Find the length of the parametric curve $x(t)=5+6t^4, quad y(t)=5+4t^6 , quad0 ≤ t ≤ 2$Finding the length of a parametric curveFind the length of the parametric curve (Difficult)arc length parametric curvesCompute the length of a parametric curve.Find the length of the parametric curveFind the exact length of the parametric curve(Not sure what I'm doing wrong)The length of a parametric curveDetermine the arc length of the following parametric curveLength of a parametric curve formula: What does the integral represent?
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Find the length of the parametric curve $x(t)=5+6t^4, quad y(t)=5+4t^6 , quad0 ≤ t ≤ 2$
Finding the length of a parametric curveFind the length of the parametric curve (Difficult)arc length parametric curvesCompute the length of a parametric curve.Find the length of the parametric curveFind the exact length of the parametric curve(Not sure what I'm doing wrong)The length of a parametric curveDetermine the arc length of the following parametric curveLength of a parametric curve formula: What does the integral represent?
$begingroup$
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
$$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
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:
$$left(fracdxdtright) = 24cdot t^3 $$
$$left(fracdydtright) = 24cdot t^5 $$
$$int_0^2sqrtleft(24cdot t^3right)^2+left(24cdot t^5right)^2dt$$
$$int_0^2sqrtleft(576cdot t^6right)+left(576cdot t^10right)dt$$
$$int_0^2sqrtleft(576cdot t^6right) cdot left(1+t^4right)dt$$
$$24+int_0^2sqrtleft(t^6right) cdot left(1+t^4right)dt$$
$$frac23cdot 17^3/2+4-frac23$$
calculus integration
edited Mar 30 at 4:55
YuiTo Cheng
2,3184937
asked Mar 30 at 1:36
curiousengcuriouseng
315
$endgroup$
|
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
$$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
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:
$$left(fracdxdtright) = 24cdot t^3 $$
$$left(fracdydtright) = 24cdot t^5 $$
$$int_0^2sqrtleft(24cdot t^3right)^2+left(24cdot t^5right)^2dt$$
$$int_0^2sqrtleft(576cdot t^6right)+left(576cdot t^10right)dt$$
$$int_0^2sqrtleft(576cdot t^6right) cdot left(1+t^4right)dt$$
$$24+int_0^2sqrtleft(t^6right) cdot left(1+t^4right)dt$$
$$frac23cdot 17^3/2+4-frac23$$
calculus integration
edited Mar 30 at 4:55
YuiTo Cheng
2,3184937
asked Mar 30 at 1:36
curiousengcuriouseng
315
$endgroup$
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
Mar 30 at 1:37
1
$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
Mar 30 at 1:45
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
Mar 30 at 1:59
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
Mar 30 at 2:30
1
$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
Mar 30 at 2:31
|
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
$$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
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:
$$left(fracdxdtright) = 24cdot t^3 $$
$$left(fracdydtright) = 24cdot t^5 $$
$$int_0^2sqrtleft(24cdot t^3right)^2+left(24cdot t^5right)^2dt$$
$$int_0^2sqrtleft(576cdot t^6right)+left(576cdot t^10right)dt$$
$$int_0^2sqrtleft(576cdot t^6right) cdot left(1+t^4right)dt$$
$$24+int_0^2sqrtleft(t^6right) cdot left(1+t^4right)dt$$
$$frac23cdot 17^3/2+4-frac23$$
calculus integration
edited Mar 30 at 4:55
YuiTo Cheng
2,3184937
asked Mar 30 at 1:36
curiousengcuriouseng
315
$endgroup$
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
$$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
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:
$$left(fracdxdtright) = 24cdot t^3 $$
$$left(fracdydtright) = 24cdot t^5 $$
$$int_0^2sqrtleft(24cdot t^3right)^2+left(24cdot t^5right)^2dt$$
$$int_0^2sqrtleft(576cdot t^6right)+left(576cdot t^10right)dt$$
$$int_0^2sqrtleft(576cdot t^6right) cdot left(1+t^4right)dt$$
$$24+int_0^2sqrtleft(t^6right) cdot left(1+t^4right)dt$$
$$frac23cdot 17^3/2+4-frac23$$
calculus integration
calculus integration
edited Mar 30 at 4:55
YuiTo Cheng
2,3184937
asked Mar 30 at 1:36
curiousengcuriouseng
315
edited Mar 30 at 4:55
YuiTo Cheng
2,3184937
asked Mar 30 at 1:36
curiousengcuriouseng
315
edited Mar 30 at 4:55
YuiTo Cheng
2,3184937
edited Mar 30 at 4:55
YuiTo Cheng
2,3184937
edited Mar 30 at 4:55
YuiTo Cheng
2,3184937
2,3184937
asked Mar 30 at 1:36
curiousengcuriouseng
315
asked Mar 30 at 1:36
curiousengcuriouseng
315
asked Mar 30 at 1:36
curiousengcuriouseng
315
315
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
Mar 30 at 1:37
1
$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
Mar 30 at 1:45
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
Mar 30 at 1:59
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
Mar 30 at 2:30
1
$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
Mar 30 at 2:31
|
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
Mar 30 at 1:37
1
$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
Mar 30 at 1:45
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
Mar 30 at 1:59
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
Mar 30 at 2:30
1
$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
Mar 30 at 2:31
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
Mar 30 at 1:37
$begingroup$
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
Mar 30 at 1:37
1
1
$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
Mar 30 at 1:45
$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
Mar 30 at 1:45
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
Mar 30 at 1:59
$begingroup$
We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
$endgroup$
– David Peterson
Mar 30 at 1:59
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
Mar 30 at 2:30
$begingroup$
@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
Mar 30 at 2:30
1
1
$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
Mar 30 at 2:31
$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
Mar 30 at 2:31
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$
Which gives us:
$$int_0^2 24sqrtt^6+t^10dt$$
Which, when integrated, gives us: $$68sqrt17-4$$
I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.
