Calculus II Question The Next CEO of Stack OverflowLength of an AstroidUnderstanding this...
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Calculus II Question
The Next CEO of Stack OverflowLength of an AstroidUnderstanding this calculus simplificationIntegration problem: $int x^{2} -x 4^{-x^{2}} dx$Finding the parametric form of a standard equationApplication of “twice the integral” even if the function is not graphically even?Find the length of the parametric curveFind the exact length of the parametric curve(Not sure what I'm doing wrong)Calculus 2 moments question.The length of a parametric curveParametric curve length - calculus
$begingroup$
Find the length of the following parametric curve.
$$x(t)=5+6t^4 ,quad y(t)=5+4t^6 ,qquad 0 ≤ t ≤ 2.$$
I used the formula
$$int_0^2sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
And I found
$$frac23cdot 17^{3/2}+4-frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.
My steps:
$$left(frac{dx}{dt}right) = 24cdot t^3 $$
$$left(frac{dy}{dt}right) = 24cdot t^5 $$
$$int_0^2sqrt{left(24cdot t^3right)^2+left(24cdot t^5right)^2}dt$$
$$int_0^2sqrt{left(576cdot t^6right)+left(576cdot t^10right)}dt$$
$$int_0^2sqrt{left(576cdot t^6right) cdot left(1+t^4right)}dt$$
$$24+int_0^2sqrt{left(t^6right) cdot left(1+t^4right)}dt$$
$$frac23cdot 17^{3/2}+4-frac23$$
calculus integration
edited 54 mins ago
rash
585116
asked 1 hour ago
curiousengcuriouseng
134
New contributor
curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 5 more comments
$begingroup$
Find the length of the following parametric curve.
$$x(t)=5+6t^4 ,quad y(t)=5+4t^6 ,qquad 0 ≤ t ≤ 2.$$
I used the formula
$$int_0^2sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
And I found
$$frac23cdot 17^{3/2}+4-frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.
My steps:
$$left(frac{dx}{dt}right) = 24cdot t^3 $$
$$left(frac{dy}{dt}right) = 24cdot t^5 $$
$$int_0^2sqrt{left(24cdot t^3right)^2+left(24cdot t^5right)^2}dt$$
$$int_0^2sqrt{left(576cdot t^6right)+left(576cdot t^10right)}dt$$
$$int_0^2sqrt{left(576cdot t^6right) cdot left(1+t^4right)}dt$$
$$24+int_0^2sqrt{left(t^6right) cdot left(1+t^4right)}dt$$
$$frac23cdot 17^{3/2}+4-frac23$$
calculus integration
edited 54 mins ago
rash
585116
asked 1 hour ago
curiousengcuriouseng
134
New contributor
curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
$endgroup$
– Ross Millikan
1 hour ago
1
$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
1 hour ago
1
$begingroup$
We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
$endgroup$
– David Peterson
1 hour ago
1
$begingroup$
@curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
$endgroup$
– John Omielan
59 mins ago
1
$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
58 mins ago
|
show 5 more comments
$begingroup$
Find the length of the following parametric curve.
$$x(t)=5+6t^4 ,quad y(t)=5+4t^6 ,qquad 0 ≤ t ≤ 2.$$
I used the formula
$$int_0^2sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
And I found
$$frac23cdot 17^{3/2}+4-frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.
My steps:
$$left(frac{dx}{dt}right) = 24cdot t^3 $$
$$left(frac{dy}{dt}right) = 24cdot t^5 $$
$$int_0^2sqrt{left(24cdot t^3right)^2+left(24cdot t^5right)^2}dt$$
$$int_0^2sqrt{left(576cdot t^6right)+left(576cdot t^10right)}dt$$
$$int_0^2sqrt{left(576cdot t^6right) cdot left(1+t^4right)}dt$$
$$24+int_0^2sqrt{left(t^6right) cdot left(1+t^4right)}dt$$
$$frac23cdot 17^{3/2}+4-frac23$$
calculus integration
edited 54 mins ago
rash
585116
asked 1 hour ago
curiousengcuriouseng
134
New contributor
curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Find the length of the following parametric curve.
$$x(t)=5+6t^4 ,quad y(t)=5+4t^6 ,qquad 0 ≤ t ≤ 2.$$
I used the formula
$$int_0^2sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
And I found
$$frac23cdot 17^{3/2}+4-frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.
