Math Problem Statement

At a point P in the xy-plane, the directional derivative of f(x,y) in the direction of v1 = 2i - j is sqrt(5) and the directional derivative of f(x,y) in the direction of v2 = i + 2j is 3sqrt(5). What is the maximum directional derivative of f(x,y) at P?

Solution

To solve the problem, let’s carefully analyze the information provided.


Given:

  1. The directional derivative of f(x,y)f(x, y) at PP in the direction of v1=2ij\mathbf{v}_1 = 2\mathbf{i} - \mathbf{j} is 5\sqrt{5}.
  2. The directional derivative of f(x,y)f(x, y) at PP in the direction of v2=i+2j\mathbf{v}_2 = \mathbf{i} + 2\mathbf{j} is 353\sqrt{5}.
  3. We are asked for the maximum directional derivative of f(x,y)f(x, y) at PP.

Approach:

  1. The directional derivative in the direction of a unit vector u\mathbf{u} is given by: Duf=fuD_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} where f\nabla f is the gradient vector of f(x,y)f(x, y) and u\mathbf{u} is a unit vector.

  2. The magnitude of the maximum directional derivative is equal to the magnitude of the gradient: Maximum directional derivative=f\text{Maximum directional derivative} = \|\nabla f\|

  3. Use the given information to find f\nabla f:

    • v1\mathbf{v}_1 and v2\mathbf{v}_2 are not necessarily unit vectors, so we need their magnitudes: v1=22+(1)2=5,v2=12+22=5\|\mathbf{v}_1\| = \sqrt{2^2 + (-1)^2} = \sqrt{5}, \quad \|\mathbf{v}_2\| = \sqrt{1^2 + 2^2} = \sqrt{5} The unit vectors in these directions are: u1=v1v1=2ij5,u2=v2v2=i+2j5\mathbf{u}_1 = \frac{\mathbf{v}_1}{\|\mathbf{v}_1\|} = \frac{2\mathbf{i} - \mathbf{j}}{\sqrt{5}}, \quad \mathbf{u}_2 = \frac{\mathbf{v}_2}{\|\mathbf{v}_2\|} = \frac{\mathbf{i} + 2\mathbf{j}}{\sqrt{5}}

    • The directional derivatives are given as: Du1f=fu1=5,Du2f=fu2=35D_{\mathbf{u}_1} f = \nabla f \cdot \mathbf{u}_1 = \sqrt{5}, \quad D_{\mathbf{u}_2} f = \nabla f \cdot \mathbf{u}_2 = 3\sqrt{5}

  4. Set up a system of equations using the dot product: Let f=a,b\nabla f = \langle a, b \rangle. Then: \frac{1}{\sqrt{5}}(2a - b) = \sqrt{5} \implies 2a - b = 5 \tag{1} \frac{1}{\sqrt{5}}(a + 2b) = 3\sqrt{5} \implies a + 2b = 15 \tag{2}

  5. Solve for aa and bb: From (1): b=2a5b = 2a - 5. Substitute into (2): a+2(2a5)=15    a+4a10=15    5a=25    a=5a + 2(2a - 5) = 15 \implies a + 4a - 10 = 15 \implies 5a = 25 \implies a = 5 Substituting a=5a = 5 into b=2a5b = 2a - 5: b=2(5)5=5b = 2(5) - 5 = 5

  6. Find f\|\nabla f\|: f=a2+b2=52+52=50=52\|\nabla f\| = \sqrt{a^2 + b^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}


Final Answer:

The maximum directional derivative is: 52\boxed{5\sqrt{2}}


Would you like additional clarification or further steps explained? Here are 5 related questions to deepen your understanding:

  1. What does the gradient vector represent geometrically?
  2. How can you confirm that u1\mathbf{u}_1 and u2\mathbf{u}_2 are unit vectors?
  3. Why is the magnitude of f\nabla f the maximum directional derivative?
  4. How would the solution change if the directional derivatives were given in different directions?
  5. Can this method generalize to 3D or higher-dimensional problems?

Tip: The maximum directional derivative always occurs in the direction of f\nabla f.

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Math Problem Analysis

Mathematical Concepts

Directional Derivative
Gradient Vector
Dot Product
Linear Algebra

Formulas

Directional Derivative: D_u f = ∇f · u
Gradient Magnitude: ||∇f|| = max directional derivative
Vector Magnitude: ||v|| = sqrt(v_x^2 + v_y^2)

Theorems

The maximum directional derivative occurs in the direction of the gradient vector.

Suitable Grade Level

Undergraduate Calculus (Multivariable Calculus)