Partial Derivative Guidance.












0












$begingroup$


I was given this tricky problem in which I'm confuse on how to solve it and I would like some help:



Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$. Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives.



My attempt (which is wrong):



Using chain rule:





  • $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$

  • $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$


So,




  • $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$

  • $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$


Thanks,






calculus multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    When posting a question on this site, it is helpful to show us your work and specifically where you got stuck. This will help us help you.
    $endgroup$
    – Nicholas Roberts
    Jan 19 at 19:31










  • $begingroup$
    Sorry, I've edited my post with my answer.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 19:37
















0












$begingroup$


I was given this tricky problem in which I'm confuse on how to solve it and I would like some help:



Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$. Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives.



My attempt (which is wrong):



Using chain rule:





  • $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$

  • $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$


So,




  • $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$

  • $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$


Thanks,






calculus multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    When posting a question on this site, it is helpful to show us your work and specifically where you got stuck. This will help us help you.
    $endgroup$
    – Nicholas Roberts
    Jan 19 at 19:31










  • $begingroup$
    Sorry, I've edited my post with my answer.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 19:37














0












0








0





$begingroup$


I was given this tricky problem in which I'm confuse on how to solve it and I would like some help:



Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$. Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives.



My attempt (which is wrong):



Using chain rule:





  • $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$

  • $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$


So,




  • $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$

  • $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$


Thanks,






calculus multivariable-calculus







share|cite|improve this question











$endgroup$




I was given this tricky problem in which I'm confuse on how to solve it and I would like some help:



Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$. Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives.



My attempt (which is wrong):



Using chain rule:





  • $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$

  • $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$


So,




  • $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$

  • $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$


Thanks,






calculus multivariable-calculus




calculus multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 19:38









Kevin Long

3,57121431




3,57121431










asked Jan 19 at 19:28









Lorenz ClarkLorenz Clark

53




53








  • 1




    $begingroup$
    When posting a question on this site, it is helpful to show us your work and specifically where you got stuck. This will help us help you.
    $endgroup$
    – Nicholas Roberts
    Jan 19 at 19:31










  • $begingroup$
    Sorry, I've edited my post with my answer.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 19:37














  • 1




    $begingroup$
    When posting a question on this site, it is helpful to show us your work and specifically where you got stuck. This will help us help you.
    $endgroup$
    – Nicholas Roberts
    Jan 19 at 19:31










  • $begingroup$
    Sorry, I've edited my post with my answer.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 19:37








1




1




$begingroup$
When posting a question on this site, it is helpful to show us your work and specifically where you got stuck. This will help us help you.
$endgroup$
– Nicholas Roberts
Jan 19 at 19:31




$begingroup$
When posting a question on this site, it is helpful to show us your work and specifically where you got stuck. This will help us help you.
$endgroup$
– Nicholas Roberts
Jan 19 at 19:31












$begingroup$
Sorry, I've edited my post with my answer.
$endgroup$
– Lorenz Clark
Jan 19 at 19:37




$begingroup$
Sorry, I've edited my post with my answer.
$endgroup$
– Lorenz Clark
Jan 19 at 19:37










1 Answer
1






active

oldest

votes


















1












$begingroup$

I think you're confusing yourself with the $x$ and $y$ here. They aren't variables in $u(s,t)$, they're just variables in $v(x,y)$. In the definition of $u(s,t)$, you just input $s-2t$ for $x$ and $t^2$ for $y$, so this is really all in $s$ and $t$. Hence, you can just write $partial u/partial s=sin(s)*frac{partial v(s-2t, t^2)}{partial s}+frac{partial sin(s)}{partial s}v(s-2t,t^2)$ by the product rule. Then you can just evaluate $frac{partial v(s-2t,t^2)}{partial s}$ by the chain rule. You can then do the same for $partial v/partial y$. Try that out by yourself.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I will try now.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 19:56










  • $begingroup$
    Well, by inputting $x = s - 2t$ and $y = t^2$ and calculating $dv(s−2t,t2)/ds$ using the chain rule I get $dv/ds$ = dv/dx * dx/ds$ and $dv/dt = dv/dx * dx/dt + dv/dy * dy/dt$ I get that $dv/dx= 0$ and same for $dv/dy$ or am I doing something wrong again.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 20:16










  • $begingroup$
    I still need some help. I'm not sure if that value's suppose to give 0.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 21:16










  • $begingroup$
    You don't know what $v$ or its partial derivatives are, so you can't say that they're zero. The problem tells you to just leave the answer in terms of $v$ and its partial derivatives, so if you get something like $5v-3partial v/partial x$ (not the answer, just an example), you can leave it like that, because you can't say anything more.
    $endgroup$
    – Kevin Long
    Jan 19 at 22:01










  • $begingroup$
    Got it!! since we don't know the values that the function v might take, we just express the derivative of the function times the derivative of x with respect of s and so on! THANKS a lot!!
    $endgroup$
    – Lorenz Clark
    Jan 19 at 22:01











