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Find the directional derivative of h(x,y,z) = cos(xy) + e^(yz) + ln(zx) at P0(1, 0, 1/2) in the direction of u = <1, 2, 2>.

Mathematics

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2

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To find the directional derivative of h(x,y,z)h(x,y,z) at P0P_0 in the direction of u\mathbf{u}, we use the formula Duh(P0)=∇h(P0)⋅u^D_{\mathbf{u}}h(P_0) = \nabla h(P_0) \cdot \hat{\mathbf{u}}, where u^\hat{\mathbf{u}} is the unit vector in the direction of u\mathbf{u}.

  1. Find the partial derivatives of h(x,y,z)h(x,y,z) with respect to xx, yy, and zz.

    Given h(x,y,z)=cos⁡(xy)+eyz+ln⁡(zx)h(x,y,z) = \cos(xy) + e^{yz} + \ln(zx).

    ∂h∂x=∂∂x(cos⁡(xy))+∂∂x(eyz)+∂∂x(ln⁡(zx))\frac{\partial h}{\partial x} = \frac{\partial}{\partial x}(\cos(xy)) + \frac{\partial}{\partial x}(e^{yz}) + \frac{\partial}{\partial x}(\ln(zx)) ∂h∂x=−ysin⁡(xy)+0+1zx⋅z=−ysin⁡(xy)+1x\frac{\partial h}{\partial x} = -y\sin(xy) + 0 + \frac{1}{zx} \cdot z = -y\sin(xy) + \frac{1}{x} ∂h∂y=∂∂y(cos⁡(xy))+∂∂y(eyz)+∂∂y(ln⁡(zx))\frac{\partial h}{\partial y} = \frac{\partial}{\partial y}(\cos(xy)) + \frac{\partial}{\partial y}(e^{yz}) + \frac{\partial}{\partial y}(\ln(zx)) ∂h∂y=−xsin⁡(xy)+eyz⋅z+0=−xsin⁡(xy)+zeyz\frac{\partial h}{\partial y} = -x\sin(xy) + e^{yz} \cdot z + 0 = -x\sin(xy) + ze^{yz} ∂h∂z=∂∂z(cos⁡(xy))+∂∂z(eyz)+∂∂z(ln⁡(zx))\frac{\partial h}{\partial z} = \frac{\partial}{\partial z}(\cos(xy)) + \frac{\partial}{\partial z}(e^{yz}) + \frac{\partial}{\partial z}(\ln(zx)) ∂h∂z=0+eyz⋅y+1zx⋅x=yeyz+1z\frac{\partial h}{\partial z} = 0 + e^{yz} \cdot y + \frac{1}{zx} \cdot x = ye^{yz} + \frac{1}{z}

  2. Form the gradient vector ∇h(x,y,z)\nabla h(x,y,z).

    ∇h(x,y,z)=⟨−ysin⁡(xy)+1x,−xsin⁡(xy)+zeyz,yeyz+1z⟩\nabla h(x,y,z) = \left\langle -y\sin(xy) + \frac{1}{x}, -x\sin(xy) + ze^{yz}, ye^{yz} + \frac{1}{z} \right\rangle

  3. Evaluate the gradient at the given point P0(1,0,1/2)P_0(1, 0, 1/2).

    Substitute x=1x=1, y=0y=0, z=1/2z=1/2 into the gradient vector:

    ∂h∂x(1,0,1/2)=−(0)sin⁡(1⋅0)+11=0+1=1\frac{\partial h}{\partial x}(1,0,1/2) = -(0)\sin(1 \cdot 0) + \frac{1}{1} = 0 + 1 = 1 ∂h∂y(1,0,1/2)=−(1)sin⁡(1⋅0)+(12)e(0)(1/2)=0+12e0=12(1)=12\frac{\partial h}{\partial y}(1,0,1/2) = -(1)\sin(1 \cdot 0) + \left(\frac{1}{2}\right)e^{(0)(1/2)} = 0 + \frac{1}{2}e^0 = \frac{1}{2}(1) = \frac{1}{2} ∂h∂z(1,0,1/2)=(0)e(0)(1/2)+11/2=0+2=2\frac{\partial h}{\partial z}(1,0,1/2) = (0)e^{(0)(1/2)} + \frac{1}{1/2} = 0 + 2 = 2

    So, the gradient at P0P_0 is:

    ∇h(1,0,1/2)=⟨1,12,2⟩\nabla h(1, 0, 1/2) = \left\langle 1, \frac{1}{2}, 2 \right\rangle

  4. Find the unit vector u^\hat{\mathbf{u}} in the direction of u=i+2j+2k\mathbf{u} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}.

    First, find the magnitude of u\mathbf{u}:

    ∣∣u∣∣=12+22+22=1+4+4=9=3||\mathbf{u}|| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3

    Now, find the unit vector:

    u^=u∣∣u∣∣=13(i+2j+2k)=⟨13,23,23⟩\hat{\mathbf{u}} = \frac{\mathbf{u}}{||\mathbf{u}||} = \frac{1}{3}(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = \left\langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right\rangle

  5. Calculate the directional derivative by taking the dot product of ∇h(P0)\nabla h(P_0) and u^\hat{\mathbf{u}}.

    Du^h(P0)=∇h(1,0,1/2)⋅u^D_{\hat{\mathbf{u}}}h(P_0) = \nabla h(1, 0, 1/2) \cdot \hat{\mathbf{u}} Du^h(P0)=⟨1,12,2⟩⋅⟨13,23,23⟩D_{\hat{\mathbf{u}}}h(P_0) = \left\langle 1, \frac{1}{2}, 2 \right\rangle \cdot \left\langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right\rangle Du^h(P0)=(1)(13)+(12)(23)+(2)(23)D_{\hat{\mathbf{u}}}h(P_0) = (1)\left(\frac{1}{3}\right) + \left(\frac{1}{2}\right)\left(\frac{2}{3}\right) + (2)\left(\frac{2}{3}\right) Du^h(P0)=13+26+43D_{\hat{\mathbf{u}}}h(P_0) = \frac{1}{3} + \frac{2}{6} + \frac{4}{3} Du^h(P0)=13+13+43D_{\hat{\mathbf{u}}}h(P_0) = \frac{1}{3} + \frac{1}{3} + \frac{4}{3} Du^h(P0)=1+1+43=63=2D_{\hat{\mathbf{u}}}h(P_0) = \frac{1+1+4}{3} = \frac{6}{3} = 2

    The directional derivative is 2.

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