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, so compute ∇z for z = sqrt(2x^2 + 2y^2).

1. Relevant concepts
- For a scalar function z(x,y), the vector of partial derivatives ∇z = (∂z/∂x, ∂z/∂y) points in the direction of greatest increase (steepest ascent).
- Any positive scalar multiple of the gradient is the same direction; sometimes a unit vector is used, but the multiple (1,1) already indicates the direction.

2. Step-by-step differentiation
Let u = 2x^2 + 2y^2, so z = u^(1/2).
- ∂u/∂x = d(2x^2)/dx + d(2y^2)/dx = 4x + 0 = 4x.
- ∂z/∂x = (1/2) u^(-1/2) * ∂u/∂x = (1/2)(1/√u)(4x) = (4x)/(2√u) = (2x)/√u.
Similarly for y:
- ∂u/∂y = 4y.
- ∂z/∂y = (2y)/√u.

3. Intermediate evaluation at (1,1)
At (1,1): u = 2(1)^2 + 2(1)^2 = 4, so √u = 2.
- ∂z/∂x (1,1) = (2·1)/2 = 1.
- ∂z/∂y (1,1) = (2·1)/2 = 1.
Thus ∇z(1,1) = (1,1).

4. Why the correct answer is correct
The gradient (1,1) points exactly in the direction of steepest increase of z at (1,1). Any nonzero scalar multiple (for instance the unit vector (1/√2, 1/√2)) represents the same direction; the multiple (1,1) appears among the choices and is therefore valid.

5. Why other options are incorrect
- (-1,-1) is the negative of the gradient: it points in the direction of steepest descent, not ascent.
- Options with mixed signs like (1,-1) or (-1,1) point along directions where one coordinate increases while the other decreases — these are not the gradient direction found above.
- Single-axis options like (1,0), (0,1), etc., are not aligned with the computed gradient (1,1).

6. Conclusion
The direction of steepest ascent at (1,1) is (1,1), so the correct answer is b.

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