11. Partial Derivatives and Tangent Planes

e. More Variables

Everything which has been said about functions of 22 variables, generalizes directly to functions of 33 or more variables. We will give the formulas for a functions f(x,y,z)f(x,y,z) which has 33 variables, but everything generalizes to any number of vaiables.

Partial Derivatives

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

Partial Derivatives
The zz partial derivative of f(x,y,z)f(x,y,z) is fz(x,y,z)=limh0f(x,y,z+h)f(x,y,z)h f_z(x,y,z)=\lim_{h\rightarrow 0} \dfrac{f(x,y,z+h)-f(x,y,z)}{h}

Find fxf_x, fyf_y, fzf_z for f(x,y,z)=xln(cos(xy2z))f(x,y,z)=x\ln(\cos(xy^2z)).

We use the chain rule twice in each derivative:
Partial derivative with respect to xx: fx=ln(cos(xy2z))xy2zsin(xy2z)cos(xy2z)=ln(cos(xy2z))xy2ztan(xy2z)\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 yy: fy=2x2yzsin(xy2z)cos(xy2z)=2x2yztan(xy2z) f_y=-2x^2yz\dfrac{\sin(xy^2z)}{\cos(xy^2z)} =-2x^2yz\tan(xy^2z) Partial derivative with respect to zz: fz=x2y2sin(xy2z)cos(xy2z)=x2y2tan(xy2z) f_z=-x^2y^2\dfrac{\sin(xy^2z)}{\cos(xy^2z)} =-x^2y^2\tan(xy^2z)

Find gxg_x, gyg_y, gzg_z if g(x,y,z)=yexy2z3g(x,y,z)=y\,e^{xy^2z^3}.

Hint

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

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Answer

gx=y3z3exy2z3gy=exy2z3+2xy2z3exy2z3gz=3xy3z2exy2z3\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}

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Physics Notation

In thermodynamics, when dealing with an ideal gas, the pressure, volume and temperature are related by the Ideal Gas Law, PV=nRTPV=nRT, where PP is the pressure, VV is the volume, and TT is the temperature of nn moles af gas, where RR is the constant R=8.314JK molR=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)E(T,P), or temperature and volume, E(T,V)E(T,V), or pressure and volume, E(P,V)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, EE. In that case, there can be confusion when taking partial derivatives as to which function is being considered. Consequently, we use the notation (ET)P\left(\dfrac{\partial E}{\partial T}\right)_P to mean that EE is a regarded as a function of TT and PP and we are taking the derivative of EE with respect to TT while PP is held constant. Similarly, (ET)V\left(\dfrac{\partial E}{\partial T}\right)_V is the derivative of EE with respect to TT at constant VV.

The internal energy of one mole (n=1n=1) of an ideal gas is given by E=cPTE=cPT where c=0.9m3Kc=0.9\,\dfrac{\text{m}^3}{\text{K}}. The specific heat at constant volume is then defined to be CV=(ET)VC_V=\left(\dfrac{\partial E}{\partial T}\right)_V. Find CVC_V when V=10m3V=10\,\text{m}^3, and T=300KT=300\,\text{K}.

Hint

Use the Ideal Gas Law, PV=RTPV=RT, to convert the energy as a function of PP and TT into the energy as a function of VV and TT.

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Answer

CV=448.96JKC_V=448.96\,\dfrac{\text{J}}{\text{K}}.

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Solution

Since E=cPTE=cPT and P=nRTVP=\dfrac{nRT}{V} and n=1n=1, we have E=cRT2VE=\dfrac{cRT^2}{V}. Consequently, CV=(ET)V=2cRTV=20.98.31430010=448.96JK\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 JK\dfrac{\text{J}}{\text{K}} because in the derivative ET\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.

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Tangent Hyperplanes

Although you cannot visualize the graph of w=f(x,y,z)w=f(x,y,z) as a surface in 44 dimensional space, the zz-trace is still the curve obtained by letting zz vary while holding all the other variables constant and its slope is still the zz-partial derivative. The notion of a tangent plane to a surface in R3\mathbb{R}^3 generalizes to a tangent hyperplane to a hypersurface in R4\mathbb{R}^4. The formula is:

Equation of a Tangent HyperPlane to a Graph
The equation of the tangent hyperplane to the graph of the function w=f(x,y,z)w=f(x,y,z) at (x,y,z)=(a,b,c)(x,y,z)=(a,b,c) is: w=ftan(x,y,z)f(a,b,c)+fx(a,b,c)(xa)+fy(a,b,c)(yb)+fz(a,b,c)(zc).\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: w=ftan(x,y,z)f(a,b,c)+fx(a,b,c)(xa)+fy(a,b,c)(yb)+fz(a,b,c)(zc).\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=x2y2z3w=x^2y^2z^3 at the point (2,1,1)(2,1,1).

Let f(x,y,z)=x2y2z3f(x,y,z)=x^2y^2z^3. We first find the value of the function at the point (2,1,1)(2,1,1). f(2,1,1)=221213=4 f(2,1,1)=2^2 1^2 1^3=4 Next we calculate the partial derivatives, fxf_x, fyf_y, and fzf_z, and their values at (2,1,1)(2,1,1). fx=2xy2z3fx(2,1,1)=4fy=2x2yz3fy(2,1,1)=8fz=3x2y2z2fz(2,1,1)=12\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. w=ftan(x,y,z)=f(2,1,1)+fx(2,1,1)(x2)+fy(2,1,1)(y1)+fz(2,1,1)(z1)=4+4(x2)+8(y1)+12(z1)=4x+8y+12z24\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 ww-intercept is 24-24.

Find the tangent hyperplane to the graph of the function w=5+x29+y216+z24w=5+\dfrac{x^2}{9}+\dfrac{y^2}{16}+\dfrac{z^2}{4} at (3,4,2)(3,4,2).

Answer

The tangent hyperplane is: w=23x+12y+z+2w=\dfrac{2}{3}x+\dfrac{1}{2}y+z+2.

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Solution

We evaluate the function and its partial derivatives at (3,4,2)(3,4,2). f=5+x29+y216+z24f(3,4,2)=8fx=2x9fx(3,4,2)=23fy=2y16fy(3,4,2)=12fz=2z4fz(3,4,2)=1\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. w=ftan(x,y,z)=f(3,4,2)+fx(3,4,2)(x3)+fy(3,4,2)(y4)+fz(3,4,2)(z2)=8+23(x3)+12(y4)+1(z2)=23x+12y+z+2\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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