Find surface area when this function is rotated around the y-axis. $y = frac{1}{3} x^{frac{3}{2}}$ ...

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Find surface area when this function is rotated around the y-axis. $y = frac{1}{3} x^{frac{3}{2}}$



The 2019 Stack Overflow Developer Survey Results Are InFind area of the surface obtained by rotating curve around x-axis?Compute surface area about the $x$-axisWhat substitution can be used to evaluate the integral giving the area of the surface of revolution of the curve $x = sqrt[3]{y}$?Finding the surface area of the solid formed by a revolution of the function $f(y)=x$ when rotated about the line $y=0$.How to find the surface area of revolution of an ellipsoid from ellipse rotating about y-axisSurface area of revolution formula for $x$ as a function of $y$ about the $x$ axisRotating area around axis, when bounds are not givenWhy do the limits of integration change when finding the surface area of the curve rotated about the y-axis?How do I find the surface area of this function $y = e^{-x^2}$ when it's rotated around the y-axis?How do I find the surface area of this function: $y = tan^{-1}x$












0












$begingroup$


I am gonna rotate this function around the y-axis, and I have to find the surface area.



Quick question: the reason we can use either definition (either derivative with respect to x or y) of arc length is because both definitions of arc length give you the... arc length unsurprisingly. We just need the arc length. It feels odd to find the surface area here with respect to dy, but find the arc length using a derivative with respect to x... but it shouldnt right?



Anyway, onto the problem:



$$y = frac{1}{3}x^{frac{3}{2}}$$ when $0 leq x leq 12$



$$frac{dy}{dx} = frac{1}{2} x^{frac{1}{2}}$$



$$frac{dy^2}{dx} = frac{1}{4} x$$



$$SA = 2 pi int_0^{12} x sqrt{1 + frac{1}{4}x} dx$$



But now I'm stuck here...I can't usub? Can I just multiply out converting $x$ to $sqrt{x^2}$?






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Bring the $x$ inside the radical ($xge 0$ here), complete the square, and integrate.
    $endgroup$
    – rogerl
    Mar 21 at 15:57










  • $begingroup$
    Set $x=dfrac{1}{4}cot^2theta$.
    $endgroup$
    – Paras Khosla
    Mar 21 at 16:21












  • $begingroup$
    Mind showing me? I'm a bit lost as to what you mean? Especially the trig sub...
    $endgroup$
    – Kitty Capital
    Mar 21 at 18:01










  • $begingroup$
    @rogerl mind showing a bit?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:16
















0












$begingroup$


I am gonna rotate this function around the y-axis, and I have to find the surface area.



Quick question: the reason we can use either definition (either derivative with respect to x or y) of arc length is because both definitions of arc length give you the... arc length unsurprisingly. We just need the arc length. It feels odd to find the surface area here with respect to dy, but find the arc length using a derivative with respect to x... but it shouldnt right?



Anyway, onto the problem:



$$y = frac{1}{3}x^{frac{3}{2}}$$ when $0 leq x leq 12$



$$frac{dy}{dx} = frac{1}{2} x^{frac{1}{2}}$$



$$frac{dy^2}{dx} = frac{1}{4} x$$



$$SA = 2 pi int_0^{12} x sqrt{1 + frac{1}{4}x} dx$$



But now I'm stuck here...I can't usub? Can I just multiply out converting $x$ to $sqrt{x^2}$?






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Bring the $x$ inside the radical ($xge 0$ here), complete the square, and integrate.
    $endgroup$
    – rogerl
    Mar 21 at 15:57










  • $begingroup$
    Set $x=dfrac{1}{4}cot^2theta$.
    $endgroup$
    – Paras Khosla
    Mar 21 at 16:21












  • $begingroup$
    Mind showing me? I'm a bit lost as to what you mean? Especially the trig sub...
    $endgroup$
    – Kitty Capital
    Mar 21 at 18:01










  • $begingroup$
    @rogerl mind showing a bit?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:16














0












0








0





$begingroup$


I am gonna rotate this function around the y-axis, and I have to find the surface area.



