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Given the scalar function (x, y, z) = e^xy - x²yz².

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Given the scalar function ϕ(x,y,z)=exy−x2yz2\phi(x, y, z) = e^{xy} - x^2yz^2.

The gradient of a scalar function ϕ(x,y,z)\phi(x, y, z) is given by:

∇ϕ=∂ϕ∂xi+∂ϕ∂yj+∂ϕ∂zk\nabla \phi = \frac{\partial \phi}{\partial x} \mathbf{i} + \frac{\partial \phi}{\partial y} \mathbf{j} + \frac{\partial \phi}{\partial z} \mathbf{k}

Part a· 4 steps

Evaluate ∇ϕ(x,y,z)\nabla \phi (x, y, z)

  1. Calculate the partial derivative with respect to xx.

    ∂ϕ∂x=∂∂x(exy−x2yz2)=yexy−2xyz2\frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x} (e^{xy} - x^2yz^2) = y e^{xy} - 2xyz^2

  2. Calculate the partial derivative with respect to yy.

    ∂ϕ∂y=∂∂y(exy−x2yz2)=xexy−x2z2\frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} (e^{xy} - x^2yz^2) = x e^{xy} - x^2z^2

  3. Calculate the partial derivative with respect to zz.

    ∂ϕ∂z=∂∂z(exy−x2yz2)=−x2y(2z)=−2x2yz\frac{\partial \phi}{\partial z} = \frac{\partial}{\partial z} (e^{xy} - x^2yz^2) = -x^2y(2z) = -2x^2yz

  4. Combine the partial derivatives to form the gradient vector.

    ∇ϕ(x,y,z)=(yexy−2xyz2)i+(xexy−x2z2)j+(−2x2yz)k\nabla \phi (x, y, z) = (y e^{xy} - 2xyz^2) \mathbf{i} + (x e^{xy} - x^2z^2) \mathbf{j} + (-2x^2yz) \mathbf{k}

    The gradient is:

    ∇ϕ(x,y,z)=(yexy−2xyz2)i+(xexy−x2z2)j−2x2yzk\boxed{\nabla \phi (x, y, z) = (y e^{xy} - 2xyz^2) \mathbf{i} + (x e^{xy} - x^2z^2) \mathbf{j} - 2x^2yz \mathbf{k}}

Part b· 4 steps

Evaluate ∇ϕ(2,−1,1)\nabla \phi (2, -1, 1)

Substitute x=2x=2, y=−1y=-1, and z=1z=1 into the expression for ∇ϕ(x,y,z)\nabla \phi (x, y, z) from part (a).

  1. Calculate the i\mathbf{i} component.

    yexy−2xyz2=(−1)e(2)(−1)−2(2)(−1)(1)2=−e−2+4y e^{xy} - 2xyz^2 = (-1) e^{(2)(-1)} - 2(2)(-1)(1)^2 = -e^{-2} + 4

  2. Calculate the j\mathbf{j} component.

    xexy−x2z2=(2)e(2)(−1)−(2)2(1)2=2e−2−4x e^{xy} - x^2z^2 = (2) e^{(2)(-1)} - (2)^2(1)^2 = 2e^{-2} - 4

  3. Calculate the k\mathbf{k} component.

    −2x2yz=−2(2)2(−1)(1)=−2(4)(−1)(1)=8-2x^2yz = -2(2)^2(-1)(1) = -2(4)(-1)(1) = 8

  4. Combine the components.

    ∇ϕ(2,−1,1)=(−e−2+4)i+(2e−2−4)j+8k\nabla \phi (2, -1, 1) = (-e^{-2} + 4) \mathbf{i} + (2e^{-2} - 4) \mathbf{j} + 8 \mathbf{k}

    The gradient at (2,−1,1)(2, -1, 1) is:

    ∇ϕ(2,−1,1)=(4−e−2)i+(2e−2−4)j+8k\boxed{\nabla \phi (2, -1, 1) = (4 - e^{-2}) \mathbf{i} + (2e^{-2} - 4) \mathbf{j} + 8 \mathbf{k}}

Part c· 5 steps

Evaluate ∣∇ϕ(1,1,−1)∣|\nabla \phi (1, 1, -1)|

First, evaluate ∇ϕ(1,1,−1)\nabla \phi (1, 1, -1) by substituting x=1x=1, y=1y=1, and z=−1z=-1 into the expression for ∇ϕ(x,y,z)\nabla \phi (x, y, z).

  1. Calculate the i\mathbf{i} component.

    yexy−2xyz2=(1)e(1)(1)−2(1)(1)(−1)2=e−2(1)(1)=e−2y e^{xy} - 2xyz^2 = (1) e^{(1)(1)} - 2(1)(1)(-1)^2 = e - 2(1)(1) = e - 2

  2. Calculate the j\mathbf{j} component.

    xexy−x2z2=(1)e(1)(1)−(1)2(−1)2=e−1(1)=e−1x e^{xy} - x^2z^2 = (1) e^{(1)(1)} - (1)^2(-1)^2 = e - 1(1) = e - 1

  3. Calculate the k\mathbf{k} component.

    −2x2yz=−2(1)2(1)(−1)=−2(1)(−1)=2-2x^2yz = -2(1)^2(1)(-1) = -2(1)(-1) = 2

  4. Form the gradient vector at (1,1,−1)(1, 1, -1).

    ∇ϕ(1,1,−1)=(e−2)i+(e−1)j+2k\nabla \phi (1, 1, -1) = (e - 2) \mathbf{i} + (e - 1) \mathbf{j} + 2 \mathbf{k}

  5. Calculate the magnitude of the vector.

    The magnitude of a vector Ai+Bj+CkA\mathbf{i} + B\mathbf{j} + C\mathbf{k} is A2+B2+C2\sqrt{A^2 + B^2 + C^2}.

    ∣∇ϕ(1,1,−1)∣=(e−2)2+(e−1)2+(2)2|\nabla \phi (1, 1, -1)| = \sqrt{(e-2)^2 + (e-1)^2 + (2)^2} ∣∇ϕ(1,1,−1)∣=(e2−4e+4)+(e2−2e+1)+4|\nabla \phi (1, 1, -1)| = \sqrt{(e^2 - 4e + 4) + (e^2 - 2e + 1) + 4} ∣∇ϕ(1,1,−1)∣=e2−4e+4+e2−2e+1+4|\nabla \phi (1, 1, -1)| = \sqrt{e^2 - 4e + 4 + e^2 - 2e + 1 + 4} ∣∇ϕ(1,1,−1)∣=2e2−6e+9|\nabla \phi (1, 1, -1)| = \sqrt{2e^2 - 6e + 9}

    The magnitude of the gradient at (1,1,−1)(1, 1, -1) is:

    ∣∇ϕ(1,1,−1)∣=2e2−6e+9\boxed{|\nabla \phi (1, 1, -1)| = \sqrt{2e^2 - 6e + 9}}
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