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

I think you're confusing yourself with the $x$ and $y$ here. They aren't variables in $u(s,t)$, they're just variables in $v(x,y)$. In the definition of $u(s,t)$, you just input $s-2t$ for $x$ and $t^2$ for $y$, so this is really all in $s$ and $t$. Hence, you can just write $partial u/partial s=sin(s)*frac{partial v(s-2t, t^2)}{partial s}+frac{partial sin(s)}{partial s}v(s-2t,t^2)$ by the product rule. Then you can just evaluate $frac{partial v(s-2t,t^2)}{partial s}$ by the chain rule. You can then do the same for $partial v/partial y$. Try that out by yourself.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I will try now.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 19:56










  • $begingroup$
    Well, by inputting $x = s - 2t$ and $y = t^2$ and calculating $dv(s−2t,t2)/ds$ using the chain rule I get $dv/ds$ = dv/dx * dx/ds$ and $dv/dt = dv/dx * dx/dt + dv/dy * dy/dt$ I get that $dv/dx= 0$ and same for $dv/dy$ or am I doing something wrong again.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 20:16










  • $begingroup$
    I still need some help. I'm not sure if that value's suppose to give 0.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 21:16










  • $begingroup$
    You don't know what $v$ or its partial derivatives are, so you can't say that they're zero. The problem tells you to just leave the answer in terms of $v$ and its partial derivatives, so if you get something like $5v-3partial v/partial x$ (not the answer, just an example), you can leave it like that, because you can't say anything more.
    $endgroup$
    – Kevin Long
    Jan 19 at 22:01










  • $begingroup$
    Got it!! since we don't know the values that the function v might take, we just express the derivative of the function times the derivative of x with respect of s and so on! THANKS a lot!!
    $endgroup$
    – Lorenz Clark
    Jan 19 at 22:01
















1












$begingroup$

I think you're confusing yourself with the $x$ and $y$ here. They aren't variables in $u(s,t)$, they're just variables in $v(x,y)$. In the definition of $u(s,t)$, you just input $s-2t$ for $x$ and $t^2$ for $y$, so this is really all in $s$ and $t$. Hence, you can just write $partial u/partial s=sin(s)*frac{partial v(s-2t, t^2)}{partial s}+frac{partial sin(s)}{partial s}v(s-2t,t^2)$ by the product rule. Then you can just evaluate $frac{partial v(s-2t,t^2)}{partial s}$ by the chain rule. You can then do the same for $partial v/partial y$. Try that out by yourself.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I will try now.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 19:56










  • $begingroup$
    Well, by inputting $x = s - 2t$ and $y = t^2$ and calculating $dv(s−2t,t2)/ds$ using the chain rule I get $dv/ds$ = dv/dx * dx/ds$ and $dv/dt = dv/dx * dx/dt + dv/dy * dy/dt$ I get that $dv/dx= 0$ and same for $dv/dy$ or am I doing something wrong again.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 20:16










  • $begingroup$
    I still need some help. I'm not sure if that value's suppose to give 0.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 21:16










  • $begingroup$
    You don't know what $v$ or its partial derivatives are, so you can't say that they're zero. The problem tells you to just leave the answer in terms of $v$ and its partial derivatives, so if you get something like $5v-3partial v/partial x$ (not the answer, just an example), you can leave it like that, because you can't say anything more.
    $endgroup$
    – Kevin Long
    Jan 19 at 22:01










  • $begingroup$
    Got it!! since we don't know the values that the function v might take, we just express the derivative of the function times the derivative of x with respect of s and so on! THANKS a lot!!
    $endgroup$
    – Lorenz Clark
    Jan 19 at 22:01














1












1








1





$begingroup$

I think you're confusing yourself with the $x$ and $y$ here. They aren't variables in $u(s,t)$, they're just variables in $v(x,y)$. In the definition of $u(s,t)$, you just input $s-2t$ for $x$ and $t^2$ for $y$, so this is really all in $s$ and $t$. Hence, you can just write $partial u/partial s=sin(s)*frac{partial v(s-2t, t^2)}{partial s}+frac{partial sin(s)}{partial s}v(s-2t,t^2)$ by the product rule. Then you can just evaluate $frac{partial v(s-2t,t^2)}{partial s}$ by the chain rule. You can then do the same for $partial v/partial y$. Try that out by yourself.






share|cite|improve this answer









$endgroup$



I think you're confusing yourself with the $x$ and $y$ here. They aren't variables in $u(s,t)$, they're just variables in $v(x,y)$. In the definition of $u(s,t)$, you just input $s-2t$ for $x$ and $t^2$ for $y$, so this is really all in $s$ and $t$. Hence, you can just write $partial u/partial s=sin(s)*frac{partial v(s-2t, t^2)}{partial s}+frac{partial sin(s)}{partial s}v(s-2t,t^2)$ by the product rule. Then you can just evaluate $frac{partial v(s-2t,t^2)}{partial s}$ by the chain rule. You can then do the same for $partial v/partial y$. Try that out by yourself.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 19 at 19:51









Kevin LongKevin Long

3,57121431




3,57121431












  • $begingroup$
    I will try now.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 19:56