Quick question: the reason we can use either definition (either derivative with respect to x or y) of arc length is because both definitions of arc length give you the... arc length unsurprisingly. We just need the arc length. It feels odd to find the surface area here with respect to dy, but find the arc length using a derivative with respect to x... but it shouldnt right?



Anyway, onto the problem:



$$y = frac{1}{3}x^{frac{3}{2}}$$ when $0 leq x leq 12$



$$frac{dy}{dx} = frac{1}{2} x^{frac{1}{2}}$$



$$frac{dy^2}{dx} = frac{1}{4} x$$



$$SA = 2 pi int_0^{12} x sqrt{1 + frac{1}{4}x} dx$$



But now I'm stuck here...I can't usub? Can I just multiply out converting $x$ to $sqrt{x^2}$?






integration







share|cite|improve this question











$endgroup$




I am gonna rotate this function around the y-axis, and I have to find the surface area.



Quick question: the reason we can use either definition (either derivative with respect to x or y) of arc length is because both definitions of arc length give you the... arc length unsurprisingly. We just need the arc length. It feels odd to find the surface area here with respect to dy, but find the arc length using a derivative with respect to x... but it shouldnt right?



Anyway, onto the problem:



$$y = frac{1}{3}x^{frac{3}{2}}$$ when $0 leq x leq 12$



$$frac{dy}{dx} = frac{1}{2} x^{frac{1}{2}}$$



$$frac{dy^2}{dx} = frac{1}{4} x$$



$$SA = 2 pi int_0^{12} x sqrt{1 + frac{1}{4}x} dx$$



But now I'm stuck here...I can't usub? Can I just multiply out converting $x$ to $sqrt{x^2}$?






integration




integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 21 at 19:07







Kitty Capital

















asked Mar 21 at 15:51









Kitty CapitalKitty Capital

1166




1166












  • $begingroup$
    Bring the $x$ inside the radical ($xge 0$ here), complete the square, and integrate.
    $endgroup$
    – rogerl
    Mar 21 at 15:57










  • $begingroup$
    Set $x=dfrac{1}{4}cot^2theta$.
    $endgroup$
    – Paras Khosla
    Mar 21 at 16:21












  • $begingroup$
    Mind showing me? I'm a bit lost as to what you mean? Especially the trig sub...
    $endgroup$
    – Kitty Capital
    Mar 21 at 18:01










  • $begingroup$
    @rogerl mind showing a bit?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:16


















  • $begingroup$
    Bring the $x$ inside the radical ($xge 0$ here), complete the square, and integrate.
    $endgroup$
    – rogerl
    Mar 21 at 15:57










  • $begingroup$
    Set $x=dfrac{1}{4}cot^2theta$.
    $endgroup$
    – Paras Khosla
    Mar 21 at 16:21












  • $begingroup$
    Mind showing me? I'm a bit lost as to what you mean? Especially the trig sub...
    $endgroup$
    – Kitty Capital
    Mar 21 at 18:01










  • $begingroup$
    @rogerl mind showing a bit?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:16
















$begingroup$
Bring the $x$ inside the radical ($xge 0$ here), complete the square, and integrate.
$endgroup$
– rogerl
Mar 21 at 15:57




$begingroup$
Bring the $x$ inside the radical ($xge 0$ here), complete the square, and integrate.
$endgroup$
– rogerl
Mar 21 at 15:57












$begingroup$
Set $x=dfrac{1}{4}cot^2theta$.
$endgroup$
– Paras Khosla
Mar 21 at 16:21






$begingroup$
Set $x=dfrac{1}{4}cot^2theta$.
$endgroup$
– Paras Khosla
Mar 21 at 16:21














$begingroup$
Mind showing me? I'm a bit lost as to what you mean? Especially the trig sub...
$endgroup$
– Kitty Capital
Mar 21 at 18:01




$begingroup$
Mind showing me? I'm a bit lost as to what you mean? Especially the trig sub...
$endgroup$
– Kitty Capital
Mar 21 at 18:01












$begingroup$
@rogerl mind showing a bit?
$endgroup$
– Kitty Capital
Mar 21 at 19:16