My steps:
$$left(frac{dx}{dt}right) = 24cdot t^3 $$
$$left(frac{dy}{dt}right) = 24cdot t^5 $$
$$int_0^2sqrt{left(24cdot t^3right)^2+left(24cdot t^5right)^2}dt$$
$$int_0^2sqrt{left(576cdot t^6right)+left(576cdot t^10right)}dt$$
$$int_0^2sqrt{left(576cdot t^6right) cdot left(1+t^4right)}dt$$
$$24+int_0^2sqrt{left(t^6right) cdot left(1+t^4right)}dt$$
$$frac23cdot 17^{3/2}+4-frac23$$
calculus integration
calculus integration
edited 54 mins ago
rash
585116
asked 1 hour ago
curiousengcuriouseng
134
New contributor
curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 54 mins ago
rash
585116
asked 1 hour ago
curiousengcuriouseng
134
New contributor
curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 54 mins ago
rash
585116
edited 54 mins ago
rash
585116
edited 54 mins ago
rash
585116
585116
asked 1 hour ago
curiousengcuriouseng
134
New contributor
curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
curiousengcuriouseng
134
asked 1 hour ago
curiousengcuriouseng
134
134
New contributor
curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
$endgroup$
– Ross Millikan
1 hour ago
1
$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
1 hour ago
1
$begingroup$
We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
$endgroup$
– David Peterson
1 hour ago
1
$begingroup$
@curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
$endgroup$
– John Omielan
59 mins ago
1
$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
58 mins ago
|
show 5 more comments
3
$begingroup$
What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
$endgroup$
– Ross Millikan
1 hour ago
1
$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
1 hour ago
1
$begingroup$
We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
$endgroup$
– David Peterson
1 hour ago
1
$begingroup$
@curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
$endgroup$
– John Omielan
59 mins ago
1
$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
58 mins ago
3
3
$begingroup$
What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
$endgroup$
– Ross Millikan
1 hour ago
$begingroup$
What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
$endgroup$
– Ross Millikan
1 hour ago
1
1
$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
1 hour ago
$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
1 hour ago
1
1
$begingroup$
We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
$endgroup$
– David Peterson
1 hour ago
$begingroup$
We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
$endgroup$
– David Peterson
1 hour ago
1
1
$begingroup$
@curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
$endgroup$
– John Omielan
59 mins ago
$begingroup$
@curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
$endgroup$
– John Omielan
59 mins ago
1
1
$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
58 mins ago
$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
58 mins ago
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$
Which gives us:
$$int_0^2 24sqrt{t^6+t^{10}}dt$$
Which, when integrated, gives us: $$68sqrt{17}-4$$
I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.
answered 1 hour ago
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
527217
$endgroup$
$begingroup$
Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
$endgroup$
– curiouseng
1 hour ago
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
1 hour ago
add a comment |
$begingroup$
Line 4 should read $$int_{t=0}^2 sqrt{576 t^6 + 576 t^{10}} , dt.$$ This is a typesetting error.
Line 5 is correct.
Line 6 should read $$24 int_{t=0}^2 sqrt{t^6 (1+t^4)} , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.
You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
$$begin{align*}
24 int_{t=0}^2 sqrt{t^6(1+t^4)} , dt
&= 24 int_{t=0}^2 t^3 sqrt{1+t^4} , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_{u=1}^{17} sqrt{u} , du \
&= 6 left[frac{2u^{3/2}}{3} right]_{u=0}^{17} \
&= 4 (17^{3/2} - 1) \
&= 68 sqrt{17} - 4.
end{align*}$$
answered 33 mins ago
heropupheropup
64.8k764103
$endgroup$
add a comment |
Your Answer
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2 Answers
2
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2 Answers
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$begingroup$
Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$
Which gives us:
$$int_0^2 24sqrt{t^6+t^{10}}dt$$
Which, when integrated, gives us: $$68sqrt{17}-4$$
I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.
answered 1 hour ago
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
527217
$endgroup$
$begingroup$
Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
$endgroup$
– curiouseng
1 hour ago
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
1 hour ago
add a comment |
$begingroup$
Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$
Which gives us:
$$int_0^2 24sqrt{t^6+t^{10}}dt$$
Which, when integrated, gives us: $$68sqrt{17}-4$$
I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.
answered 1 hour ago
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
527217
$endgroup$
$begingroup$
Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
$endgroup$
– curiouseng
1 hour ago
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
1 hour ago
add a comment |
$begingroup$
Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$
Which gives us:
$$int_0^2 24sqrt{t^6+t^{10}}dt$$
Which, when integrated, gives us: $$68sqrt{17}-4$$
I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.
answered 1 hour ago
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
527217
$endgroup$
Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$
Which gives us:
$$int_0^2 24sqrt{t^6+t^{10}}dt$$
Which, when integrated, gives us: $$68sqrt{17}-4$$
I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.
answered 1 hour ago
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
527217
answered 1 hour ago
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
527217
answered 1 hour ago
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
527217
answered 1 hour ago
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
527217
527217
$begingroup$
Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
$endgroup$
– curiouseng
1 hour ago
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
1 hour ago
add a comment |
$begingroup$
Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
$endgroup$
– curiouseng
1 hour ago
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
1 hour ago
$begingroup$
Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
$endgroup$
– curiouseng
1 hour ago
$begingroup$
Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
$endgroup$
– curiouseng
1 hour ago
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
1 hour ago
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
1 hour ago
add a comment |
$begingroup$
Line 4 should read $$int_{t=0}^2 sqrt{576 t^6 + 576 t^{10}} , dt.$$ This is a typesetting error.