answered Mar 30 at 2:03
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
536217
$endgroup$
$begingroup$
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
Mar 30 at 2:28
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 2:29
add a comment |
$begingroup$
Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.
Line 5 is correct.
Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ 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:
$$beginalign*
24 int_t=0^2 sqrtt^6(1+t^4) , dt
&= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_u=1^17 sqrtu , du \
&= 6 left[frac2u^3/23 right]_u=0^17 \
&= 4 (17^3/2 - 1) \
&= 68 sqrt17 - 4.
endalign*$$
answered Mar 30 at 2:57
heropupheropup
65.4k865104
$endgroup$
add a comment |
$begingroup$
Alternatively:
$$begincasesx=5+6t^4\ y=5+4t^6endcases Rightarrow begincasest^2=left(fracx-56right)^1/2\ t^2=left(fracy-54right)^1/3endcases Rightarrow y=4left(fracx-56right)^3/2+5\
0le tle 2 Rightarrow 5le xle 101$$
Hence:
$$S=int_a^b sqrt1+y'(x) dx= int_5^101 sqrt1+left(fracx-56right) dx=\
=int_5^101 sqrtfracx+16 dx=4cdot fracx+16cdot sqrtfracx+16bigg_5^101=\
=68sqrt17-4.$$
answered Mar 30 at 7:14
farruhotafarruhota
21.9k2842
$endgroup$
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
$begingroup$
Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$
Which gives us:
$$int_0^2 24sqrtt^6+t^10dt$$
Which, when integrated, gives us: $$68sqrt17-4$$
I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.
answered Mar 30 at 2:03
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
536217
$endgroup$
$begingroup$
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
Mar 30 at 2:28
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 2:29
add a comment |
$begingroup$
Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$
Which gives us:
$$int_0^2 24sqrtt^6+t^10dt$$
Which, when integrated, gives us: $$68sqrt17-4$$
I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.
answered Mar 30 at 2:03
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
536217
$endgroup$
$begingroup$
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
Mar 30 at 2:28
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 2:29
add a comment |
$begingroup$
Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$
Which gives us:
$$int_0^2 24sqrtt^6+t^10dt$$
Which, when integrated, gives us: $$68sqrt17-4$$
I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.
answered Mar 30 at 2:03
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
536217
$endgroup$
Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$
Which gives us:
$$int_0^2 24sqrtt^6+t^10dt$$
Which, when integrated, gives us: $$68sqrt17-4$$
I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.
answered Mar 30 at 2:03
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
536217
answered Mar 30 at 2:03
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
536217
answered Mar 30 at 2:03
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
536217
answered Mar 30 at 2:03
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
536217
536217
$begingroup$
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
Mar 30 at 2:28
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 2:29
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
Mar 30 at 2:28
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 2:29
$begingroup$
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
Mar 30 at 2:28
$begingroup$
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
Mar 30 at 2:28
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 2:29
$begingroup$
@curiouseng You are very welcome, regards!
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 2:29
add a comment |
$begingroup$
Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.
Line 5 is correct.
Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ 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:
$$beginalign*
24 int_t=0^2 sqrtt^6(1+t^4) , dt
&= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_u=1^17 sqrtu , du \
&= 6 left[frac2u^3/23 right]_u=0^17 \
&= 4 (17^3/2 - 1) \
&= 68 sqrt17 - 4.
endalign*$$
answered Mar 30 at 2:57
heropupheropup
65.4k865104
$endgroup$
add a comment |
$begingroup$
Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.
Line 5 is correct.
Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ 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:
$$beginalign*
24 int_t=0^2 sqrtt^6(1+t^4) , dt
&= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_u=1^17 sqrtu , du \
&= 6 left[frac2u^3/23 right]_u=0^17 \
&= 4 (17^3/2 - 1) \
&= 68 sqrt17 - 4.
endalign*$$
answered Mar 30 at 2:57
heropupheropup
65.4k865104
$endgroup$
add a comment |
$begingroup$
Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.
Line 5 is correct.
Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ 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:
$$beginalign*
24 int_t=0^2 sqrtt^6(1+t^4) , dt
&= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_u=1^17 sqrtu , du \
&= 6 left[frac2u^3/23 right]_u=0^17 \
&= 4 (17^3/2 - 1) \
&= 68 sqrt17 - 4.
endalign*$$
answered Mar 30 at 2:57
heropupheropup
65.4k865104
$endgroup$
Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.
Line 5 is correct.
Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ 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:
$$beginalign*
24 int_t=0^2 sqrtt^6(1+t^4) , dt
&= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_u=1^17 sqrtu , du \
&= 6 left[frac2u^3/23 right]_u=0^17 \
&= 4 (17^3/2 - 1) \
&= 68 sqrt17 - 4.
endalign*$$
answered Mar 30 at 2:57
heropupheropup
65.4k865104
answered Mar 30 at 2:57
heropupheropup
65.4k865104
answered Mar 30 at 2:57
heropupheropup
65.4k865104
answered Mar 30 at 2:57
heropupheropup
65.4k865104
65.4k865104
add a comment |
add a comment |
$begingroup$
Alternatively:
$$begincasesx=5+6t^4\ y=5+4t^6endcases Rightarrow begincasest^2=left(fracx-56right)^1/2\ t^2=left(fracy-54right)^1/3endcases Rightarrow y=4left(fracx-56right)^3/2+5\
0le tle 2 Rightarrow 5le xle 101$$
Hence:
$$S=int_a^b sqrt1+y'(x) dx= int_5^101 sqrt1+left(fracx-56right) dx=\
=int_5^101 sqrtfracx+16 dx=4cdot fracx+16cdot sqrtfracx+16bigg_5^101=\
=68sqrt17-4.$$
answered Mar 30 at 7:14
farruhotafarruhota
21.9k2842
$endgroup$
add a comment |
$begingroup$
Alternatively:
$$begincasesx=5+6t^4\ y=5+4t^6endcases Rightarrow begincasest^2=left(fracx-56right)^1/2\ t^2=left(fracy-54right)^1/3endcases Rightarrow y=4left(fracx-56right)^3/2+5\
0le tle 2 Rightarrow 5le xle 101$$
Hence:
$$S=int_a^b sqrt1+y'(x) dx= int_5^101 sqrt1+left(fracx-56right) dx=\
=int_5^101 sqrtfracx+16 dx=4cdot fracx+16cdot sqrtfracx+16bigg_5^101=\
=68sqrt17-4.$$
answered Mar 30 at 7:14
farruhotafarruhota
21.9k2842
$endgroup$
add a comment |
$begingroup$
Alternatively:
$$begincasesx=5+6t^4\ y=5+4t^6endcases Rightarrow begincasest^2=left(fracx-56right)^1/2\ t^2=left(fracy-54right)^1/3endcases Rightarrow y=4left(fracx-56right)^3/2+5\
0le tle 2 Rightarrow 5le xle 101$$
Hence:
$$S=int_a^b sqrt1+y'(x) dx= int_5^101 sqrt1+left(fracx-56right) dx=\
=int_5^101 sqrtfracx+16 dx=4cdot fracx+16cdot sqrtfracx+16bigg_5^101=\
=68sqrt17-4.$$
answered Mar 30 at 7:14
farruhotafarruhota
21.9k2842
$endgroup$
Alternatively:
$$begincasesx=5+6t^4\ y=5+4t^6endcases Rightarrow begincasest^2=left(fracx-56right)^1/2\ t^2=left(fracy-54right)^1/3endcases Rightarrow y=4left(fracx-56right)^3/2+5\
0le tle 2 Rightarrow 5le xle 101$$
Hence:
$$S=int_a^b sqrt1+y'(x) dx= int_5^101 sqrt1+left(fracx-56right) dx=\
=int_5^101 sqrtfracx+16 dx=4cdot fracx+16cdot sqrtfracx+16bigg_5^101=\
=68sqrt17-4.$$
answered Mar 30 at 7:14
farruhotafarruhota
21.9k2842
answered Mar 30 at 7:14
farruhotafarruhota
21.9k2842
answered Mar 30 at 7:14
farruhotafarruhota
21.9k2842
answered Mar 30 at 7:14
farruhotafarruhota
21.9k2842
21.9k2842
add a comment |
add a comment |
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$begingroup$
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
Mar 30 at 1:37
1
$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
Mar 30 at 1:45
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
Mar 30 at 1:59
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
Mar 30 at 2:30
1
$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
Mar 30 at 2:31