$begingroup$
@rogerl mind showing a bit?
$endgroup$
– Kitty Capital
Mar 21 at 19:16










2 Answers
2






active

oldest

votes


















0












$begingroup$

$$text{Let }begin{bmatrix}x \ mathrm dxend{bmatrix}=begin{bmatrix}1/4cdotcot^2theta\ -1/2cdot cotthetacsc^2thetamathrm dthetaend{bmatrix}$$



$$begin{aligned}int xsqrt{1+dfrac{1}{4x}}mathrm dx&=-dfrac{1}{8}intcot^2theta sectheta cotthetacsc^2thetamathrm dtheta\&=-dfrac{1}{8}intdfrac{cos^2theta}{sin^2theta}cdotdfrac{1}{costheta}cdotdfrac{costheta}{sintheta}cdotdfrac{1}{sin^2theta}mathrm dtheta\ &=-dfrac{1}{8}intcot^2thetacsc^2thetamathrm dthetaend{aligned}$$



Now simply let $u=cotthetaimplies mathrm du=-csc^2thetamathrm dtheta$. Can you proceed?




Post OP's edit



$$begin{aligned}int xsqrt{1+dfrac{1}{4}x}mathrm dx&=dfrac{1}{2}int xsqrt{x+4}mathrm dxend{aligned}$$



Let $u=x+4iff x=u-4implies mathrm du=mathrm dx$



$$begin{aligned}dfrac{1}{2}int xsqrt{x+4}mathrm dx&=dfrac{1}{2}int sqrt{u}(u-4)mathrm du\&=dfrac{1}{2}int left(u^{3/2}-4sqrt{u}right)mathrm duend{aligned}$$



Can you proceed?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I think the 1/4x is wrong... I made a mistake, the x should not be in the denominator?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:11












  • $begingroup$
    Either way... where did the sec ceom from?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:12










  • $begingroup$
    That is because $1+tan^2theta=sec^2theta$.
    $endgroup$
    – Paras Khosla
    Mar 21 at 19:23










  • $begingroup$
    Ohhh.. This is a pretty innovative approach. Do you see how to complete the square? I'd like to see that approach but got stuck as well. Any idea?
    $endgroup$
    – Kitty Capital
    Mar 21 at 21:20






  • 1




    $begingroup$
    You're right. With substitution.
    $endgroup$
    – Chris Custer
    Mar 22 at 13:38



















0












$begingroup$

Hint: Use integration by parts. We have $int xsqrt{1+frac14 x}operatorname {dx}=x(frac83)(1+frac14 x)^{frac32} -(frac83) int (1+frac14 x)^{frac32}operatorname {dx}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Where did the 8/3 come from?
    $endgroup$
    – Kitty Capital
    Mar 22 at 13:59










  • $begingroup$
    To correct for the $frac32$ and $frac14 $ when you differentiate.
    $endgroup$
    – Chris Custer
    Mar 22 at 14:02












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

$$text{Let }begin{bmatrix}x \ mathrm dxend{bmatrix}=begin{bmatrix}1/4cdotcot^2theta\ -1/2cdot cotthetacsc^2thetamathrm dthetaend{bmatrix}$$



$$begin{aligned}int xsqrt{1+dfrac{1}{4x}}mathrm dx&=-dfrac{1}{8}intcot^2theta sectheta cotthetacsc^2thetamathrm dtheta\&=-dfrac{1}{8}intdfrac{cos^2theta}{sin^2theta}cdotdfrac{1}{costheta}cdotdfrac{costheta}{sintheta}cdotdfrac{1}{sin^2theta}mathrm dtheta\ &=-dfrac{1}{8}intcot^2thetacsc^2thetamathrm dthetaend{aligned}$$



Now simply let $u=cotthetaimplies mathrm du=-csc^2thetamathrm dtheta$. Can you proceed?




