Rate of change $frac{dV}{dt}$












0












$begingroup$


I have the following task: "Poiscuille's Law: $V=frac{P}{4Lv} (R^2 - r^2)$. Assume that $r$ is a constant as well as $P,L,v$. Find the rate of change $frac{dV}{dt}$ in terms of $R$ and $frac{dR}{dt}$ when $L=1$mm, $p=100$, $v=0.05$."



I cannot understand how can how can diff that, what is the $t$ here? I can write $V = 500 (R^2 - r^2)$. But what after? And what is "find the rate of change in terms of...? Something like: $frac{dV}{dt} = frac{dV}{dR}frac{dR}{dt}$? I am right?






real-analysis derivatives







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    0












    $begingroup$


    I have the following task: "Poiscuille's Law: $V=frac{P}{4Lv} (R^2 - r^2)$. Assume that $r$ is a constant as well as $P,L,v$. Find the rate of change $frac{dV}{dt}$ in terms of $R$ and $frac{dR}{dt}$ when $L=1$mm, $p=100$, $v=0.05$."



    I cannot understand how can how can diff that, what is the $t$ here? I can write $V = 500 (R^2 - r^2)$. But what after? And what is "find the rate of change in terms of...? Something like: $frac{dV}{dt} = frac{dV}{dR}frac{dR}{dt}$? I am right?






    real-analysis derivatives







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have the following task: "Poiscuille's Law: $V=frac{P}{4Lv} (R^2 - r^2)$. Assume that $r$ is a constant as well as $P,L,v$. Find the rate of change $frac{dV}{dt}$ in terms of $R$ and $frac{dR}{dt}$ when $L=1$mm, $p=100$, $v=0.05$."



      I cannot understand how can how can diff that, what is the $t$ here? I can write $V = 500 (R^2 - r^2)$. But what after? And what is "find the rate of change in terms of...? Something like: $frac{dV}{dt} = frac{dV}{dR}frac{dR}{dt}$? I am right?






      real-analysis derivatives







      share|cite|improve this question









      $endgroup$




      I have the following task: "Poiscuille's Law: $V=frac{P}{4Lv} (R^2 - r^2)$. Assume that $r$ is a constant as well as $P,L,v$. Find the rate of change $frac{dV}{dt}$ in terms of $R$ and $frac{dR}{dt}$ when $L=1$mm, $p=100$, $v=0.05$."



      I cannot understand how can how can diff that, what is the $t$ here? I can write $V = 500 (R^2 - r^2)$. But what after? And what is "find the rate of change in terms of...? Something like: $frac{dV}{dt} = frac{dV}{dR}frac{dR}{dt}$? I am right?






      real-analysis derivatives




      real-analysis derivatives






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 20 '18 at 13:06









      Just do itJust do it

      17218




      17218






















          1 Answer
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          1












          $begingroup$

          You have a function
          $$V(R)=frac{p}{4Lv}(R^2-r^2)$$
          But $R$ is a function of the time, so you have that
          $$V(t)=V(R(t))=frac{p}{4Lv}(R^2(t)-r^2)$$
          So you are right, you should just use the chain rule.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ou, $R$ is function of the time? This was not written in the task...
            $endgroup$
            – Just do it
            Dec 20 '18 at 13:12






          • 1




            $begingroup$
            @Arsenii For me (and I think for the author of the question as well), the appearance of $frac{mathrm{d}R}{mathrm{d}t}$ implies that $R$ is a function of $t$.
            $endgroup$
            – Botond
            Dec 20 '18 at 13:15






          • 2




            $begingroup$
            @Arsenii Or you can "feel" it from the fact that everything except the $R$ is a constant, so $R$ must be time-dependent to make some sense. But of course, it is not a mathematically correct way to state a problem, but it's common in physics.
            $endgroup$
            – Botond
            Dec 20 '18 at 13:19










          • $begingroup$
            Okey, thx) It was not obvious to me)
            $endgroup$
            – Just do it
            Dec 20 '18 at 13:19











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          You have a function
          $$V(R)=frac{p}{4Lv}(R^2-r^2)$$
          But $R$ is a function of the time, so you have that
          $$V(t)=V(R(t))=frac{p}{4Lv}(R^2(t)-r^2)$$
          So you are right, you should just use the chain rule.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ou, $R$ is function of the time? This was not written in the task...
            $endgroup$
            – Just do it
            Dec 20 '18 at 13:12






          • 1




            $begingroup$
            @Arsenii For me (and I think for the author of the question as well), the appearance of $frac{mathrm{d}R}{mathrm{d}t}$ implies that $R$ is a function of $t$.
            $endgroup$
            – Botond
            Dec 20 '18 at 13:15






          • 2




            $begingroup$
            @Arsenii Or you can "feel" it from the fact that everything except the $R$ is a constant, so $R$ must be time-dependent to make some sense. But of course, it is not a mathematically correct way to state a problem, but it's common in physics.
            $endgroup$
            – Botond
            Dec 20 '18 at 13:19










          • $begingroup$
            Okey, thx) It was not obvious to me)
            $endgroup$
            – Just do it
            Dec 20 '18 at 13:19
















