MATH 251: Calculus 3, SET8
14: Partial Derivatives
14.3: Partial Derivatives
These problems are done with the CAS. See Hand Solutions for details.
We estimate the partial derivatives from the contour plot using difference quotients.
f_x_2_1 = (14-10) / (3-2)
As well as ordinary derivatives (from Calculus 1), the diff command computes partial derivatives as well.
fx = diff(f,x), fy = diff(f,y)
Recall Part 2 of the Fundamental Theorem of Calculus and Properties of Integrals.
F(alpha,beta) = int(sqrt(t^3 + 1), t, alpha, beta);
F_alpha = simplify(diff(F,alpha)), F_beta = simplify(diff(F,beta))
Here we do an experiment with a specific number of variables, .
This may help you to see the general pattern, which is obtained via the Chain Rule.
In general, whence .
u = sin(X(1) + 2*X(2) + 3*X(3) + 4*X(4) + 5*X(5)) % sample with n = 5
M = [M; [j diff(f, X(j))]];
M % column 1: j; column 2: partial derivative of u w.r.t. the jth variable
f(x,y,z) = x^(y*z), fz = diff(f,z), e = exp(sym(1))
Here we look ahead to the end of Section 14.5: The Chain Rule. There we find a much quicker way to do implicit differentiation. See page 943 of textbook.
( It sure beats the Calc 1 way of doing things since its a one-step formula! )
F = y*z + x*log(y) - z^2 % Let F = LHS - RHS of the equation given, understood to be set to zero.
zx = -diff(F,x) / diff(F,z)
zy = -diff(F,y) / diff(F,z)
By hand, you would compute a first, second, and third partial derivative in succession.
That's what the diff command is doing internally and just displaying the final result.
g = exp(r) * sin(s*t), g_rst = simplify( diff(g, r,s,t) )
Let's check all six functions at once! We see that 1st and 3rd functions are not solutions of Laplace's equation ; the others are solutions.
u = [x^2+y^2 x^2-y^2 x^3+3*x*y^2 1/2*log(x^2+y^2) sin(x)*cosh(y)+cos(x)*sinh(y) exp(-x)*cos(y)-exp(-y)*cos(x)].'
L = simplify( diff(u,x,2) + diff(u,y,2) )
The requisite partial derivatives are evaluated to determine rates of change of temperature in the specified directions.
T(x,y) = 60 / (1+x^2+y^2)
Tx = diff(T,x), Ty = diff(T,y)
Tx_P = Tx(2,1), Ty_P = Ty(2,1)
Parameterize the curve of intersection of the paraboloid and the plane as . The point corresponds to . Now proceed in the usual manner.
r(t) = [1 t 4-2*t^2], Dr = diff(r,t), s1 = sym(1)
fsurf(z, [-2 4 -2 4], 'MeshDensity', 16); hold on
fplot3(s1, t, 4-2*t^2, [-2 4], 'm', 'LineWidth', 3)
fplot3(s1, u+2, -8*u-4, [-5 3], 'r', 'LineWidth', 3)
xlabel('x'); ylabel('y'); zlabel('z')