11. Partial Derivatives and Tangent Planes

e. More Variables

Everything which has been said about functions of \(2\) variables, generalizes directly to functions of \(3\) or more variables. We will give the formulas for a functions \(f(x,y,z)\) which has \(3\) variables, but everything generalizes to any number of vaiables.

Partial Derivatives

A partial derivative with respect to one variable, say \(z\), is computed by differentiating with respect to \(z\) while holding all the other variables constant. In particular,

The \(z\) partial derivative of \(f(x,y,z)\) is \[ f_z(x,y,z)=\lim_{h\rightarrow 0} \dfrac{f(x,y,z+h)-f(x,y,z)}{h} \]

Find \(f_x\), \(f_y\), \(f_z\) for \(f(x,y,z)=x\ln(\cos(xy^2z))\).

We use the chain rule twice in each derivative:
Partial derivative with respect to \(x\): \[\begin{aligned} f_x&=\ln(\cos(xy^2z))-xy^2z\dfrac{\sin(xy^2z)}{\cos(xy^2z)} \\ &=\ln(\cos(xy^2z))-xy^2z\tan(xy^2z) \end{aligned}\] Partial derivative with respect to \(y\): \[ f_y=-2x^2yz\dfrac{\sin(xy^2z)}{\cos(xy^2z)} =-2x^2yz\tan(xy^2z) \] Partial derivative with respect to \(z\): \[ f_z=-x^2y^2\dfrac{\sin(xy^2z)}{\cos(xy^2z)} =-x^2y^2\tan(xy^2z) \]

Find \(g_x\), \(g_y\), \(g_z\) if \(g(x,y,z)=y\,e^{xy^2z^3}\).

As you take each partial derivative, regard the other variables as constants.

\(\begin{aligned} g_x&=y^3z^3e^{xy^2z^3} \\ g_y&=e^{xy^2z^3}+2xy^2z^3e^{xy^2z^3} \\ g_z&=3xy^3z^2e^{xy^2z^3} \end{aligned}\)

Physics Notation

In thermodynamics, when dealing with an ideal gas, the pressure, volume and temperature are related by the Ideal Gas Law, \(PV=nRT\), where \(P\) is the pressure, \(V\) is the volume, and \(T\) is the temperature of \(n\) moles af gas, where \(R\) is the constant \(R=8.314\,\dfrac{\text{J}}{\text{K mol}}\). So quantities such as energy (or the Helmholtz or Gibbs functions), can be regarded as functions either of temperature and pressure, \(E(T,P)\), or temperature and volume, \(E(T,V)\), or pressure and volume, \(E(P,V)\). Mathematicians would give three different names to these versions of the energy and go back and forth by constructing compositions with the Ideal Gas Law. Derivatives are then computed using a chain rule. However, physicists prefer to call them all energy and use the same symbol, \(E\). In that case, there can be confusion when taking partial derivatives as to which function is being considered. Consequently, we use the notation \(\left(\dfrac{\partial E}{\partial T}\right)_P\) to mean that \(E\) is a regarded as a function of \(T\) and \(P\) and we are taking the derivative of \(E\) with respect to \(T\) while \(P\) is held constant. Similarly, \(\left(\dfrac{\partial E}{\partial T}\right)_V\) is the derivative of \(E\) with respect to \(T\) at constant \(V\).

The internal energy of one mole (\(n=1\)) of an ideal gas is given by \(E=cPT\) where \(c=0.9\,\dfrac{\text{m}^3}{\text{K}}\). The specific heat at constant volume is then defined to be \(C_V=\left(\dfrac{\partial E}{\partial T}\right)_V\). Find \(C_V\) when \(V=10\,\text{m}^3\), and \(T=300\,\text{K}\).

Use the Ideal Gas Law, \(PV=RT\), to convert the energy as a function of \(P\) and \(T\) into the energy as a function of \(V\) and \(T\).

\(C_V=448.96\,\dfrac{\text{J}}{\text{K}}\).

Since \(E=cPT\) and \(P=\dfrac{nRT}{V}\) and \(n=1\), we have \(E=\dfrac{cRT^2}{V}\). Consequently, \[\begin{aligned} C_V&=\left(\dfrac{\partial E}{\partial T}\right)_V =\dfrac{2cRT}{V} \\ &=\dfrac{2\cdot0.9\cdot8.314\cdot300}{10}=448.96\,\dfrac{\text{J}}{\text{K}} \end{aligned}\] The units are \(\dfrac{\text{J}}{\text{K}}\) because in the derivative \(\dfrac{\partial E}{\partial T}\), the energy is in Joules and the temperature is in Kelvins.
Later we will learn how to solve this problem using the chain rule.