          1












          $begingroup$

          You have a function
          $$V(R)=frac{p}{4Lv}(R^2-r^2)$$
          But $R$ is a function of the time, so you have that
          $$V(t)=V(R(t))=frac{p}{4Lv}(R^2(t)-r^2)$$
          So you are right, you should just use the chain rule.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ou, $R$ is function of the time? This was not written in the task...
            $endgroup$
            – Just do it
            Dec 20 '18 at 13:12






          • 1




            $begingroup$
            @Arsenii For me (and I think for the author of the question as well), the appearance of $frac{mathrm{d}R}{mathrm{d}t}$ implies that $R$ is a function of $t$.
            $endgroup$
            – Botond
            Dec 20 '18 at 13:15






          • 2




            $begingroup$
            @Arsenii Or you can "feel" it from the fact that everything except the $R$ is a constant, so $R$ must be time-dependent to make some sense. But of course, it is not a mathematically correct way to state a problem, but it's common in physics.
            $endgroup$
            – Botond
            Dec 20 '18 at 13:19










          • $begingroup$
            Okey, thx) It was not obvious to me)
            $endgroup$
            – Just do it
            Dec 20 '18 at 13:19














          1












          1








          1





          $begingroup$

          You have a function
          $$V(R)=frac{p}{4Lv}(R^2-r^2)$$
          But $R$ is a function of the time, so you have that
          $$V(t)=V(R(t))=frac{p}{4Lv}(R^2(t)-r^2)$$
          So you are right, you should just use the chain rule.






          share|cite|improve this answer









          $endgroup$



          You have a function
          $$V(R)=frac{p}{4Lv}(R^2-r^2)$$
          But $R$ is a function of the time, so you have that
          $$V(t)=V(R(t))=frac{p}{4Lv}(R^2(t)-r^2)$$
          So you are right, you should just use the chain rule.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 13:10









          BotondBotond

          5,6822732




          5,6822732












          • $begingroup$
            Ou, $R$ is function of the time? This was not written in the task...
            $endgroup$
            – Just do it
            Dec 20 '18 at 13:12






          • 1




            $begingroup$
            @Arsenii For me (and I think for the author of the question as well), the appearance of $frac{mathrm{d}R}{mathrm{d}t}$ implies that $R$ is a function of $t$.
            $endgroup$
            – Botond
            Dec 20 '18 at 13:15






          • 2




            $begingroup$
            @Arsenii Or you can "feel" it from the fact that everything except the $R$ is a constant, so $R$ must be time-dependent to make some sense. But of course, it is not a mathematically correct way to state a problem, but it's common in physics.
            $endgroup$
            – Botond
            Dec 20 '18 at 13:19










          • $begingroup$
            Okey, thx) It was not obvious to me)
            $endgroup$
            – Just do it
            Dec 20 '18 at 13:19


















          • $begingroup$
            Ou, $R$ is function of the time? This was not written in the task...
            $endgroup$
            – Just do it
            Dec 20 '18 at 13:12






          • 1




            $begingroup$
            @Arsenii For me (and I think for the author of the question as well), the appearance of $frac{mathrm{d}R}{mathrm{d}t}$ implies that $R$ is a function of $t$.
            $endgroup$
            – Botond
            Dec 20 '18 at 13:15






          • 2




            $begingroup$
            @Arsenii Or you can "feel" it from the fact that everything except the $R$ is a constant, so $R$ must be time-dependent to make some sense. But of course, it is not a mathematically correct way to state a problem, but it's common in physics.
            $endgroup$
            – Botond
            Dec 20 '18 at 13:19










          • $begingroup$
            Okey, thx) It was not obvious to me)
            $endgroup$
            – Just do it
            Dec 20 '18 at 13:19
















          $begingroup$
          Ou, $R$ is function of the time? This was not written in the task...
          $endgroup$
          – Just do it
          Dec 20 '18 at 13:12




          $begingroup$
          Ou, $R$ is function of the time? This was not written in the task...
          $endgroup$
          – Just do it
          Dec 20 '18 at 13:12




          1




          1




          $begingroup$
          @Arsenii For me (and I think for the author of the question as well), the appearance of $frac{mathrm{d}R}{mathrm{d}t}$ implies that $R$ is a function of $t$.
          $endgroup$
          – Botond
          Dec 20 '18 at 13:15




          $begingroup$
          @Arsenii For me (and I think for the author of the question as well), the appearance of $frac{mathrm{d}R}{mathrm{d}t}$ implies that $R$ is a function of $t$.
          $endgroup$
          – Botond
          Dec 20 '18 at 13:15




          2




          2




          $begingroup$
          @Arsenii Or you can "feel" it from the fact that everything except the $R$ is a constant, so $R$ must be time-dependent to make some sense. But of course, it is not a mathematically correct way to state a problem, but it's common in physics.
          $endgroup$
          – Botond
          Dec 20 '18 at 13:19




          $begingroup$
          @Arsenii Or you can "feel" it from the fact that everything except the $R$ is a constant, so $R$ must be time-dependent to make some sense. But of course, it is not a mathematically correct way to state a problem, but it's common in physics.
          $endgroup$
          – Botond
          Dec 20 '18 at 13:19












          $begingroup$
          Okey, thx) It was not obvious to me)
          $endgroup$
          – Just do it
          Dec 20 '18 at 13:19




          $begingroup$
          Okey, thx) It was not obvious to me)
          $endgroup$
          – Just do it
          Dec 20 '18 at 13:19


















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