Line 5 is correct.
Line 6 should read $$24 int_{t=0}^2 sqrt{t^6 (1+t^4)} , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.
You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
$$begin{align*}
24 int_{t=0}^2 sqrt{t^6(1+t^4)} , dt
&= 24 int_{t=0}^2 t^3 sqrt{1+t^4} , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_{u=1}^{17} sqrt{u} , du \
&= 6 left[frac{2u^{3/2}}{3} right]_{u=0}^{17} \
&= 4 (17^{3/2} - 1) \
&= 68 sqrt{17} - 4.
end{align*}$$
answered 33 mins ago
heropupheropup
64.8k764103
$endgroup$
add a comment |
$begingroup$
Line 4 should read $$int_{t=0}^2 sqrt{576 t^6 + 576 t^{10}} , dt.$$ This is a typesetting error.
Line 5 is correct.
Line 6 should read $$24 int_{t=0}^2 sqrt{t^6 (1+t^4)} , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.
You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
$$begin{align*}
24 int_{t=0}^2 sqrt{t^6(1+t^4)} , dt
&= 24 int_{t=0}^2 t^3 sqrt{1+t^4} , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_{u=1}^{17} sqrt{u} , du \
&= 6 left[frac{2u^{3/2}}{3} right]_{u=0}^{17} \
&= 4 (17^{3/2} - 1) \
&= 68 sqrt{17} - 4.
end{align*}$$
answered 33 mins ago
heropupheropup
64.8k764103
$endgroup$
add a comment |
$begingroup$
Line 4 should read $$int_{t=0}^2 sqrt{576 t^6 + 576 t^{10}} , dt.$$ This is a typesetting error.
Line 5 is correct.
Line 6 should read $$24 int_{t=0}^2 sqrt{t^6 (1+t^4)} , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.
You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
$$begin{align*}
24 int_{t=0}^2 sqrt{t^6(1+t^4)} , dt
&= 24 int_{t=0}^2 t^3 sqrt{1+t^4} , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_{u=1}^{17} sqrt{u} , du \
&= 6 left[frac{2u^{3/2}}{3} right]_{u=0}^{17} \
&= 4 (17^{3/2} - 1) \
&= 68 sqrt{17} - 4.
end{align*}$$
answered 33 mins ago
heropupheropup
64.8k764103
$endgroup$
Line 4 should read $$int_{t=0}^2 sqrt{576 t^6 + 576 t^{10}} , dt.$$ This is a typesetting error.
Line 5 is correct.
Line 6 should read $$24 int_{t=0}^2 sqrt{t^6 (1+t^4)} , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.
You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
$$begin{align*}
24 int_{t=0}^2 sqrt{t^6(1+t^4)} , dt
&= 24 int_{t=0}^2 t^3 sqrt{1+t^4} , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_{u=1}^{17} sqrt{u} , du \
&= 6 left[frac{2u^{3/2}}{3} right]_{u=0}^{17} \
&= 4 (17^{3/2} - 1) \
&= 68 sqrt{17} - 4.
end{align*}$$
answered 33 mins ago
heropupheropup
64.8k764103
answered 33 mins ago
heropupheropup
64.8k764103
answered 33 mins ago
heropupheropup
64.8k764103
answered 33 mins ago
heropupheropup
64.8k764103
64.8k764103
add a comment |
add a comment |
curiouseng is a new contributor. Be nice, and check out our Code of Conduct.
curiouseng is a new contributor. Be nice, and check out our Code of Conduct.
curiouseng is a new contributor. Be nice, and check out our Code of Conduct.
curiouseng is a new contributor. Be nice, and check out our Code of Conduct.
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3
$begingroup$
What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
$endgroup$
– Ross Millikan
1 hour ago
1
$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
1 hour ago
1
$begingroup$
We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
$endgroup$
– David Peterson
1 hour ago
1
$begingroup$
@curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
$endgroup$
– John Omielan
59 mins ago
1
$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
58 mins ago