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Partial Derivatives

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Miguel Holgado

on 25 May 2011

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Transcript of Partial Derivatives

artial ∂erivatives Miguel
-College: Texas A&M
-Majors:
Nuclear Engineering
Mechanical Engineering Brandon
-College: Texas A&M
-Major: Biomedical Engineering Recall that given a function of one variable, the derivative, represents the rate of change of the function as x changes. But what if the functions depends on several variables such as x and y?
For example: z = 9 - x^2 - y^2 Partial derivatives can be used for:
-tangent plane approximations
-gradient vectors
-relative maxs/mins
-absolute maxs/mins
-Lagrange multipliers Functions of several variables often result in a 3-D surface. Taking the derivative at a certain point on the surface would be pointless because there would be an infinite number of orientations that the tangent line could lie on the point. This is why we take a partial derivative in which we take the derivative of the function with respect to a specific variable while treating other variables constant. We can take the partial derivative of z with respect to x while treating y as a constant.
z = x^2 + xy + y^2
∂z/∂x = 2x + y Higher-order Partial Derivatives END Suppose that we wish to find the partial derivative of the function with respect to x at y =1 at f(1,1). Electrical Circuits: Changes in Current A circuit with an electromotive force/voltage (V) and a resistance (R) has a current (I) that depends on both the voltage and the resistance.
I = V/R = V(R^-1) The partial derivative of the current with respect to the voltage while keeping the resistance constant is:
∂I/∂V = 1/R when V = 120 and R = 15
∂I/∂V = 1/15 = 0.067 mho At the instant when V=120 volts and R=15 ohms , what is the rate of change of current with respect to voltage? What is the rate of change of current with respect to resistance? Verbal interpretation in context: If the resistance is fixed at 15 ohms, the current is increasing with respect to voltage at the rate of 0.0667 amperes per volt when the EMF is 120 volts. The partial derivative of the current with respect to resistance while keeping voltage constant:
∂I/∂R = -E/(R^2) when V = 120 and R = 15
∂I/∂V = -120/225 = -0.533 ampere/ohm Verbal interpretation in context: if the EMF is fixed at 120 volts, the current is decreasing with respect to resistance at the rate of 0.533 amperes per ohm when the resistance is 15 ohms. ∂ ∂ z = 9 - x^2 - y^2
∂z/∂x = -2x
∂z/∂x evaluated at f(1,1)
∂z/∂x = -2
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