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Question: What is the \( \nabla \Phi \) at the point (0, 1, 0) of a scalar function \( \Phi \) if \( \Phi = 2{...

What is the ∇Φ\nabla \Phi at the point (0, 1, 0) of a scalar function Φ\Phi if Φ=2x2+y2+3z2\Phi = 2{x^2} + {y^2} + 3{z^2} ?

  1. 2j^2\hat j
  2. 3j^3\hat j
  3. 4i^+2j^4\hat i + 2\hat j
  4. 3i^+3j^3\hat i + 3\hat j
Explanation

Solution

Scalar products and vector products are two ways of multiplying two different vectors. Here, we are given the function equation Φ\Phi and we need to find the value of ∇Φ\nabla \Phi at the point (0, 1, 0) . We know the value of ∇=i^.∂∂x+j^.∂∂y+k^.∂∂z\nabla = \hat i.\dfrac{\partial }{{\partial x}} + \hat j.\dfrac{\partial }{{\partial y}} + \hat k.\dfrac{\partial }{{\partial z}} where i^\hat i , j^\hat j and k^\hat k are unit vectors along the x-axis, y-axis and z-axis respectively. After finding the value, we will substitute the value of all the unit vectors and then we will get the final output.

Complete Step By Step Answer:
Given that,
Φ=2x2+y2+3z2\Phi = 2{x^2} + {y^2} + 3{z^2}
We know that, ∇=i^.∂∂x+j^.∂∂y+k^.∂∂z\nabla = \hat i.\dfrac{\partial }{{\partial x}} + \hat j.\dfrac{\partial }{{\partial y}} + \hat k.\dfrac{\partial }{{\partial z}} where,
i^\hat i = unit vector along x-axis,
j^\hat j = unit vector along y-axis,
k^\hat k =unit vector along the z-axis.
So,
∴∇Φ=(i^.∂∂x+j^.∂∂y+k^.∂∂z)(2x2+y2+3z2)\therefore \nabla \Phi = \left( {\hat i.\dfrac{\partial }{{\partial x}} + \hat j.\dfrac{\partial }{{\partial y}} + \hat k.\dfrac{\partial }{{\partial z}}} \right)\left( {2{x^2} + {y^2} + 3{z^2}} \right)
=i^.∂∂x(2x2+y2+3z2)+j^.∂∂y(2x2+y2+3z2)+k^.∂∂z(2x2+y2+3z2)= \hat i.\dfrac{\partial }{{\partial x}}(2{x^2} + {y^2} + 3{z^2}) + \hat j.\dfrac{\partial }{{\partial y}}(2{x^2} + {y^2} + 3{z^2}) + \hat k.\dfrac{\partial }{{\partial z}}(2{x^2} + {y^2} + 3{z^2})
=4x.i^+2y.j^+6z.k^= 4x.\hat i + 2y.\hat j + 6z.\hat k
Now, we need to find the value of ∇Φ\nabla \Phi at the point (0, 1, 0) is as below:
Substituting the above values, we will get,
=4.0.i^+2.1.j^+6.0.k^= 4.0.\hat i + 2.1.\hat j + 6.0.\hat k
On evaluating this, we will get,
=0+2j^+0= 0 + 2\hat j + 0
=2j^= 2\hat j
Hence, for the given Φ=2x2+y2+3z2\Phi = 2{x^2} + {y^2} + 3{z^2} the value of ∇Φ=2j^\nabla \Phi = 2\hat j .

Note:
The scalar product is also termed as the dot product or inner product and remember that scalar multiplication is always denoted by a dot. The scalar product of two vectors is defined as the product of the magnitudes of the two vectors and the cosine of the angles between them. The magnitude vector product of two given vectors can be found by taking the product of the magnitudes of the vectors times the sine of the angle between them. Application of scalar and vector products are countless especially in situations where there are two forces acting on a body in a different direction.