4}{t}sin^{2}{t}+9sin^{4}{t}cos^{2}{t}} dt\
&= int_{0}^{2pi}sqrt{9cos^{2}{t}sin^{2}{t}(sin^{2}{t}+cos^{2}{t})} dt\
&= int_{0}^{2pi}3|cos{t}sin{t}| dt\
&= int_{0}^{2pi}frac{3}{2}|sin{2t}| dt
end{align}
You can easily verify that $sin{2t}
geq 0$ when $tin[0,frac{pi}{2}]cup[pi,frac{3pi}{2}]$ and that $sin{2t}leq0$ when $tin[frac{pi}{2},pi]cup[frac{3pi}{2},2pi].$ Therefore
$$L=frac{3}{2}left[int_{0}^{pi/2}sin{2t} dt + int_{pi}^{3pi/2}sin{2t} dt - int_{pi/2}^{pi}sin{2t} dt - int_{3pi/2}^{2pi}sin{2t} dt right]$$
You can also check that the first two integrals are $1$ and the last two are $-1$. Therefore
$$L=frac{3}{2}(4)=6$$
answered Dec 25 '18 at 22:22
pwerthpwerth
3,340417
$endgroup$
add a comment |
$begingroup$
The length of the curve, known as arc length, is given by
$$L = int_{0}^{2pi}sqrt{x'(t)^{2}+y'(t)^{2}} dt$$
In this case, this gives
begin{align}
L&=int_{0}^{2pi}sqrt{(3cos^{2}{t}(-sin{t}))^{2}+(3sin^{2}{t}(cos{t}))^{2}} dt \
&= int_{0}^{2pi}sqrt{9cos^{4}{t}sin^{2}{t}+9sin^{4}{t}cos^{2}{t}} dt\
&= int_{0}^{2pi}sqrt{9cos^{2}{t}sin^{2}{t}(sin^{2}{t}+cos^{2}{t})} dt\
&= int_{0}^{2pi}3|cos{t}sin{t}| dt\
&= int_{0}^{2pi}frac{3}{2}|sin{2t}| dt
end{align}
You can easily verify that $sin{2t}
geq 0$ when $tin[0,frac{pi}{2}]cup[pi,frac{3pi}{2}]$ and that $sin{2t}leq0$ when $tin[frac{pi}{2},pi]cup[frac{3pi}{2},2pi].$ Therefore
$$L=frac{3}{2}left[int_{0}^{pi/2}sin{2t} dt + int_{pi}^{3pi/2}sin{2t} dt - int_{pi/2}^{pi}sin{2t} dt - int_{3pi/2}^{2pi}sin{2t} dt right]$$
You can also check that the first two integrals are $1$ and the last two are $-1$. Therefore
$$L=frac{3}{2}(4)=6$$
answered Dec 25 '18 at 22:22
pwerthpwerth
3,340417
$endgroup$
add a comment |
$begingroup$
The length of the curve, known as arc length, is given by
$$L = int_{0}^{2pi}sqrt{x'(t)^{2}+y'(t)^{2}} dt$$
In this case, this gives
begin{align}
L&=int_{0}^{2pi}sqrt{(3cos^{2}{t}(-sin{t}))^{2}+(3sin^{2}{t}(cos{t}))^{2}} dt \
&= int_{0}^{2pi}sqrt{9cos^{4}{t}sin^{2}{t}+9sin^{4}{t}cos^{2}{t}} dt\
&= int_{0}^{2pi}sqrt{9cos^{2}{t}sin^{2}{t}(sin^{2}{t}+cos^{2}{t})} dt\
&= int_{0}^{2pi}3|cos{t}sin{t}| dt\
&= int_{0}^{2pi}frac{3}{2}|sin{2t}| dt
end{align}
You can easily verify that $sin{2t}
geq 0$ when $tin[0,frac{pi}{2}]cup[pi,frac{3pi}{2}]$ and that $sin{2t}leq0$ when $tin[frac{pi}{2},pi]cup[frac{3pi}{2},2pi].$ Therefore
$$L=frac{3}{2}left[int_{0}^{pi/2}sin{2t} dt + int_{pi}^{3pi/2}sin{2t} dt - int_{pi/2}^{pi}sin{2t} dt - int_{3pi/2}^{2pi}sin{2t} dt right]$$
You can also check that the first two integrals are $1$ and the last two are $-1$. Therefore
$$L=frac{3}{2}(4)=6$$
answered Dec 25 '18 at 22:22
pwerthpwerth
3,340417
$endgroup$
The length of the curve, known as arc length, is given by
$$L = int_{0}^{2pi}sqrt{x'(t)^{2}+y'(t)^{2}} dt$$
In this case, this gives
begin{align}
L&=int_{0}^{2pi}sqrt{(3cos^{2}{t}(-sin{t}))^{2}+(3sin^{2}{t}(cos{t}))^{2}} dt \
