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

Gradient of z gives the direction of steepest ascent. Compute the gradient of z(x,y) = (2x^2 + 2y^2)^{1/2} using the chain rule.

1) Compute partial derivatives:
- Let g(x,y) = 2x^2 + 2y^2, so z = g^{1/2}. ∂z/∂x = (1/2) g^{-1/2} · ∂g/∂x.
- ∂g/∂x = 4x, 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).

2) Evaluate at (1,1):
- Inside: 2(1)^2 + 2(1)^2 = 4, so √(...) = 2.
- ∂z/∂x (1,1) = (2·1)/2 = 1.
- ∂z/∂y (1,1) = (2·1)/2 = 1.

Thus ∇z(1,1) = (1,1). The gradient vector points in the direction of steepest ascent; any positive scalar multiple (for example the unit vector (1,1)/√2) gives the same direction. Therefore the direction of steepest ascent at (1,1) is along (1,1).

Why other options are incorrect:
- Vectors like (1,-1), (-1,1), (-1,-1) point in different quadrants and do not match the computed gradient components.
- Coordinate-axis vectors (1,0), (0,1), (0,-1), (-1,0) have zero in one component and cannot represent the gradient (1,1).
- Only (1,1) (or any positive multiple) matches the gradient direction computed.

Conclusion: The correct direction of steepest ascent at (1,1) is choice h. (1,1), matching the provided answer.

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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