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Engineering Mathematics → Vector Calculus → Directional Derivatives
Last updated 5 September 2026
Question A function is defined in Cartesian coordinate system as
f ( x , y ) = x e y f(x, y) = xe^y f ( x , y ) = x e y . The value of the directional derivative of the function (
in integer ) at the point
( 2 , 0 ) (2, 0) ( 2 , 0 ) along the direction of the straight line segment from point
( 2 , 0 ) (2, 0) ( 2 , 0 ) to point
( 1 2 , 2 ) (\frac{1}{2}, 2) ( 2 1 , 2 ) is
____________ .
Solution 1. Find the gradient of f ( x , y ) = x e y f(x, y) = xe^y f ( x , y ) = x e y ∇ f = ( ∂ f ∂ x , ∂ f ∂ y ) = ( e y , x e y ) \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (e^y, xe^y) ∇ f = ( ∂ x ∂ f , ∂ y ∂ f ) = ( e y , x e y ) 2. Evaluate the gradient at the point P ( 2 , 0 ) P(2, 0) P ( 2 , 0 ) ∇ f ( 2 , 0 ) = ( e 0 , 2 e 0 ) = ( 1 , 2 ) \nabla f(2, 0) = (e^0, 2e^0) = (1, 2) ∇ f ( 2 , 0 ) = ( e 0 , 2 e 0 ) = ( 1 , 2 ) 3. Find the direction vector v ⃗ \vec{v} v P ( 2 , 0 ) P(2, 0) P ( 2 , 0 ) Q ( 1 2 , 2 ) Q(\frac{1}{2}, 2) Q ( 2 1 , 2 ) v ⃗ = Q − P = ( 1 2 − 2 , 2 − 0 ) = ( − 3 2 , 2 ) \vec{v} = Q - P = \left( \frac{1}{2} - 2, 2 - 0 \right) = \left( -\frac{3}{2}, 2 \right) v = Q − P = ( 2 1 − 2 , 2 − 0 ) = ( − 2 3 , 2 ) 4. Find the unit vector u ^ \hat{u} u ^ v ⃗ \vec{v} v ∣ v ⃗ ∣ = ( − 3 2 ) 2 + 2 2 = 9 4 + 4 = 25 4 = 5 2 |\vec{v}| = \sqrt{\left( -\frac{3}{2} \right)^2 + 2^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2} ∣ v ∣ = ( − 2 3 ) 2 + 2 2 = 4 9 + 4 = 4 25 = 2 5 u ^ = v ⃗ ∣ v ⃗ ∣ = ( − 3 / 2 , 2 ) 5 / 2 = ( − 3 5 , 4 5 ) \hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{(-3/2, 2)}{5/2} = \left( -\frac{3}{5}, \frac{4}{5} \right) u ^ = ∣ v ∣ v = 5/2 ( − 3/2 , 2 ) = ( − 5 3 , 5 4 ) 5. Calculate the directional derivative D u ^ f D_{\hat{u}} f D u ^ f D u ^ f = ∇ f ⋅ u ^ = ( 1 , 2 ) ⋅ ( − 3 5 , 4 5 ) = 1 ( − 3 5 ) + 2 ( 4 5 ) = − 3 5 + 8 5 = 5 5 = 1 D_{\hat{u}} f = \nabla f \cdot \hat{u} = (1, 2) \cdot \left( -\frac{3}{5}, \frac{4}{5} \right) = 1 \left( -\frac{3}{5} \right) + 2 \left( \frac{4}{5} \right) = -\frac{3}{5} + \frac{8}{5} = \frac{5}{5} = 1 D u ^ f = ∇ f ⋅ u ^ = ( 1 , 2 ) ⋅ ( − 5 3 , 5 4 ) = 1 ( − 5 3 ) + 2 ( 5 4 ) = − 5 3 + 5 8 = 5 5 = 1 The directional derivative is 1.
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Solution 1. Find the gradient of f ( x , y ) = x e y f(x, y) = xe^y f ( x , y ) = x e y ∇ f = ( ∂ f ∂ x , ∂ f ∂ y ) = ( e y , x e y ) \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (e^y, xe^y) ∇ f = ( ∂ x ∂ f , ∂ y ∂ f ) = ( e y , x e y ) 2. Evaluate the gradient at the point P ( 2 , 0 ) P(2, 0) P ( 2 , 0 ) ∇ f ( 2 , 0 ) = ( e 0 , 2 e 0 ) = ( 1 , 2 ) \nabla f(2, 0) = (e^0, 2e^0) = (1, 2) ∇ f ( 2 , 0 ) = ( e 0 , 2 e 0 ) = ( 1 , 2 ) 3. Find the direction vector v ⃗ \vec{v} v P ( 2 , 0 ) P(2, 0) P ( 2 , 0 ) Q ( 1 2 , 2 ) Q(\frac{1}{2}, 2) Q ( 2 1 , 2 ) v ⃗ = Q − P = ( 1 2 − 2 , 2 − 0 ) = ( − 3 2 , 2 ) \vec{v} = Q - P = \left( \frac{1}{2} - 2, 2 - 0 \right) = \left( -\frac{3}{2}, 2 \right) v = Q − P = ( 2 1 − 2 , 2 − 0 ) = ( − 2 3 , 2 ) 4. Find the unit vector u ^ \hat{u} u ^ v ⃗ \vec{v} v ∣ v ⃗ ∣ = ( − 3 2 ) 2 + 2 2 = 9 4 + 4 = 25 4 = 5 2 |\vec{v}| = \sqrt{\left( -\frac{3}{2} \right)^2 + 2^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2} ∣ v ∣ = ( − 2 3 ) 2 + 2 2 = 4 9 + 4 = 4 25 = 2 5 u ^ = v ⃗ ∣ v ⃗ ∣ = ( − 3 / 2 , 2 ) 5 / 2 = ( − 3 5 , 4 5 ) \hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{(-3/2, 2)}{5/2} = \left( -\frac{3}{5}, \frac{4}{5} \right) u ^ = ∣ v ∣ v = 5/2 ( − 3/2 , 2 ) = ( − 5 3 , 5 4 ) 5. Calculate the directional derivative D u ^ f D_{\hat{u}} f D u ^ f D u ^ f = ∇ f ⋅ u ^ = ( 1 , 2 ) ⋅ ( − 3 5 , 4 5 ) = 1 ( − 3 5 ) + 2 ( 4 5 ) = − 3 5 + 8 5 = 5 5 = 1 D_{\hat{u}} f = \nabla f \cdot \hat{u} = (1, 2) \cdot \left( -\frac{3}{5}, \frac{4}{5} \right) = 1 \left( -\frac{3}{5} \right) + 2 \left( \frac{4}{5} \right) = -\frac{3}{5} + \frac{8}{5} = \frac{5}{5} = 1 D u ^ f = ∇ f ⋅ u ^ = ( 1 , 2 ) ⋅ ( − 5 3 , 5 4 ) = 1 ( − 5 3 ) + 2 ( 5 4 ) = − 5 3 + 5 8 = 5 5 = 1 The directional derivative is 1.
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