Question

A unit vector that maximizes the directional derivative of the function \(f(x,y)=g(2x+y)\), where \(g'(3)=3\), at the point (1,1) is ____.

• A. \(\dfrac{1}{\sqrt5}(2\hat i+\hat j)\)
• B. \(\dfrac{1}{\sqrt5}(-2\hat i+\hat j)\)
• C. \(\dfrac{1}{\sqrt5}(2\hat i-\hat j)\)
• D. \(-\dfrac{1}{\sqrt5}(2\hat i+\hat j)\)

Hint:

The directional derivative is maximized in the direction of the gradient vector, so normalize the gradient at the given point.

Answer 1

The Correct Option is A

Step 1: Compute the gradient using the chain rule. Let \(u=2x+y\), so \(f(x,y)=g(u)\). Then \[f_x=g'(u)\cdot 2,\qquad f_y=g'(u)\cdot 1.\] At the point (1,1), \(u=2(1)+1=3\), and we are given \(g'(3)=3\). So \[\nabla f(1,1)=(2\cdot 3,\ 1\cdot 3)=(6,3).\]

Step 2: The direction of maximum increase. The directional derivative in the direction of a unit vector \(\hat u\) is \(D_{\hat u}f=\nabla f\cdot \hat u\), and this is largest when \(\hat u\) points along \(\nabla f\) (by the Cauchy-Schwarz inequality). So the maximizing unit vector is \[\hat u=\frac{\nabla f(1,1)}{|\nabla f(1,1)|}.\]

Step 3: Normalize. \[|\nabla f(1,1)|=\sqrt{6^2+3^2}=\sqrt{45}=3\sqrt5.\] \[\hat u=\frac{(6,3)}{3\sqrt5}=\frac{(2,1)}{\sqrt5}=\frac{1}{\sqrt5}(2\hat i+\hat j).\] \[\boxed{\hat u=\frac{1}{\sqrt5}(2\hat i+\hat j)}\]