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Given D = e-xsin y i – e-xcos y j
Find divergence of D.
Find divergence of D.
- a)3
- b)2
- c)1
- d)0
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Given D = e-xsin y i – e-xcos y jFind divergence of D.a)3b)2c)1d...
Answer: d
Explanation: Div (D) = Dx(e-xsin y) + Dy(-e-xcos y ) = -e-xsin y + e-xsin y = 0.
Explanation: Div (D) = Dx(e-xsin y) + Dy(-e-xcos y ) = -e-xsin y + e-xsin y = 0.
Most Upvoted Answer
Given D = e-xsin y i – e-xcos y jFind divergence of D.a)3b)2c)1d...
To compute ∇D, we need to take the partial derivatives of each component of D with respect to x, y, and z.
∂D/∂x = ∂(e^(-x)sin(y))/∂x = -e^(-x)sin(y)
∂D/∂y = ∂(e^(-x)sin(y))/∂y = e^(-x)cos(y)
∂D/∂z = ∂(e^(-x)sin(y))/∂z = 0
Therefore, ∇D = -e^(-x)sin(y)i + e^(-x)cos(y)j + 0k = -e^(-x)sin(y)i + e^(-x)cos(y)j.
∂D/∂x = ∂(e^(-x)sin(y))/∂x = -e^(-x)sin(y)
∂D/∂y = ∂(e^(-x)sin(y))/∂y = e^(-x)cos(y)
∂D/∂z = ∂(e^(-x)sin(y))/∂z = 0
Therefore, ∇D = -e^(-x)sin(y)i + e^(-x)cos(y)j + 0k = -e^(-x)sin(y)i + e^(-x)cos(y)j.
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Given D = e-xsin y i – e-xcos y jFind divergence of D.a)3b)2c)1d...
Given d = e-xsin y i – e-xcos y j find divergence of d.with full explanation
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