Post OP's edit



$$begin{aligned}int xsqrt{1+dfrac{1}{4}x}mathrm dx&=dfrac{1}{2}int xsqrt{x+4}mathrm dxend{aligned}$$



Let $u=x+4iff x=u-4implies mathrm du=mathrm dx$



$$begin{aligned}dfrac{1}{2}int xsqrt{x+4}mathrm dx&=dfrac{1}{2}int sqrt{u}(u-4)mathrm du\&=dfrac{1}{2}int left(u^{3/2}-4sqrt{u}right)mathrm duend{aligned}$$



Can you proceed?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I think the 1/4x is wrong... I made a mistake, the x should not be in the denominator?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:11












  • $begingroup$
    Either way... where did the sec ceom from?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:12










  • $begingroup$
    That is because $1+tan^2theta=sec^2theta$.
    $endgroup$
    – Paras Khosla
    Mar 21 at 19:23










  • $begingroup$
    Ohhh.. This is a pretty innovative approach. Do you see how to complete the square? I'd like to see that approach but got stuck as well. Any idea?
    $endgroup$
    – Kitty Capital
    Mar 21 at 21:20






  • 1




    $begingroup$
    You're right. With substitution.
    $endgroup$
    – Chris Custer
    Mar 22 at 13:38
















0












$begingroup$

$$text{Let }begin{bmatrix}x \ mathrm dxend{bmatrix}=begin{bmatrix}1/4cdotcot^2theta\ -1/2cdot cotthetacsc^2thetamathrm dthetaend{bmatrix}$$



$$begin{aligned}int xsqrt{1+dfrac{1}{4x}}mathrm dx&=-dfrac{1}{8}intcot^2theta sectheta cotthetacsc^2thetamathrm dtheta\&=-dfrac{1}{8}intdfrac{cos^2theta}{sin^2theta}cdotdfrac{1}{costheta}cdotdfrac{costheta}{sintheta}cdotdfrac{1}{sin^2theta}mathrm dtheta\ &=-dfrac{1}{8}intcot^2thetacsc^2thetamathrm dthetaend{aligned}$$



Now simply let $u=cotthetaimplies mathrm du=-csc^2thetamathrm dtheta$. Can you proceed?




Post OP's edit



$$begin{aligned}int xsqrt{1+dfrac{1}{4}x}mathrm dx&=dfrac{1}{2}int xsqrt{x+4}mathrm dxend{aligned}$$



Let $u=x+4iff x=u-4implies mathrm du=mathrm dx$



$$begin{aligned}dfrac{1}{2}int xsqrt{x+4}mathrm dx&=dfrac{1}{2}int sqrt{u}(u-4)mathrm du\&=dfrac{1}{2}int left(u^{3/2}-4sqrt{u}right)mathrm duend{aligned}$$



Can you proceed?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I think the 1/4x is wrong... I made a mistake, the x should not be in the denominator?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:11












  • $begingroup$
    Either way... where did the sec ceom from?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:12










  • $begingroup$
    That is because $1+tan^2theta=sec^2theta$.
    $endgroup$
    – Paras Khosla
    Mar 21 at 19:23










  • $begingroup$
    Ohhh.. This is a pretty innovative approach. Do you see how to complete the square? I'd like to see that approach but got stuck as well. Any idea?
    $endgroup$
    – Kitty Capital
    Mar 21 at 21:20






  • 1




    $begingroup$
    You're right. With substitution.
    $endgroup$
    – Chris Custer
    Mar 22 at 13:38














0












0








0





$begingroup$

$$text{Let }begin{bmatrix}x \ mathrm dxend{bmatrix}=begin{bmatrix}1/4cdotcot^2theta\ -1/2cdot cotthetacsc^2thetamathrm dthetaend{bmatrix}$$



$$begin{aligned}int xsqrt{1+dfrac{1}{4x}}mathrm dx&=-dfrac{1}{8}intcot^2theta sectheta cotthetacsc^2thetamathrm dtheta\&=-dfrac{1}{8}intdfrac{cos^2theta}{sin^2theta}cdotdfrac{1}{costheta}cdotdfrac{costheta}{sintheta}cdotdfrac{1}{sin^2theta}mathrm dtheta\ &=-dfrac{1}{8}intcot^2thetacsc^2thetamathrm dthetaend{aligned}$$



Now simply let $u=cotthetaimplies mathrm du=-csc^2thetamathrm dtheta$. Can you proceed?