&= int_{0}^{2pi}sqrt{9cos^{4}{t}sin^{2}{t}+9sin^{4}{t}cos^{2}{t}} dt\
&= int_{0}^{2pi}sqrt{9cos^{2}{t}sin^{2}{t}(sin^{2}{t}+cos^{2}{t})} dt\
&= int_{0}^{2pi}3|cos{t}sin{t}| dt\
&= int_{0}^{2pi}frac{3}{2}|sin{2t}| dt
end{align}
You can easily verify that $sin{2t}
geq 0$ when $tin[0,frac{pi}{2}]cup[pi,frac{3pi}{2}]$ and that $sin{2t}leq0$ when $tin[frac{pi}{2},pi]cup[frac{3pi}{2},2pi].$ Therefore
$$L=frac{3}{2}left[int_{0}^{pi/2}sin{2t} dt + int_{pi}^{3pi/2}sin{2t} dt - int_{pi/2}^{pi}sin{2t} dt - int_{3pi/2}^{2pi}sin{2t} dt right]$$
You can also check that the first two integrals are $1$ and the last two are $-1$. Therefore
$$L=frac{3}{2}(4)=6$$
answered Dec 25 '18 at 22:22
pwerthpwerth
3,340417
answered Dec 25 '18 at 22:22
pwerthpwerth
3,340417
answered Dec 25 '18 at 22:22
pwerthpwerth
3,340417
answered Dec 25 '18 at 22:22
pwerthpwerth
3,340417
3,340417
add a comment |
add a comment |
$begingroup$
If you don't know the formula imagine a little particle moving along this path. Then at each small slice of time, the particle moves its total speed times time. That is, if the particle's path is described a vector-valued function of time $$overrightarrow{r}(t) = left langle begin{matrix} cos ^3 (t) \ sin^3 (t)end{matrix} right rangle $$ then we have that $$ text{d}s = | r'(t) | text{d}t $$ so that $$s = int_{t_1}^{t_2} sqrt{(x'(t))^2 + (y'(t))^2} text{d}t$$ From here on, the solution is identical to pwerth's excellent derivation, so in your case the arc length is $$ s= int_0^{2pi}sqrt{left (frac{text{d}}{text{d}x} cos^3 (t) right)^2 + left (frac{text{d}}{text{d}x} sin^3 (t) right)^2 } text{d}t $$ $$= int_0^{2pi} sqrt{left(3 cos^2(t) cdot (- sin(t)) right)^2 + left(3 sin^2(t) cos(t) right)^2} text{d}t $$ $$= int_0^{2pi} sqrt{9 cos^4(t)sin^2(t) + 9 sin^4(t) cos^2(t)} text{d}t $$ $$=int_0^{2pi} |3sin(t) cos(t)| sqrt{sin^2(t) + cos^2(t)} text{d}t $$ $$ = 3 int_0^{2pi} frac{1}{2} |sin(2t)| text{d}t = frac{3}{2} cdot 4 = 6 $$
answered Dec 25 '18 at 22:27
Cade ReinbergerCade Reinberger
38317
$endgroup$
add a comment |
$begingroup$
If you don't know the formula imagine a little particle moving along this path. Then at each small slice of time, the particle moves its total speed times time. That is, if the particle's path is described a vector-valued function of time $$overrightarrow{r}(t) = left langle begin{matrix} cos ^3 (t) \ sin^3 (t)end{matrix} right rangle $$ then we have that $$ text{d}s = | r'(t) | text{d}t $$ so that $$s = int_{t_1}^{t_2} sqrt{(x'(t))^2 + (y'(t))^2} text{d}t$$ From here on, the solution is identical to pwerth's excellent derivation, so in your case the arc length is $$ s= int_0^{2pi}sqrt{left (frac{text{d}}{text{d}x} cos^3 (t) right)^2 + left (frac{text{d}}{text{d}x} sin^3 (t) right)^2 } text{d}t $$ $$= int_0^{2pi} sqrt{left(3 cos^2(t) cdot (- sin(t)) right)^2 + left(3 sin^2(t) cos(t) right)^2} text{d}t $$ $$= int_0^{2pi} sqrt{9 cos^4(t)sin^2(t) + 9 sin^4(t) cos^2(t)} text{d}t $$ $$=int_0^{2pi} |3sin(t) cos(t)| sqrt{sin^2(t) + cos^2(t)} text{d}t $$ $$ = 3 int_0^{2pi} frac{1}{2} |sin(2t)| text{d}t = frac{3}{2} cdot 4 = 6 $$
