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,
Partial Derivatives
The z partial derivative of f(x,y,z) is
fz(x,y,z)=h→0limhf(x,y,z+h)−f(x,y,z)
Find fx, fy, fz for f(x,y,z)=xln(cos(xy2z)).
We use the chain rule twice in each derivative:
Partial derivative with respect to x:
fx=ln(cos(xy2z))−xy2zcos(xy2z)sin(xy2z)=ln(cos(xy2z))−xy2ztan(xy2z)
Partial derivative with respect to y:
fy=−2x2yzcos(xy2z)sin(xy2z)=−2x2yztan(xy2z)
Partial derivative with respect to z:
fz=−x2y2cos(xy2z)sin(xy2z)=−x2y2tan(xy2z)
Find gx, gy, gz if g(x,y,z)=yexy2z3.
Hint
As you take each partial derivative, regard the other variables as constants.
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=8.314K molJ. 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 (∂T∂E)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, (∂T∂E)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=0.9Km3. The specific
heat at constant volume is then defined to be
CV=(∂T∂E)V.
Find CV when V=10m3, and T=300K.
Hint
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.
Since E=cPT and P=VnRT and n=1, we have
E=VcRT2. Consequently,
CV=(∂T∂E)V=V2cRT=102⋅0.9⋅8.314⋅300=448.96KJ
The units are KJ because in the derivative
∂T∂E, the energy is in Joules and the
temperature is in Kelvins.
Later we will learn how to solve this problem using the
chain rule.
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 R3 generalizes
to a tangent hyperplane to a hypersurface in R4. 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) at (x,y,z)=(a,b,c) is:
w=ftan(x,y,z)≡f(a,b,c)+fx(a,b,c)(x−a)+fy(a,b,c)(y−b)+fz(a,b,c)(z−c).
In differential notation, this is:
w=ftan(x,y,z)≡f(a,b,c)+∂x∂f∣∣∣∣(a,b,c)(x−a)+∂y∂f∣∣∣∣(a,b,c)(y−b)+∂z∂f∣∣∣∣(a,b,c)(z−c).
Find the tangent hyperplane to the graph of the function w=x2y2z3
at the point (2,1,1).
Let f(x,y,z)=x2y2z3. We first find the value of the function at the
point (2,1,1).
f(2,1,1)=221213=4
Next we calculate the partial derivatives, fx, fy, and fz,
and their values at (2,1,1).
fxfyfz=2xy2z3=2x2yz3=3x2y2z2fx(2,1,1)fy(2,1,1)fz(2,1,1)=4=8=12
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)(x−2)+fy(2,1,1)(y−1)+fz(2,1,1)(z−1)=4+4(x−2)+8(y−1)+12(z−1)=4x+8y+12z−24
If we need it, the w-intercept is −24.
Find the tangent hyperplane to the graph of the function
w=5+9x2+16y2+4z2 at (3,4,2).
We evaluate the function and its partial derivatives at (3,4,2).
ffxfyfz=5+9x2+16y2+4z2=92x=162y=42zf(3,4,2)fx(3,4,2)fy(3,4,2)fz(3,4,2)=8=32=21=1
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)(x−3)+fy(3,4,2)(y−4)+fz(3,4,2)(z−2)=8+32(x−3)+21(y−4)+1(z−2)=32x+21y+z+2
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