Parametric curve length - calculus The Next CEO of Stack OverflowDetect “cusp” in...
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Parametric curve length - calculus
The Next CEO of Stack OverflowDetect “cusp” in parametric curveFind the length of the parametric curve (Difficult)Parametric curve parametriced by lengthCompute the length of a parametric curve.Arc Length parametric curveSampling a curve (parametric)Arc Length with Parametric EquationsFind the length of the parametric curveDetermine the length of the Parametric Curve given by the set of parametric equations.Length of a parametric curve formula: What does the integral represent?
$begingroup$
Find the length of the following parametric curve.
$x = 5 + frac92 t^3$, $y = 4 + 3 t^{frac92}$, $0 leq t leq 2$.
I used integration and after some point I got lost :( What are the steps?
calculus parametric
$endgroup$
add a comment |
$begingroup$
Find the length of the following parametric curve.
$x = 5 + frac92 t^3$, $y = 4 + 3 t^{frac92}$, $0 leq t leq 2$.
I used integration and after some point I got lost :( What are the steps?
calculus parametric
$endgroup$
$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
6 hours ago
add a comment |
$begingroup$
Find the length of the following parametric curve.
$x = 5 + frac92 t^3$, $y = 4 + 3 t^{frac92}$, $0 leq t leq 2$.
I used integration and after some point I got lost :( What are the steps?
calculus parametric
$endgroup$
Find the length of the following parametric curve.
$x = 5 + frac92 t^3$, $y = 4 + 3 t^{frac92}$, $0 leq t leq 2$.
I used integration and after some point I got lost :( What are the steps?
calculus parametric
calculus parametric
edited 41 mins ago
Peter Mortensen
565310
565310
asked 6 hours ago
McAMcA
204
204
$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
6 hours ago
add a comment |
$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
6 hours ago
$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
6 hours ago
$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
6 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Apply the formula for arc length, we get
$$
int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrt{v} dv = 78.
$$
New contributor
$endgroup$
add a comment |
$begingroup$
begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}
Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^{frac32}$ is an antiderivative of $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.
$endgroup$
add a comment |
$begingroup$
You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
$$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Apply the formula for arc length, we get
$$
int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrt{v} dv = 78.
$$
New contributor
$endgroup$
add a comment |
$begingroup$
Apply the formula for arc length, we get
$$
int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrt{v} dv = 78.
$$
New contributor
$endgroup$
add a comment |
$begingroup$
Apply the formula for arc length, we get
$$
int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrt{v} dv = 78.
$$
New contributor
$endgroup$
Apply the formula for arc length, we get
$$
int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrt{v} dv = 78.
$$
New contributor
New contributor
answered 6 hours ago
EagleToLearnEagleToLearn
233
233
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}
Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^{frac32}$ is an antiderivative of $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.
$endgroup$
add a comment |
$begingroup$
begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}
Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^{frac32}$ is an antiderivative of $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.
$endgroup$
add a comment |
$begingroup$
begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}
Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^{frac32}$ is an antiderivative of $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.
$endgroup$
begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}
Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^{frac32}$ is an antiderivative of $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.
edited 5 hours ago
answered 6 hours ago
Matt A PeltoMatt A Pelto
2,667621
2,667621
add a comment |
add a comment |
$begingroup$
You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
$$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$
$endgroup$
add a comment |
$begingroup$
You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
$$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$
$endgroup$
add a comment |
$begingroup$
You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
$$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$
$endgroup$
You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
$$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$
answered 6 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
78.2k42867
add a comment |
add a comment |
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$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
6 hours ago