# Partial derivative problems

1. Feb 8, 2005

### tandoorichicken

Two homework problems I can't get.

(1) The question is find the first partial derivatives of the function. The problem is that the function in this problem is
$$f(x, y) = \int_{y}^{x} \cos{t^2} dt$$
The main obstacle is getting past this function. I can't integrate it and neither can my calculator. Is there a way to do this problem without integrating, or how do you integrate this function?

(2) The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasing as the resistor heats up. Use Ohm's Law, V = IR, to find how the current I is changing at the moment when R = 400$\Omega$, I = 0.08 A, $\frac{dV}{dt}$ = -0.01 V/s, and $\frac{dR}{dt}$ = 0.03 $\Omega$/s.

2. Feb 8, 2005

### dextercioby

1.How about applying the fundamental theorem of calculus...??Both parts of it.

2.Differentiate wrt time the Ohm's law and plug in tht #-s.

Daniel.

3. Feb 9, 2005

### xanthym

{#1}

NO integration needed here. Use the Fundamental Theorem of Calculus:

$$f(x, y) = \int_{y}^{x} \cos{t^2} dt$$

$$\frac {\partial f(x,y)} {\partial x} = \cos{x^2}$$

$$\frac {\partial f(x,y)} {\partial y} = -\cos{y^2}$$

Incidentally, in the more general case, we have:

$$f(x, y) = \int_{y}^{g(x)} h(t) dt$$

$$\frac {\partial f(x,y)} {\partial x} = h(g(x))*g'(x)$$

{#2}

V = I*R
-----> I = V/R
Using the Quotient Rule:
(dI/dt) = {(dV/dt)*R - V*(dR/dt)}/(R^2)
(dI/dt) = {(dV/dt)*R - (I*R)*(dR/dt)}/(R^2)
(dI/dt) = {(dV/dt) - (I)*(dR/dt)}/(R)

Placing given values into the above equation, we get:
(dI/dt) = {(-0.01 V/s) - (0.08 A)*(0.03 ohm/s)}/(400 ohm)
(dI/dt) = (-3.1)x10^(-5) Amps/sec
~

Last edited: Feb 9, 2005
4. Feb 9, 2005

### dextercioby

Well,the general idea of this specific forum is not to solve his problems entirely,but to help him solve them by himself.
I hope u'l get it.

Daniel.