answered Dec 25 '18 at 22:27
Cade ReinbergerCade Reinberger
38317
$endgroup$
add a comment |
$begingroup$
If you don't know the formula imagine a little particle moving along this path. Then at each small slice of time, the particle moves its total speed times time. That is, if the particle's path is described a vector-valued function of time $$overrightarrow{r}(t) = left langle begin{matrix} cos ^3 (t) \ sin^3 (t)end{matrix} right rangle $$ then we have that $$ text{d}s = | r'(t) | text{d}t $$ so that $$s = int_{t_1}^{t_2} sqrt{(x'(t))^2 + (y'(t))^2} text{d}t$$ From here on, the solution is identical to pwerth's excellent derivation, so in your case the arc length is $$ s= int_0^{2pi}sqrt{left (frac{text{d}}{text{d}x} cos^3 (t) right)^2 + left (frac{text{d}}{text{d}x} sin^3 (t) right)^2 } text{d}t $$ $$= int_0^{2pi} sqrt{left(3 cos^2(t) cdot (- sin(t)) right)^2 + left(3 sin^2(t) cos(t) right)^2} text{d}t $$ $$= int_0^{2pi} sqrt{9 cos^4(t)sin^2(t) + 9 sin^4(t) cos^2(t)} text{d}t $$ $$=int_0^{2pi} |3sin(t) cos(t)| sqrt{sin^2(t) + cos^2(t)} text{d}t $$ $$ = 3 int_0^{2pi} frac{1}{2} |sin(2t)| text{d}t = frac{3}{2} cdot 4 = 6 $$
answered Dec 25 '18 at 22:27
Cade ReinbergerCade Reinberger
38317
$endgroup$
If you don't know the formula imagine a little particle moving along this path. Then at each small slice of time, the particle moves its total speed times time. That is, if the particle's path is described a vector-valued function of time $$overrightarrow{r}(t) = left langle begin{matrix} cos ^3 (t) \ sin^3 (t)end{matrix} right rangle $$ then we have that $$ text{d}s = | r'(t) | text{d}t $$ so that $$s = int_{t_1}^{t_2} sqrt{(x'(t))^2 + (y'(t))^2} text{d}t$$ From here on, the solution is identical to pwerth's excellent derivation, so in your case the arc length is $$ s= int_0^{2pi}sqrt{left (frac{text{d}}{text{d}x} cos^3 (t) right)^2 + left (frac{text{d}}{text{d}x} sin^3 (t) right)^2 } text{d}t $$ $$= int_0^{2pi} sqrt{left(3 cos^2(t) cdot (- sin(t)) right)^2 + left(3 sin^2(t) cos(t) right)^2} text{d}t $$ $$= int_0^{2pi} sqrt{9 cos^4(t)sin^2(t) + 9 sin^4(t) cos^2(t)} text{d}t $$ $$=int_0^{2pi} |3sin(t) cos(t)| sqrt{sin^2(t) + cos^2(t)} text{d}t $$ $$ = 3 int_0^{2pi} frac{1}{2} |sin(2t)| text{d}t = frac{3}{2} cdot 4 = 6 $$
answered Dec 25 '18 at 22:27
Cade ReinbergerCade Reinberger
38317
answered Dec 25 '18 at 22:27
Cade ReinbergerCade Reinberger
38317
answered Dec 25 '18 at 22:27
Cade ReinbergerCade Reinberger
38317
answered Dec 25 '18 at 22:27
Cade ReinbergerCade Reinberger
38317
38317
add a comment |
add a comment |
