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Math  /  Calculus

Questionb. Find the direction derivatives of ϕ=x2yz+4xz2\phi=x^{2} y z+4 x z^{2} at (1,−2,−1)(1,-2,-1) in the direction 2i−j−2k2 i-j-2 k

Studdy Solution

STEP 1

Assumptions
1. The function is given by ϕ(x,y,z)=x2yz+4xz2\phi(x, y, z) = x^2 y z + 4 x z^2.
2. We need to find the directional derivative of ϕ\phi at the point (1,−2,−1)(1, -2, -1).
3. The direction vector is given by v=2i−j−2k\mathbf{v} = 2\mathbf{i} - \mathbf{j} - 2\mathbf{k}.
4. The directional derivative is calculated using the gradient of ϕ\phi and the unit vector in the direction of v\mathbf{v}.

STEP 2

First, compute the gradient of ϕ\phi. The gradient ∇ϕ\nabla \phi is given by:
∇ϕ=(∂ϕ∂x,∂ϕ∂y,∂ϕ∂z)\nabla \phi = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right)

STEP 3

Calculate ∂ϕ∂x\frac{\partial \phi}{\partial x}:
∂ϕ∂x=∂∂x(x2yz+4xz2)=2xyz+4z2\frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x} (x^2 y z + 4 x z^2) = 2 x y z + 4 z^2

STEP 4

Calculate ∂ϕ∂y\frac{\partial \phi}{\partial y}:
∂ϕ∂y=∂∂y(x2yz+4xz2)=x2z\frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} (x^2 y z + 4 x z^2) = x^2 z

STEP 5

Calculate ∂ϕ∂z\frac{\partial \phi}{\partial z}:
∂ϕ∂z=∂∂z(x2yz+4xz2)=x2y+8xz\frac{\partial \phi}{\partial z} = \frac{\partial}{\partial z} (x^2 y z + 4 x z^2) = x^2 y + 8 x z

STEP 6

Evaluate the gradient ∇ϕ\nabla \phi at the point (1,−2,−1)(1, -2, -1):
∇ϕ(1,−2,−1)=(2(1)(−2)(−1)+4(−1)2,(1)2(−1),(1)2(−2)+8(1)(−1))\nabla \phi (1, -2, -1) = \left( 2(1)(-2)(-1) + 4(-1)^2, (1)^2(-1), (1)^2(-2) + 8(1)(-1) \right)

STEP 7

Simplify the components of the gradient at (1,−2,−1)(1, -2, -1):
∂ϕ∂x∣(1,−2,−1)=2(1)(−2)(−1)+4(−1)2=4+4=8\frac{\partial \phi}{\partial x} \bigg|_{(1, -2, -1)} = 2(1)(-2)(-1) + 4(-1)^2 = 4 + 4 = 8
∂ϕ∂y∣(1,−2,−1)=(1)2(−1)=−1\frac{\partial \phi}{\partial y} \bigg|_{(1, -2, -1)} = (1)^2(-1) = -1
∂ϕ∂z∣(1,−2,−1)=(1)2(−2)+8(1)(−1)=−2−8=−10\frac{\partial \phi}{\partial z} \bigg|_{(1, -2, -1)} = (1)^2(-2) + 8(1)(-1) = -2 - 8 = -10

STEP 8

Thus, the gradient at (1,−2,−1)(1, -2, -1) is:
∇ϕ(1,−2,−1)=(8,−1,−10)\nabla \phi (1, -2, -1) = (8, -1, -10)

STEP 9

Normalize the direction vector v=2i−j−2k\mathbf{v} = 2\mathbf{i} - \mathbf{j} - 2\mathbf{k}. The unit vector u\mathbf{u} in the direction of v\mathbf{v} is given by:
u=v∥v∥\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|}

STEP 10

Calculate the magnitude of v\mathbf{v}:
∥v∥=22+(−1)2+(−2)2=4+1+4=9=3\|\mathbf{v}\| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3

STEP 11

Normalize v\mathbf{v}:
u=13(2,−1,−2)=(23,−13,−23)\mathbf{u} = \frac{1}{3}(2, -1, -2) = \left( \frac{2}{3}, -\frac{1}{3}, -\frac{2}{3} \right)

STEP 12

The directional derivative of ϕ\phi in the direction of v\mathbf{v} at (1,−2,−1)(1, -2, -1) is given by the dot product of ∇ϕ(1,−2,−1)\nabla \phi (1, -2, -1) and u\mathbf{u}:
Duϕ(1,−2,−1)=∇ϕ(1,−2,−1)⋅uD_{\mathbf{u}} \phi (1, -2, -1) = \nabla \phi (1, -2, -1) \cdot \mathbf{u}

STEP 13

Compute the dot product:
Duϕ(1,−2,−1)=(8,−1,−10)⋅(23,−13,−23)D_{\mathbf{u}} \phi (1, -2, -1) = (8, -1, -10) \cdot \left( \frac{2}{3}, -\frac{1}{3}, -\frac{2}{3} \right)

STEP 14

Calculate the dot product:
Duϕ(1,−2,−1)=8⋅23+(−1)⋅(−13)+(−10)⋅(−23)D_{\mathbf{u}} \phi (1, -2, -1) = 8 \cdot \frac{2}{3} + (-1) \cdot \left( -\frac{1}{3} \right) + (-10) \cdot \left( -\frac{2}{3} \right)

STEP 15

Simplify the expression:
Duϕ(1,−2,−1)=163+13+203D_{\mathbf{u}} \phi (1, -2, -1) = \frac{16}{3} + \frac{1}{3} + \frac{20}{3}

STEP 16

Combine the terms:
Duϕ(1,−2,−1)=16+1+203=373D_{\mathbf{u}} \phi (1, -2, -1) = \frac{16 + 1 + 20}{3} = \frac{37}{3}

STEP 17

Thus, the directional derivative of ϕ\phi at (1,−2,−1)(1, -2, -1) in the direction 2i−j−2k2\mathbf{i} - \mathbf{j} - 2\mathbf{k} is:
373\boxed{\frac{37}{3}}

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