I have a function T(x,y) where x and y are functions of t. I am given the two first order partial derivatives (dt/dx and dt/dy - the d is meant to be the partial symbol) and I am asked to use the total derivative to find a minimum value of t.

I have found the total derivative using the chain rule but I am unsure how this helps me find a minimum for t?

edit: all information given: dt/dx=6x-3y dt/dy=6y-3x x=cos(t) y=sin(t) *t between 0 and pi/2

  • $\begingroup$ Could you write in all information you have? $\endgroup$ – Mark Fantini Dec 9 '13 at 19:46
  • $\begingroup$ dt/dx=6x-3y dt/dy=6y-3x x=cost y=sint 0<t<pi/2 $\endgroup$ – maths Dec 9 '13 at 19:47
  • $\begingroup$ That would be best done in the question itself. =) Also, the $t$ you are constantly referring is the function $T(x,y)$ or the parameter $t$? $\endgroup$ – Mark Fantini Dec 9 '13 at 19:47
  • $\begingroup$ it is the parameter $\endgroup$ – maths Dec 9 '13 at 19:49
  • $\begingroup$ I think what you are writing as $$\frac{dt}{dx}$$ is actually $$\frac{\partial T}{\partial x}.$$ $\endgroup$ – Mark Fantini Dec 9 '13 at 19:51

You have $T(x,y)$ where $x$ and $y$ are functions of a parameter $t$. In other words, we can abuse notation and write $T(x,y) = T(x(t), y(t)) = T(t)$. Using the chain rule, we get

$$\frac{dT}{dt} = \frac{\partial T}{\partial x} \frac{dx}{dt} + \frac{\partial T}{\partial y} \frac{dy}{dt}.$$

You have all terms involved, and since you want to find extreme values all you have to do is equate this to zero, find the values, and then compute the second derivative to find whether it is a minimum or not.


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