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Home , Directional derivative, Gradient

Math 314 Lecture #16 §14.6: Directional and the Vector

Outcome A: Compute the directional of a of several variables at a given point in a given direction.

The partial derivatives fx(x0, y0) and fy(x0, y0) measure the rate of change of f in the x and y directions respectively, i.e., in the ~i and ~j direction. To measure the rate of change in another unit direction ~u = ha, bi we look at the curve obtained by the intersection of z = f(x, y) and the vertical plane containing the point (x0, y0, z0) and parallel to ~u. Here is a picture of this intersection.

For a small constant h, a small displacement from the point (x0, y0) in the direction of ~u is the vector h~u, that is we move from (x0, y0) to the point (x0 + ha, y0 + hb). The “rise” over “run” of f for this displacement is the ratio ∆z z − z f(x + ha, y + hb) − f(x , y ) = 0 = 0 0 0 0 . h h h

The of f at the point (x0, y0) in the direction of the ~u is the scalar quantity

f(x0 + ha, y0 + hb) − f(x0, y0) D~uf(x0, y0) = lim , h→0 h if the exists. Theorem. If f(x, y) is differentiable, then f has directional derivative in the direction of every unit vector ~u = ha, bi, and

D~uf(x, y) = fx(x, y)a + fy(x, y)b. Notice that the directional derivative is nothing more than the of the vector hfx(x, y), fy(x, y)i with ~u.

Proof. For x = x0 + ha and y = y0 + hb, the composition g(h) = f(x0 + ha, y0 + hb) is a differentiable function of h by the , and so direct differentiation gives

0 g(h) − g(0) f(x0 + ha, y0 + hb) − f(x0, y0) g (0) = lim = lim = D~uf(x0, y0). h→0 h h→0 h On the other hand, the Chain Rule gives an equal expression for the derivative of g with respect to h, ∂f dx ∂f dy g0(h) = + = f (x + ha, y + hb)a + f (x + ha, y + hb)b, ∂x dh ∂y dt x 0 0 y 0 0 in which by setting h = 0 we get

0 g (0) = fx(x0, y0)a + fy(x0, y0)b.

0 We have two expressions for g (0), and so the two expressions are equal.  For a differentiable function f of three variables x, y, z, the directional derivative at a point (x0, y0, z0) in the direction of a unit vector ~u = ha, b, ci is the scalar

D~uf(x0, y0, z0) = hfx(x0, y0, z0), fy(x0, y0, z0), fz(x0, y0, z0)i · ha, b, ci.

2 Example. Find√ the directional derivative of f(x, y) = y /x at the point (1, 2) in the direction of h2, 5i. Notice that the given direction vector IS NOT a unit vector, so we normalize it first to get * √ + 2 5 ~u = , . 3 3

2 2 With fx = −y /x and fy = 2y/x, the directional derivative of f at (1, 2) in the direction ~u is √ √ 4 5 − 8 D f(1, 2) = −4(2/3) + 4( 5/3) = > 0. ~u 3

Outcome B: Find the gradient of a function of several variables.

The vector hfx, fyi that appears in the computation of the directional derivative of a function of two variables, is an example of a vector field, i.e., at each point (x, y) we attach the vector hfx, fyi. The gradient of a function f(x, y) is the vector field given by

∇f(x, y) = hfx(x, y), fy(x, y)i. The gradient of a function f(x, y, z) is the vector field given by

∇f(x, y, z) = hfx(x, y, z), fy(x, y, z), fz(x, y, z)i. The derivative of a differentiable function f of two or more variables is the gradient ∇f, i.e., it appears in the definition of differentiability of a function of two variables as

∆z = ∇f(x, y) · h∆x, ∆yi + 1∆x + 2∆y. Example. The gradient (derivative) of hyperbolic paraboloid z = f(x, y) = x2 − y2 is ∇f(x, y) = h2x, −2yi. Outcome C: Find the maximum rate of change of a function at a given point and the direction in which it occurs. Theorem. If f is a differentiable function of two or more variables, then the maximum value of the directional derivative is k∇fk and it occurs in the direction of ∇f.

Proof. The direction derivative is the dot product D~uf = ∇f · u for a unit vector ~u. Recall that ~a ·~b = k~ak kbk cos θ where θ is the angle between ~a and ~b.

Thus the directional derivative is D~uf = k∇fk k~uk cos θ = k∇fk cos θ.

The maximum value of D~uf occurs when cos θ = 1, i.e., when the angle between ∇f and ~u is 0.

The maximum value of D~uf occurs when ~u = ∇f/k∇fk, and is k∇fk.  Example. The gradient of w = f(x, y, z) = (x + y)/z is

∇f = h1/z, 1/z, −(x + y)/z2i.

The gradient of f at the point (1, 1, −1) is the vector

∇f(1, 1, −1) = h−1, −1, −2i.

Moving in the direction of h−1, −1, −2i from the point (1, 1, −1) gives the maximum rate of change in the values f which is √ k∇f(1, 1, −1)k = kh−1, −1, −2ik = 6.

Outcome D: Find equations of the plane and the line to a at a point. Suppose a surface S is given implicitly by F (x, y, z) = k for a differentiable function F , i.e., the level surface of a differentiable function of three variables. Any surface of the form z = f(x, y) can be put in this form as f(x, y) − z = 0 for a differentiable f, i.e., F (x, y, z) = f(x, y) − z and k = 0. A smooth curve ~r(t) = hx(t), y(t), z(t)i that lies on S satisfies the equation

F (~r(t)) = k.

Applying the Chain Rule to this equation gives dx dy dz 0 = F ~r(t) + F ~r(t) + F ~r(t) = ∇F ~r(t) · ~r 0(t). x dt y dt z dt

Since ~r(t) is a smooth curve that lies in S, the tangent line of ~r(t) at t = t0 lies in the tangent plane of S at ~r(t0). 0 With hx0, y0, z0i = ~r(t0), the equation ∇F ·~r = 0 implies that ∇F (x0, y0, z0) is orthogonal to the tangent plane of S at (x0, y0, z0). Thus another equation for the tangent plane of S at (x0, y0, z0) is

Fx(x0, y0, z0)(x − x0) + Fy(x0, y0, z0)(y − y0) + Fz(x0, y0, z0)(z − z0) = 0.

When F (x, y, z) = f(x, y) − z, then Fz = −1 and this equation for the tangent line becomes the familiar

z − z0 = fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0).

An equation for the normal line to S at (x0, y0, z0) is

hx, y, zi = hx0, y0, z0i + t∇F (x0, y0, z0).

Example. If F (x, y, z) = x2 + y2 + 2z2 − 4 = 0, then

∇F (x, y, z) = h2x, 2y, 4zi.

An equation for the tangent plane at the point (1, 1, 1) on this ellipsoid is

2(x − 1) + 2(y − 1) − 4(z − 1) = 0 or z − 1 = 2(x − 1) + 4(y − 1).

Here is a picture of the ellipsoid and its tangent plane at the point (1, 1, 1).

An equation for the normal line to this surface at the point (2, 1, 3) is

hx, y, zi = h1, 1, 1i + th2, 2, 4i.