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Find the directional derivative of f(x,y) = 2x2 + y2 at P0(-1, 1) in the direction of u = 3i − 4j.

Mathematics

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Answer

-4

Step-by-step solution

5 steps

To find the directional derivative of f(x,y)f(x,y) at P0P_0 in the direction of u\mathbf{u}, we use the formula Duf(P0)=∇f(P0)⋅u^D_{\mathbf{u}}f(P_0) = \nabla f(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 f(x,y)f(x,y) with respect to xx and yy.

    Given f(x,y)=2x2+y2f(x,y) = 2x^2 + y^2.

    ∂f∂x=∂∂x(2x2+y2)=4x\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(2x^2 + y^2) = 4x ∂f∂y=∂∂y(2x2+y2)=2y\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(2x^2 + y^2) = 2y
  2. Form the gradient vector ∇f(x,y)\nabla f(x,y).

    ∇f(x,y)=⟨∂f∂x,∂f∂y⟩=⟨4x,2y⟩\nabla f(x,y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle = \langle 4x, 2y \rangle
  3. Evaluate the gradient at the given point P0(−1,1)P_0(-1, 1).

    ∇f(−1,1)=⟨4(−1),2(1)⟩=⟨−4,2⟩\nabla f(-1, 1) = \langle 4(-1), 2(1) \rangle = \langle -4, 2 \rangle
  4. Find the unit vector u^\hat{\mathbf{u}} in the direction of u=3i−4j\mathbf{u} = 3\mathbf{i} - 4\mathbf{j}.

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

    ∣∣u∣∣=32+(−4)2=9+16=25=5||\mathbf{u}|| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

    Now, find the unit vector:

    u^=u∣∣u∣∣=15(3i−4j)=⟨35,−45⟩\hat{\mathbf{u}} = \frac{\mathbf{u}}{||\mathbf{u}||} = \frac{1}{5}(3\mathbf{i} - 4\mathbf{j}) = \left\langle \frac{3}{5}, -\frac{4}{5} \right\rangle
  5. Calculate the directional derivative by taking the dot product of ∇f(−1,1)\nabla f(-1, 1) and u^\hat{\mathbf{u}}.

    Du^f(−1,1)=∇f(−1,1)⋅u^D_{\hat{\mathbf{u}}}f(-1, 1) = \nabla f(-1, 1) \cdot \hat{\mathbf{u}} Du^f(−1,1)=⟨−4,2⟩⋅⟨35,−45⟩D_{\hat{\mathbf{u}}}f(-1, 1) = \langle -4, 2 \rangle \cdot \left\langle \frac{3}{5}, -\frac{4}{5} \right\rangle Du^f(−1,1)=(−4)(35)+(2)(−45)D_{\hat{\mathbf{u}}}f(-1, 1) = (-4)\left(\frac{3}{5}\right) + (2)\left(-\frac{4}{5}\right) Du^f(−1,1)=−125−85D_{\hat{\mathbf{u}}}f(-1, 1) = -\frac{12}{5} - \frac{8}{5} Du^f(−1,1)=−205D_{\hat{\mathbf{u}}}f(-1, 1) = -\frac{20}{5} Du^f(−1,1)=−4D_{\hat{\mathbf{u}}}f(-1, 1) = -4

    The directional derivative is −4-4.

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