Post OP's edit



$$begin{aligned}int xsqrt{1+dfrac{1}{4}x}mathrm dx&=dfrac{1}{2}int xsqrt{x+4}mathrm dxend{aligned}$$



Let $u=x+4iff x=u-4implies mathrm du=mathrm dx$



$$begin{aligned}dfrac{1}{2}int xsqrt{x+4}mathrm dx&=dfrac{1}{2}int sqrt{u}(u-4)mathrm du\&=dfrac{1}{2}int left(u^{3/2}-4sqrt{u}right)mathrm duend{aligned}$$



Can you proceed?






share|cite|improve this answer











$endgroup$



$$text{Let }begin{bmatrix}x \ mathrm dxend{bmatrix}=begin{bmatrix}1/4cdotcot^2theta\ -1/2cdot cotthetacsc^2thetamathrm dthetaend{bmatrix}$$



$$begin{aligned}int xsqrt{1+dfrac{1}{4x}}mathrm dx&=-dfrac{1}{8}intcot^2theta sectheta cotthetacsc^2thetamathrm dtheta\&=-dfrac{1}{8}intdfrac{cos^2theta}{sin^2theta}cdotdfrac{1}{costheta}cdotdfrac{costheta}{sintheta}cdotdfrac{1}{sin^2theta}mathrm dtheta\ &=-dfrac{1}{8}intcot^2thetacsc^2thetamathrm dthetaend{aligned}$$



Now simply let $u=cotthetaimplies mathrm du=-csc^2thetamathrm dtheta$. Can you proceed?




Post OP's edit



$$begin{aligned}int xsqrt{1+dfrac{1}{4}x}mathrm dx&=dfrac{1}{2}int xsqrt{x+4}mathrm dxend{aligned}$$



Let $u=x+4iff x=u-4implies mathrm du=mathrm dx$



$$begin{aligned}dfrac{1}{2}int xsqrt{x+4}mathrm dx&=dfrac{1}{2}int sqrt{u}(u-4)mathrm du\&=dfrac{1}{2}int left(u^{3/2}-4sqrt{u}right)mathrm duend{aligned}$$



Can you proceed?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 22 at 4:03

























answered Mar 21 at 18:29









Paras KhoslaParas Khosla

3,206626




3,206626












  • $begingroup$
    I think the 1/4x is wrong... I made a mistake, the x should not be in the denominator?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:11












  • $begingroup$
    Either way... where did the sec ceom from?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:12










  • $begingroup$
    That is because $1+tan^2theta=sec^2theta$.
    $endgroup$
    – Paras Khosla
    Mar 21 at 19:23










  • $begingroup$
    Ohhh.. This is a pretty innovative approach. Do you see how to complete the square? I'd like to see that approach but got stuck as well. Any idea?
    $endgroup$
    – Kitty Capital
    Mar 21 at 21:20






  • 1




    $begingroup$
    You're right. With substitution.
    $endgroup$
    – Chris Custer
    Mar 22 at 13:38


















  • $begingroup$
    I think the 1/4x is wrong... I made a mistake, the x should not be in the denominator?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:11












  • $begingroup$
    Either way... where did the sec ceom from?
    $endgroup$
    – Kitty Capital
    Mar 21 at 19:12










  • $begingroup$
    That is because $1+tan^2theta=sec^2theta$.
    $endgroup$
    – Paras Khosla
    Mar 21 at 19:23










  • $begingroup$
    Ohhh.. This is a pretty innovative approach. Do you see how to complete the square? I'd like to see that approach but got stuck as well. Any idea?
    $endgroup$
    – Kitty Capital
    Mar 21 at 21:20






  • 1




    $begingroup$
    You're right. With substitution.
    $endgroup$
    – Chris Custer
    Mar 22 at 13:38
















$begingroup$
I think the 1/4x is wrong... I made a mistake, the x should not be in the denominator?
$endgroup$
– Kitty Capital
Mar 21 at 19:11






$begingroup$
I think the 1/4x is wrong... I made a mistake, the x should not be in the denominator?
$endgroup$
– Kitty Capital
Mar 21 at 19:11














