Exercise · ZW-GRADIENT-007 · ★★☆ Intermediate
The level curve of \(f(x,y)=x^2+y^2\) through \((1,1)\) has a tangent vector there. Which one?
Choose one option.
Show answer
- C. \(\langle 1,-1\rangle\)
Explanation
The gradient \(\langle2,2\rangle\) is perpendicular to the level curve, and \(\langle1,-1\rangle\) is the one with \(\langle2,2\rangle\cdot\langle1,-1\rangle=0\).
Review the concept: Gradient
The gradient \(\nabla f\) of a differentiable function \(f\) is the vector that collects its partial derivatives at a point; it points where \(f\) increases fastest.
\[\nabla f=\langle f_x,\,f_y\rangle=f_x\,\mathbf{i}+f_y\,\mathbf{j}\]
More practice problems: gradient
- ZW-GRADIENT-001 Compute ∇ f(1,2) for f(x,y)=x^2y.
- ZW-GRADIENT-002 Compute ∇ f(2,0) for f(x,y)=xe^y.
- ZW-GRADIENT-003 Compute ∇ f(1,2,3) for f(x,y,z)=xyz.
- ZW-GRADIENT-004 Compute ∇ f(1,1) for f(x,y)=ln (x^2+y^2).
- ZW-GRADIENT-005 For f(x,y)=x^2+y^2 at (3,4), what is the maximum rate of change of f?
- ZW-GRADIENT-006 In which unit vector direction does f(x,y)=xy increase fastest from (2,3)?
- ZW-GRADIENT-008 If ∇ f(P)=0 and f is differentiable, what can be said at P?
- ZW-GRADIENT-009 Find the tangent line to the curve x^2+4y^2=8 at the point (2,1).