Section 14.6 Directional Derivatives and the Gradient Vector

Section 14.6 Directional Derivatives and the Gradient Vector

EX.5

f (x, y) = ye^−x⇒ f_x(x, y) = −ye^−x and f_y(x, y) = e^−x.

If u is a unit vector in the direction of θ = 2π/3, then from Equation 6, D_uf (0, 4) = f_x(0, 4) cos(^2π₃ ) + f_y(0, 4) sin(^2π₃ ) = −4 · (−¹₂) + 1 ·

√3

2 = 2 +

√3 2 .

EX.9

f (x, y, z) = xe^2yz

(a)∇f (x, y, z) = hf_x(x, y, z), f_y(x, y, z), f_z(x, y, z)i = he^2yz, 2xze^2yz, 2xye^2yzi (b)∇f (3, 0, 2) =< 1, 12, 0 >

(c)By Equation 14, D_uf (3, 0, 2) = ∇f (3, 0, 2) · u =< 1, 12, 0 > ·₂

3, −²₃,¹₃ = −²²₃ .

EX.15

f (x, y, z) = xe^y + ye^z+ ze^x ⇒

∇f (x, y, z) = he^y + ze^x, xe^y + e^z, ye^z+ e^xi , ∇f (0, 0, 0) = h1, 1, 1i ,

and a unit vector in the direction of v is u = ^√₂₅₊₁₊₄¹ h5, 1, −2i = ^√1₃₀h5, 1, −2i , so D_uf (0, 0, 0) = ∇f (0, 0, 0) · u = h1, 1, 1i · ^√1

30h5, 1, −2i = ^√4

30.

EX.23

f (x, y) = sin(xy) ⇒ ∇f (x, y) = hy cos(xy), x cos(xy)i , ∇f (1, 0) = h0, 1i . Thus the maximum rate of change is |∇f (1, 0)| = 1 in the directionh0, 1i.

EX.39

f (x, y) = x³+ 5x²y + y³ ⇒

D_uf (x, y) = ∇f (x, y) · u = h3x²+ 10xy, 5x²+ 3y²i ·₃

5,⁴₅

= ⁹₅x²+ 6xy + 4x²+ ¹²₅y² = ²⁹₅x²+ 6xy + ¹²₅y². Then D_u²f (x, y) = D_u[D_uf (x, y)] = ∇ [D_uf (x, y)] · u

=₅₈

5x + 6y, 6x + ²⁴₅y · ³₅,⁴₅ = ²⁹⁴₂₅x +¹⁸⁶₂₅y and D_u²f (2, 1) = ²⁹⁴₂₅(2) + ¹⁸⁶₂₅(1) = ⁷⁷⁴₂₅.

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EX.45

Let F (x, y, z) = x + y + z − e^xyz.Then

x + y + z = e^xyz is the level surface F (x, y, z) = 0, and

∇F (x, y, z) = h1 − yze^xyz, 1 − xze^xyz, 1 − xye^xyzi .

(a)∇F (0, 0, 1) = h1, 1, 1i is a normal vector for the tangent plane at (0, 0, 1), so an equation of the tangent plane is

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

(b) The normal line has directionh1, 1, 1i , so parametric equations are x = t, y = t, z = 1 + t, and symmetric equations are x = y = z − 1.

EX.51

∇F (x₀, y₀, z₀) =_2x0

a² ,^2y_b2⁰,^2z_c2⁰ . Thus an equation of the tangent plane at (x₀, y₀, z₀) is

2x0

a² x + ^2y_b2⁰y + ^2z_c2⁰z = 2

x²₀

a² +^y_b⁰2² +^z_c⁰2²



= 2(1) = 2 since (x0, y0, z0)is a point on the ellipsoid. Hence

x0

a²x + ^y_b⁰2y + ^z_c2⁰z = 1 is an equation of the tangent plane.

EX.56

First note that the point (1, 1, 2) is on both surfaces. The ellipsoid is a level surface of F (x, y, z) = 3x²+ 2y²+z²and ∇F (x, y, z) = h6x, 4y, 2zi . A normal vector to the surface at (1, 1, 2) is ∇F (1, 1, 2) = h6, 4, 4i and an equation of the tangent plane there is 6(x − 1) + 4(y − 1) + 4(z − 2) = 0 or 6x + 4y + 4z = 18 or 3x + 2y + 2z = 9. The sphere is a level surface of G(x, y, z) = x²+ y² + z² − 8x − 6y − 8z + 24 and ∇G(x, y, z) = h2x − 8, 2y − 6, 2z − 8i. A normal vector to the sphere at (1, 1, 2) is ∇G(1, 1, 2) = h−6, −4, −4i and the tangent plane there is −6(x − 1) − 4(y − 1) − 4(z − 2) = 0 or 3x + 2y + 2z = 9.

Since these tangent planes are identical, the surfaces are tangent to each other at the point (1, 1, 2).

EX.61

Let(x₀, y₀, z₀) be a point on the surface. Then an equation of the tangent plane at the point is

x 2√

x0 + ₂^√y_y

0 +₂^√z_z

0 =

√x0+√ y0+√

z0

2 . But √

x₀+√

y₀+√

z₀ =√

c, so the equation is

√x x0+^√y_y

0+^√z_z

0 =√

c. The x-,y-,and z-intercepts are√ cx₀,√

cy₀, and√

cz₀respectively. (The x-intercept is found by setting y = z = 0 and solving the resulting equation for x, and the y- and z-intercepts are found similarly.) So the sum of the intercepts is √

c √

x₀+√

y₀+√

z₀
= c, a constant.

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EX.65

(a)The direction of the normal line of F is given by ∇F , and that of G by ∇G. Assuming that

∇F 6= 0 6= ∇G, the two normal lines are perpendicular at P if ∇F · ∇G = 0 at P

⇔ h∂F/∂x, ∂F/∂y, ∂F/∂zi · h∂G/∂x, ∂G/∂y, ∂G/∂zi = 0 at P

⇔ FxGx+ FyGy+ FzGz = 0 at P .

(b)Here F = x²+ y²− z² and G = x²+ y²+ z²− r², so

∇F · ∇G = h2x, 2y, −2zi · h2x, 2y, 2zi = 4x² + 4y² − 4z² = 4F = 0, since the point (x, y, z) lies on the graph of F = 0. To see that this is true without using calculus, note that G = 0 is the equation of a sphere centered at the origin and F = 0 is the equation of a right circular cone with vertex at the origin (which is generated by lines through the origin). At any point of intersection, the sphere’s normal line(which passes through the origin) lies on the cone, and thus is perpendicular to the cone’s normal line. So the surfaces with equations F = 0 and G = 0 are everywhere orthogonal.

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