  • $begingroup$
    Well, by inputting $x = s - 2t$ and $y = t^2$ and calculating $dv(s−2t,t2)/ds$ using the chain rule I get $dv/ds$ = dv/dx * dx/ds$ and $dv/dt = dv/dx * dx/dt + dv/dy * dy/dt$ I get that $dv/dx= 0$ and same for $dv/dy$ or am I doing something wrong again.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 20:16










  • $begingroup$
    I still need some help. I'm not sure if that value's suppose to give 0.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 21:16










  • $begingroup$
    You don't know what $v$ or its partial derivatives are, so you can't say that they're zero. The problem tells you to just leave the answer in terms of $v$ and its partial derivatives, so if you get something like $5v-3partial v/partial x$ (not the answer, just an example), you can leave it like that, because you can't say anything more.
    $endgroup$
    – Kevin Long
    Jan 19 at 22:01










  • $begingroup$
    Got it!! since we don't know the values that the function v might take, we just express the derivative of the function times the derivative of x with respect of s and so on! THANKS a lot!!
    $endgroup$
    – Lorenz Clark
    Jan 19 at 22:01


















  • $begingroup$
    I will try now.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 19:56










  • $begingroup$
    Well, by inputting $x = s - 2t$ and $y = t^2$ and calculating $dv(s−2t,t2)/ds$ using the chain rule I get $dv/ds$ = dv/dx * dx/ds$ and $dv/dt = dv/dx * dx/dt + dv/dy * dy/dt$ I get that $dv/dx= 0$ and same for $dv/dy$ or am I doing something wrong again.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 20:16










  • $begingroup$
    I still need some help. I'm not sure if that value's suppose to give 0.
    $endgroup$
    – Lorenz Clark
    Jan 19 at 21:16










  • $begingroup$
    You don't know what $v$ or its partial derivatives are, so you can't say that they're zero. The problem tells you to just leave the answer in terms of $v$ and its partial derivatives, so if you get something like $5v-3partial v/partial x$ (not the answer, just an example), you can leave it like that, because you can't say anything more.
    $endgroup$
    – Kevin Long
    Jan 19 at 22:01










  • $begingroup$
    Got it!! since we don't know the values that the function v might take, we just express the derivative of the function times the derivative of x with respect of s and so on! THANKS a lot!!
    $endgroup$
    – Lorenz Clark
    Jan 19 at 22:01
















$begingroup$
I will try now.
$endgroup$
– Lorenz Clark
Jan 19 at 19:56




$begingroup$
I will try now.
$endgroup$
– Lorenz Clark
Jan 19 at 19:56












$begingroup$
Well, by inputting $x = s - 2t$ and $y = t^2$ and calculating $dv(s−2t,t2)/ds$ using the chain rule I get $dv/ds$ = dv/dx * dx/ds$ and $dv/dt = dv/dx * dx/dt + dv/dy * dy/dt$ I get that $dv/dx= 0$ and same for $dv/dy$ or am I doing something wrong again.
$endgroup$
– Lorenz Clark
Jan 19 at 20:16




$begingroup$
Well, by inputting $x = s - 2t$ and $y = t^2$ and calculating $dv(s−2t,t2)/ds$ using the chain rule I get $dv/ds$ = dv/dx * dx/ds$ and $dv/dt = dv/dx * dx/dt + dv/dy * dy/dt$ I get that $dv/dx= 0$ and same for $dv/dy$ or am I doing something wrong again.
$endgroup$
– Lorenz Clark
Jan 19 at 20:16












$begingroup$
I still need some help. I'm not sure if that value's suppose to give 0.
$endgroup$
– Lorenz Clark
Jan 19 at 21:16




$begingroup$
I still need some help. I'm not sure if that value's suppose to give 0.
$endgroup$
– Lorenz Clark
Jan 19 at 21:16












$begingroup$
You don't know what $v$ or its partial derivatives are, so you can't say that they're zero. The problem tells you to just leave the answer in terms of $v$ and its partial derivatives, so if you get something like $5v-3partial v/partial x$ (not the answer, just an example), you can leave it like that, because you can't say anything more.
$endgroup$
– Kevin Long
Jan 19 at 22:01




$begingroup$
You don't know what $v$ or its partial derivatives are, so you can't say that they're zero. The problem tells you to just leave the answer in terms of $v$ and its partial derivatives, so if you get something like $5v-3partial v/partial x$ (not the answer, just an example), you can leave it like that, because you can't say anything more.
$endgroup$
– Kevin Long
Jan 19 at 22:01












$begingroup$
Got it!! since we don't know the values that the function v might take, we just express the derivative of the function times the derivative of x with respect of s and so on! THANKS a lot!!
$endgroup$
– Lorenz Clark
Jan 19 at 22:01




$begingroup$
Got it!! since we don't know the values that the function v might take, we just express the derivative of the function times the derivative of x with respect of s and so on! THANKS a lot!!
$endgroup$
– Lorenz Clark
Jan 19 at 22:01


















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