Question

1. Determine the first order partial derivative of the following functions:

a. z = In (x + t^2)

b. F(x,y,z) = xy^2 e^-xz

2. Clairaut's Theorwm holds that Uxy = Uyx , show that the following equations obey Clairaut's Theorem:

a. U = In( x+2y)

b. U = e^xy sin y

3. Laplaces equation holds that Uxx + Uyy= 0, verify that the second derivative of the following equations are Laplace's equations:

a. U = In√x^2 + y^2

b. U = x^2 - y^2

Answer 1

1.

a.

$$\dfrac{\partial z}{\partial x}=\dfrac{1}{x+t^2}$$

$$\dfrac{\partial z}{\partial t}=\dfrac{2t}{x+t^2}$$

b.

$$\dfrac{\partial F}{\partial x}=y^2e^{-xz}-xy^2ze^{-xz}$$ $$\dfrac{\partial F}{\partial y}=2ye^{-xz}$$ $$\dfrac{\partial F}{\partial z}=-x^2y^2e^{-xz}$$

2.

a.

$$\dfrac{\partial U}{\partial x}=\dfrac{1}{x+2y}$$

$$\dfrac{\partial U}{\partial y\partial x}=-\dfrac{2}{(x+2y)^2}$$

$$\dfrac{\partial U}{\partial y}=\dfrac{2}{x+2y}$$

$$\dfrac{\partial U}{\partial x\partial y}=-\dfrac{2}{(x+2y)^2}$$

$$\dfrac{\partial U}{\partial y\partial x}=-\dfrac{2}{(x+2y)^2}=\dfrac{\partial U}{\partial x\partial y}, True$$

b.

$$\dfrac{\partial U}{\partial x}=ye^{xy}\sin y$$

$$\dfrac{\partial U}{\partial y\partial x}=e^{xy}\sin y+xye^{xy}\sin y+ye^{xy}\cos y$$

$$\dfrac{\partial U}{\partial y}=xe^{xy}\sin y+e^{xy}\cos y$$

$$\dfrac{\partial U}{\partial x\partial y}=e^{xy}\sin y+xye^{xy}\sin y+ye^{xy}\cos y$$

$$\dfrac{\partial U}{\partial y\partial x}=e^{xy}\sin y+xye^{xy}\sin y+ye^{xy}\cos y$$

$$=\dfrac{\partial U}{\partial x\partial y}, True$$

3.

a.

$$U = \ln\sqrt{x^2 + y^2}=\dfrac{1}{2}\ln (x^2+y^2)$$

$$U_x=\dfrac{1}{2}(\dfrac{1}{x^2+y^2})(2x)=\dfrac{x}{x^2+y^2}$$

$$U_{xx}=\dfrac{x^2+y^2-2x^2}{(x^2+y^2)^2}=\dfrac{y^2-x^2}{(x^2+y^2)^2}$$

$$U_y=\dfrac{1}{2}(\dfrac{1}{x^2+y^2})(2x)=\dfrac{y}{x^2+y^2}$$

$$U_{yy}=\dfrac{x^2+y^2-2y^2}{(x^2+y^2)^2}=\dfrac{x^2-y^2}{(x^2+y^2)^2}$$

$$U_{xx}+U_{yy}=\dfrac{y^2-x^2}{(x^2+y^2)^2}+\dfrac{x^2-y^2}{(x^2+y^2)^2}=0, True$$

b.

$$U_x=2x$$

$$U_{xx}=2$$

$$U_y=-2y$$

$$U_{yy}=-2$$

$$U_{xx}+U_{yy}=2-2=0, True$$