PYQs / GATE EC / 2014 / Set 4 / Q15 GATE EC 2014 Set 4 — Question 15 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. NAT +1 / -0 Medium Directional Derivatives Vector Calculus Engineering Mathematics
Engineering Mathematics → Vector Calculus → Directional Derivatives
Last updated 5 September 2026
Question The directional derivative of
f ( x , y ) = x y 2 ( x + y ) f(x, y) = \frac{xy}{\sqrt{2}}(x + y) f ( x , y ) = 2 x y ( x + y ) at
( 1 , 1 ) (1, 1) ( 1 , 1 ) in the direction of the unit vector at an angle of
π 4 \frac{\pi}{4} 4 π with y-axis, is given by
______ .
Correct answer 2.99 to 3.01
Solution Given
f ( x , y ) = x y 2 ( x + y ) = 1 2 ( x 2 y + x y 2 ) f(x, y) = \frac{xy}{\sqrt{2}}(x + y) = \frac{1}{\sqrt{2}}(x^2y + xy^2) f ( x , y ) = 2 x y ( x + y ) = 2 1 ( x 2 y + x y 2 ) .
1. Calculate the gradient ∇ f = ( ∂ f ∂ x , ∂ f ∂ y ) \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) ∇ f = ( ∂ x ∂ f , ∂ y ∂ f ) ∂ f ∂ x = 1 2 ( 2 x y + y 2 ) \frac{\partial f}{\partial x} = \frac{1}{\sqrt{2}}(2xy + y^2) ∂ x ∂ f = 2 1 ( 2 x y + y 2 ) . At
( 1 , 1 ) (1, 1) ( 1 , 1 ) ,
∂ f ∂ x = 3 2 \frac{\partial f}{\partial x} = \frac{3}{\sqrt{2}} ∂ x ∂ f = 2 3 .
∂ f ∂ y = 1 2 ( x 2 + 2 x y ) \frac{\partial f}{\partial y} = \frac{1}{\sqrt{2}}(x^2 + 2xy) ∂ y ∂ f = 2 1 ( x 2 + 2 x y ) . At
( 1 , 1 ) (1, 1) ( 1 , 1 ) ,
∂ f ∂ y = 3 2 \frac{\partial f}{\partial y} = \frac{3}{\sqrt{2}} ∂ y ∂ f = 2 3 .
So,
∇ f ( 1 , 1 ) = 3 2 i ^ + 3 2 j ^ \nabla f(1, 1) = \frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j} ∇ f ( 1 , 1 ) = 2 3 i ^ + 2 3 j ^ .
2. Find the unit vector u ^ \hat{u} u ^ π 4 \frac{\pi}{4} 4 π π 2 − π 4 = π 4 \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} 2 π − 4 π = 4 π u ^ = cos ( π / 4 ) i ^ + sin ( π / 4 ) j ^ = 1 2 i ^ + 1 2 j ^ \hat{u} = \cos(\pi/4) \hat{i} + \sin(\pi/4) \hat{j} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} u ^ = cos ( π /4 ) i ^ + sin ( π /4 ) j ^ = 2 1 i ^ + 2 1 j ^ .
3. The directional derivative is D u f = ∇ f ⋅ u ^ = ( 3 2 ) ( 1 2 ) + ( 3 2 ) ( 1 2 ) = 3 2 + 3 2 = 3 D_u f = \nabla f \cdot \hat{u} = \left( \frac{3}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{2}} \right) + \left( \frac{3}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{2}} \right) = \frac{3}{2} + \frac{3}{2} = 3 D u f = ∇ f ⋅ u ^ = ( 2 3 ) ( 2 1 ) + ( 2 3 ) ( 2 1 ) = 2 3 + 2 3 = 3 Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker. Review answer and solution without JavaScript Interactive answer checking needs JavaScript. The published solution is available below.
Correct answer 2.99 to 3.01
Solution Given
f ( x , y ) = x y 2 ( x + y ) = 1 2 ( x 2 y + x y 2 ) f(x, y) = \frac{xy}{\sqrt{2}}(x + y) = \frac{1}{\sqrt{2}}(x^2y + xy^2) f ( x , y ) = 2 x y ( x + y ) = 2 1 ( x 2 y + x y 2 ) .
1. Calculate the gradient ∇ f = ( ∂ f ∂ x , ∂ f ∂ y ) \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) ∇ f = ( ∂ x ∂ f , ∂ y ∂ f ) ∂ f ∂ x = 1 2 ( 2 x y + y 2 ) \frac{\partial f}{\partial x} = \frac{1}{\sqrt{2}}(2xy + y^2) ∂ x ∂ f = 2 1 ( 2 x y + y 2 ) . At
( 1 , 1 ) (1, 1) ( 1 , 1 ) ,
∂ f ∂ x = 3 2 \frac{\partial f}{\partial x} = \frac{3}{\sqrt{2}} ∂ x ∂ f = 2 3 .
∂ f ∂ y = 1 2 ( x 2 + 2 x y ) \frac{\partial f}{\partial y} = \frac{1}{\sqrt{2}}(x^2 + 2xy) ∂ y ∂ f = 2 1 ( x 2 + 2 x y ) . At
( 1 , 1 ) (1, 1) ( 1 , 1 ) ,
∂ f ∂ y = 3 2 \frac{\partial f}{\partial y} = \frac{3}{\sqrt{2}} ∂ y ∂ f = 2 3 .
So,
∇ f ( 1 , 1 ) = 3 2 i ^ + 3 2 j ^ \nabla f(1, 1) = \frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j} ∇ f ( 1 , 1 ) = 2 3 i ^ + 2 3 j ^ .
2. Find the unit vector u ^ \hat{u} u ^ π 4 \frac{\pi}{4} 4 π π 2 − π 4 = π 4 \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} 2 π − 4 π = 4 π u ^ = cos ( π / 4 ) i ^ + sin ( π / 4 ) j ^ = 1 2 i ^ + 1 2 j ^ \hat{u} = \cos(\pi/4) \hat{i} + \sin(\pi/4) \hat{j} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} u ^ = cos ( π /4 ) i ^ + sin ( π /4 ) j ^ = 2 1 i ^ + 2 1 j ^ .
3. The directional derivative is D u f = ∇ f ⋅ u ^ = ( 3 2 ) ( 1 2 ) + ( 3 2 ) ( 1 2 ) = 3 2 + 3 2 = 3 D_u f = \nabla f \cdot \hat{u} = \left( \frac{3}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{2}} \right) + \left( \frac{3}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{2}} \right) = \frac{3}{2} + \frac{3}{2} = 3 D u f = ∇ f ⋅ u ^ = ( 2 3 ) ( 2 1 ) + ( 2 3 ) ( 2 1 ) = 2 3 + 2 3 = 3 Understand the concept, then try another question Revisit Engineering Mathematics with concept notes, common mistakes and an original worked example before your next attempt.
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