QuestionSeptember 20, 2025

If f(x,y)=sin(x)+sin(y) then -sqrt (2)leqslant D_(u)fleqslant sqrt (2) True False

If f(x,y)=sin(x)+sin(y) then -sqrt (2)leqslant D_(u)fleqslant sqrt (2) True False
If f(x,y)=sin(x)+sin(y) then -sqrt (2)leqslant D_(u)fleqslant sqrt (2) True False

Solution
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Answer

True Explanation 1. Find the gradient of f(x, y) \nabla f = (\cos(x), \cos(y)) 2. Compute the magnitude of the gradient |\nabla f| = \sqrt{\cos^2(x) + \cos^2(y)} 3. Find the maximum value of |\nabla f| Maximum occurs when \cos^2(x) = 1 and \cos^2(y) = 1, so |\nabla f|_{max} = \sqrt{1+1} = \sqrt{2} 4. Range of directional derivative D_u f D_u f = \nabla f \cdot \vec{u}, where |\vec{u}|=1. Thus, -\sqrt{2} \leq D_u f \leq \sqrt{2}

Explanation

1. Find the gradient of $f(x, y)$<br /> $\nabla f = (\cos(x), \cos(y))$<br />2. Compute the magnitude of the gradient<br /> $|\nabla f| = \sqrt{\cos^2(x) + \cos^2(y)}$<br />3. Find the maximum value of $|\nabla f|$<br /> Maximum occurs when $\cos^2(x) = 1$ and $\cos^2(y) = 1$, so $|\nabla f|_{max} = \sqrt{1+1} = \sqrt{2}$<br />4. Range of directional derivative $D_u f$<br /> $D_u f = \nabla f \cdot \vec{u}$, where $|\vec{u}|=1$. Thus, $-\sqrt{2} \leq D_u f \leq \sqrt{2}$
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