Tangent Hyperplanes

Although you cannot visualize the graph of \(w=f(x,y,z)\) as a surface in \(4\) dimensional space, the \(z\)-trace is still the curve obtained by letting \(z\) vary while holding all the other variables constant and its slope is still the \(z\)-partial derivative. The notion of a tangent plane to a surface in \(\mathbb{R}^3\) generalizes to a tangent hyperplane to a hypersurface in \(\mathbb{R}^4\). The formula is:

The equation of the tangent hyperplane to the graph of the function \(w=f(x,y,z)\) at \((x,y,z)=(a,b,c)\) is: \[\begin{aligned} w&=f_{\tan}(x,y,z) \\ &\equiv f(a,b,c)+f_x(a,b,c)(x-a) \\ &\qquad+f_y(a,b,c)(y-b)+f_z(a,b,c)(z-c). \end{aligned}\] In differential notation, this is: \[\begin{aligned} w&=f_{\tan}(x,y,z) \\ &\equiv f(a,b,c) +\left.\dfrac{\partial f}{\partial x}\right|_{(a,b,c)}(x-a) \\ &\qquad+\left.\dfrac{\partial f}{\partial y}\right|_{(a,b,c)}(y-b) +\left.\dfrac{\partial f}{\partial z}\right|_{(a,b,c)}(z-c). \end{aligned}\]

Find the tangent hyperplane to the graph of the function \(w=x^2y^2z^3\) at the point \((2,1,1)\).

Let \(f(x,y,z)=x^2y^2z^3\). We first find the value of the function at the point \((2,1,1)\). \[ f(2,1,1)=2^2 1^2 1^3=4 \] Next we calculate the partial derivatives, \(f_x\), \(f_y\), and \(f_z\), and their values at \((2,1,1)\). \[\begin{aligned} f_x&=2xy^2z^3\qquad& f_x(2,1,1)&=4\\ f_y&=2x^2yz^3\qquad& f_y(2,1,1)&=8 \\ f_z&=3x^2y^2z^2\qquad& f_z(2,1,1)&=12 \\ \end{aligned}\] Finally, we plug the numbers into the equation of the tangent hyperplane. \[\begin{aligned} w&=f_{\tan}(x,y,z) \\ &=f(2,1,1)+f_x(2,1,1)(x-2) \\ &\qquad+f_y(2,1,1)(y-1)+f_z(2,1,1)(z-1) \\ &=4+4(x-2)+8(y-1)+12(z-1) \\ &=4x+8y+12z-24 \\ \end{aligned}\] If we need it, the \(w\)-intercept is \(-24\).

Find the tangent hyperplane to the graph of the function \(w=5+\dfrac{x^2}{9}+\dfrac{y^2}{16}+\dfrac{z^2}{4}\) at \((3,4,2)\).

The tangent hyperplane is: \(w=\dfrac{2}{3}x+\dfrac{1}{2}y+z+2\).

We evaluate the function and its partial derivatives at \((3,4,2)\). \[\begin{aligned} f&=5+\dfrac{x^2}{9}+\dfrac{y^2}{16}+\dfrac{z^2}{4}\quad&f(3,4,2)&=8\\ f_x&=\dfrac{2x}{9}\quad&f_x(3,4,2)&=\dfrac{2}{3}\\ f_y&=\dfrac{2y}{16}\quad&f_y(3,4,2)&=\dfrac{1}{2} \\ f_z&=\dfrac{2z}{4}\quad&f_z(3,4,2)&=1 \\ \end{aligned}\] Finally, we plug the numbers into the equation of the tangent hyperplane. \[\begin{aligned} w&=f_{\tan}(x,y,z) \\ &=f(3,4,2)+f_x(3,4,2)(x-3) \\ &\qquad+f_y(3,4,2)(y-4)+f_z(3,4,2)(z-2) \\ &=8+\dfrac{2}{3}(x-3)+\dfrac{1}{2}(y-4)+1(z-2) \\ &=\dfrac{2}{3}x+\dfrac{1}{2}y+z+2 \\ \end{aligned}\]

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