calculus parametric curve length The Next CEO of Stack OverflowFind the length of the parametric curve (Difficult)Parametric Curve Tangent EquationsParametric 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 curveTransforming quadratic parametric curve to implicit formLength of a parametric curve formula: What does the integral represent?
Would this house-rule that treats advantage as a +1 to the roll instead (and disadvantage as -1) and allows them to stack be balanced?
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calculus parametric curve length
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calculus parametric curve length
The Next CEO of Stack OverflowFind the length of the parametric curve (Difficult)Parametric Curve Tangent EquationsParametric 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 curveTransforming quadratic parametric curve to implicit formLength 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 :( Can anyone show me the steps?
calculus parametric
edited 4 hours ago
Matt A Pelto
2,667621
asked 4 hours ago
McAMcA
224
$endgroup$
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 :( Can anyone show me the steps?
calculus parametric
edited 4 hours ago
Matt A Pelto
2,667621
asked 4 hours ago
McAMcA
224
$endgroup$
$begingroup$
Is this $$x=5+frac92t^3,y=4+3t^9/2$$?
$endgroup$
– Dr. Sonnhard Graubner
4 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 :( Can anyone show me the steps?
calculus parametric
edited 4 hours ago
Matt A Pelto
2,667621
asked 4 hours ago
McAMcA
224
$endgroup$
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 :( Can anyone show me the steps?
calculus parametric
calculus parametric
edited 4 hours ago
Matt A Pelto
2,667621
asked 4 hours ago
McAMcA
224
edited 4 hours ago
Matt A Pelto
2,667621
asked 4 hours ago
McAMcA
224
edited 4 hours ago
Matt A Pelto
2,667621
edited 4 hours ago
Matt A Pelto
2,667621
edited 4 hours ago
Matt A Pelto
2,667621
2,667621
asked 4 hours ago
McAMcA
224
asked 4 hours ago
McAMcA
224
asked 4 hours ago
McAMcA
224
224
$begingroup$
Is this $$x=5+frac92t^3,y=4+3t^9/2$$?
$endgroup$
– Dr. Sonnhard Graubner
4 hours ago
add a comment |
$begingroup$
Is this $$x=5+frac92t^3,y=4+3t^9/2$$?
$endgroup$
– Dr. Sonnhard Graubner
4 hours ago
$begingroup$
Is this $$x=5+frac92t^3,y=4+3t^9/2$$?
$endgroup$
– Dr. Sonnhard Graubner
4 hours ago
$begingroup$
Is this $$x=5+frac92t^3,y=4+3t^9/2$$?
$endgroup$
– Dr. Sonnhard Graubner
4 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Apply the formula for arc length, we get
$$
int_0^2 frac27t^2,sqrtt^3+12 dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrtv dv = 78.
$$
answered 4 hours ago
EagleToLearnEagleToLearn
233
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned
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 $f(t)=frac272t^2(1+t^3)^frac12$.
answered 4 hours ago
Matt A PeltoMatt A Pelto
2,667621
$endgroup$
add a comment |
$begingroup$
You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
$$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$
answered 4 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
add a comment |
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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
$$
int_0^2 frac27t^2,sqrtt^3+12 dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrtv dv = 78.
$$
answered 4 hours ago
EagleToLearnEagleToLearn
233
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Apply the formula for arc length, we get
$$
int_0^2 frac27t^2,sqrtt^3+12 dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrtv dv = 78.
$$
answered 4 hours ago
EagleToLearnEagleToLearn
233
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Apply the formula for arc length, we get
$$
int_0^2 frac27t^2,sqrtt^3+12 dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrtv dv = 78.
$$
answered 4 hours ago
EagleToLearnEagleToLearn
233
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Apply the formula for arc length, we get
$$
int_0^2 frac27t^2,sqrtt^3+12 dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrtv dv = 78.
$$
answered 4 hours ago
EagleToLearnEagleToLearn
233
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
EagleToLearnEagleToLearn
233
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
EagleToLearnEagleToLearn
233
answered 4 hours ago
EagleToLearnEagleToLearn
233
233
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned
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 $f(t)=frac272t^2(1+t^3)^frac12$.
answered 4 hours ago
Matt A PeltoMatt A Pelto
2,667621
$endgroup$
add a comment |
$begingroup$
beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned
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 $f(t)=frac272t^2(1+t^3)^frac12$.
answered 4 hours ago
Matt A PeltoMatt A Pelto
2,667621
$endgroup$
add a comment |
$begingroup$
beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned
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 $f(t)=frac272t^2(1+t^3)^frac12$.
answered 4 hours ago
Matt A PeltoMatt A Pelto
2,667621
$endgroup$
beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned
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 $f(t)=frac272t^2(1+t^3)^frac12$.
answered 4 hours ago
Matt A PeltoMatt A Pelto
2,667621
edited 3 hours ago
answered 4 hours ago
Matt A PeltoMatt A Pelto
2,667621
answered 4 hours ago
Matt A PeltoMatt A Pelto
2,667621
answered 4 hours ago
Matt A PeltoMatt A Pelto
2,667621
2,667621
add a comment |
add a comment |
$begingroup$
You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
$$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$
answered 4 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
add a comment |
$begingroup$
You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
$$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$
answered 4 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
add a comment |
$begingroup$
You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
$$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$
answered 4 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
$$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$
answered 4 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
answered 4 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
answered 4 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
answered 4 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
78.2k42867
add a comment |
add a comment |
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$begingroup$
Is this $$x=5+frac92t^3,y=4+3t^9/2$$?
$endgroup$
– Dr. Sonnhard Graubner
4 hours ago