Math Problem Statement
Solution
To solve the problem, let’s carefully analyze the information provided.
Given:
- The directional derivative of at in the direction of is .
- The directional derivative of at in the direction of is .
- We are asked for the maximum directional derivative of at .
Approach:
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The directional derivative in the direction of a unit vector is given by: where is the gradient vector of and is a unit vector.
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The magnitude of the maximum directional derivative is equal to the magnitude of the gradient:
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Use the given information to find :
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and are not necessarily unit vectors, so we need their magnitudes: The unit vectors in these directions are:
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The directional derivatives are given as:
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Set up a system of equations using the dot product: Let . 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}
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Solve for and : From (1): . Substitute into (2): Substituting into :
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Find :
Final Answer:
The maximum directional derivative is:
Would you like additional clarification or further steps explained? Here are 5 related questions to deepen your understanding:
- What does the gradient vector represent geometrically?
- How can you confirm that and are unit vectors?
- Why is the magnitude of the maximum directional derivative?
- How would the solution change if the directional derivatives were given in different directions?
- Can this method generalize to 3D or higher-dimensional problems?
Tip: The maximum directional derivative always occurs in the direction of .
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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)
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