Derivative by Leibniz's integral rule

Undoubtedly0

Hi all. I am looking at the functions [itex]\tau(x,y)[/itex], [itex]T(x,y)[/itex], and [itex]R(T)[/itex] related by

[tex] \tau(x,y) = \int_{\pi}^{T(x,y)} R(\theta) d\theta .[/tex]

It seems that by Leibniz's integral rule

[tex] \frac{\partial \tau}{\partial x} = \int_\pi^{T} \frac{\partial R}{\partial x} d\theta + R(T)\frac{\partial T}{\partial x}. [/tex]

It seems that [itex]\partial R / \partial x [/itex] need not be zero, yet another resource tells me that

[tex] \frac{\partial \tau}{\partial x} = R(T)\frac{\partial T}{\partial x} .[/tex]

Have I gone wrong somewhere? Thanks!

lurflurf

Sometimes ∂R/∂x is zero, sometimes not. Here you have written R(θ) which is often a hint that ∂R/∂x is zero, otherwise one usually writes R(θ,x).

Undoubtedly0

[itex]\theta[/itex] is a dummy variable that stands for [itex]T = T(x,y)[/itex], which I think means that [itex]R(\theta) = R(\theta(x,y))[/itex], correct?

lurflurf

$$\dfrac{d}{dt}\int^{\mathrm{b}(t)}_{\mathrm{a}(t)} \mathrm{f}(x,t) \, \mathrm{dx}=\int^{\mathrm{b}(t)}_{\mathrm{a}(t)} \dfrac{\partial \mathrm{f}}{\partial t} \, \mathrm{dx}+\mathrm{f}(\mathrm{b}(t),t) \mathrm{b} ^ \prime (t)-\mathrm{f}(\mathrm{a}(t),t) \mathrm{a} ^ \prime (t)$$

No the dummy variable takes all the values in a set like (pi,T), it does not depend on the end point. The other term gives the effect of the boundary. Leibniz's integral rule breaks the dependence of the integral on the variable into dependence on the the interior and the boundary. Usual examples are cars on a highway or water in a pipe or electricity in a wire.

Undoubtedly0

So if we take it to be the case that

[tex] \frac{\partial \tau}{\partial x} = R(T)\frac{\partial T}{\partial x},[/tex]

then what would be the value of [itex] \partial \tau/\partial x |_{x=c} [/itex]?

[tex] \left.\frac{\partial \tau}{\partial x}\right|_{x=c} = R(T)|_{x=c} \cdot\left.\frac{\partial T}{\partial x}\right|_{x=c}[/tex]

What is meant by [itex] R(T)|_{x=c}[/itex]?

elfmotat

[tex]R(T)=R(T(x,y))[/tex]

Let's say you're given the point (a,b). Plug that point into T and it returns a number (call it c): T(a,b)=c. Then you plug this number into R to get R(T): R(T(a,b))=R(c)=d.

If you're only given x=a but y remains a variable, then T(a,y) is a function dependent only on the variable y. Thus R(T) is also a function dependent only on y. Plugging in a particular value of y will return a single number for R(T).