Consider the function $z=\sqrt{2x^2+2y^2}$. What is the direction of steepest ascent at the point $ (1,1) $?

Question

BlackTom AI · Accepted Answer

The gradient ∇z gives the direction of steepest ascent. For z = sqrt(2x^2 + 2y^2) compute the partial derivatives and evaluate at (1,1).

1. Relevant concepts
- The direction of steepest ascent of a scalar function z(x,y) is the gradient vector ∇z = (∂z/∂x, ∂z/∂y).
- The value of the gradient at a point both gives the direction and its magnitude equals the maximum directional derivative there.

2. Step-by-step differentiation
z(x,y) = (2x^2 + 2y^2)^{1/2}.
Use the chain rule:
∂z/∂x = (1/2)(2x^2 + 2y^2)^{-1/2} · d/dx(2x^2 + 2y^2).
d/dx(2x^2 + 2y^2) = 4x (since derivative of 2x^2 is 4x and y treated as constant).
So ∂z/∂x = (1/2)(2x^2 + 2y^2)^{-1/2} · 4x = (4x)/(2√(2x^2+2y^2)) = (2x)/√(2x^2+2y^2).

Similarly,
∂z/∂y = (2y)/√(2x^2+2y^2).

3. Evaluate at (1,1)
Compute denominator: √(2·1^2 + 2·1^2) = √(2 + 2) = √4 = 2.
Thus
∂z/∂x (1,1) = (2·1)/2 = 1,
∂z/∂y (1,1) = (2·1)/2 = 1.

So ∇z(1,1) = (1,1).

4. Why this is correct
The gradient vector (1,1) points in the direction where z increases fastest from (1,1). Any positive scalar multiple or the corresponding unit vector (1/√2, 1/√2) represents the same direction. The magnitude |∇z| = √2 is the maximal rate of increase.

5. Why other choices are incorrect
- (-1,-1) is the negative of the gradient and therefore points in the direction of steepest descent, not ascent.
- (0,1), (−1,0), (1,−1), etc., are different directions and do not align with ∇z(1,1).

6. Conclusion
The correct direction of steepest ascent at (1,1) is (1,1), which matches the provided answer (d).

Consider the function \(z=\sqrt{2x^2+2y^2}\). What is the direction of steepest ascent at the point \( (1,1) \)?单项选择题

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