$begingroup$
Either way... where did the sec ceom from?
$endgroup$
– Kitty Capital
Mar 21 at 19:12




$begingroup$
Either way... where did the sec ceom from?
$endgroup$
– Kitty Capital
Mar 21 at 19:12












$begingroup$
That is because $1+tan^2theta=sec^2theta$.
$endgroup$
– Paras Khosla
Mar 21 at 19:23




$begingroup$
That is because $1+tan^2theta=sec^2theta$.
$endgroup$
– Paras Khosla
Mar 21 at 19:23












$begingroup$
Ohhh.. This is a pretty innovative approach. Do you see how to complete the square? I'd like to see that approach but got stuck as well. Any idea?
$endgroup$
– Kitty Capital
Mar 21 at 21:20




$begingroup$
Ohhh.. This is a pretty innovative approach. Do you see how to complete the square? I'd like to see that approach but got stuck as well. Any idea?
$endgroup$
– Kitty Capital
Mar 21 at 21:20




1




1




$begingroup$
You're right. With substitution.
$endgroup$
– Chris Custer
Mar 22 at 13:38




$begingroup$
You're right. With substitution.
$endgroup$
– Chris Custer
Mar 22 at 13:38











0












$begingroup$

Hint: Use integration by parts. We have $int xsqrt{1+frac14 x}operatorname {dx}=x(frac83)(1+frac14 x)^{frac32} -(frac83) int (1+frac14 x)^{frac32}operatorname {dx}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Where did the 8/3 come from?
    $endgroup$
    – Kitty Capital
    Mar 22 at 13:59










  • $begingroup$
    To correct for the $frac32$ and $frac14 $ when you differentiate.
    $endgroup$
    – Chris Custer
    Mar 22 at 14:02
















0












$begingroup$

Hint: Use integration by parts. We have $int xsqrt{1+frac14 x}operatorname {dx}=x(frac83)(1+frac14 x)^{frac32} -(frac83) int (1+frac14 x)^{frac32}operatorname {dx}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Where did the 8/3 come from?
    $endgroup$
    – Kitty Capital
    Mar 22 at 13:59










  • $begingroup$
    To correct for the $frac32$ and $frac14 $ when you differentiate.
    $endgroup$
    – Chris Custer
    Mar 22 at 14:02














0












0








0





$begingroup$

Hint: Use integration by parts. We have $int xsqrt{1+frac14 x}operatorname {dx}=x(frac83)(1+frac14 x)^{frac32} -(frac83) int (1+frac14 x)^{frac32}operatorname {dx}$.






share|cite|improve this answer









$endgroup$



Hint: Use integration by parts. We have $int xsqrt{1+frac14 x}operatorname {dx}=x(frac83)(1+frac14 x)^{frac32} -(frac83) int (1+frac14 x)^{frac32}operatorname {dx}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 22 at 13:35









Chris CusterChris Custer

14.3k3827




14.3k3827












  • $begingroup$
    Where did the 8/3 come from?
    $endgroup$
    – Kitty Capital
    Mar 22 at 13:59










  • $begingroup$
    To correct for the $frac32$ and $frac14 $ when you differentiate.
    $endgroup$
    – Chris Custer
    Mar 22 at 14:02


















  • $begingroup$
    Where did the 8/3 come from?
    $endgroup$
    – Kitty Capital
    Mar 22 at 13:59










  • $begingroup$
    To correct for the $frac32$ and $frac14 $ when you differentiate.
    $endgroup$
    – Chris Custer
    Mar 22 at 14:02
















$begingroup$
Where did the 8/3 come from?
$endgroup$
– Kitty Capital
Mar 22 at 13:59




$begingroup$
Where did the 8/3 come from?
$endgroup$
– Kitty Capital
Mar 22 at 13:59












$begingroup$
To correct for the $frac32$ and $frac14 $ when you differentiate.
$endgroup$
– Chris Custer
Mar 22 at 14:02




$begingroup$
To correct for the $frac32$ and $frac14 $ when you differentiate.
$endgroup$
– Chris Custer
Mar 22 at 14:02


















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