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Find the curl of A = (y cos ax)i + (y + ex)k
- a)2i – ex j – cos ax k
- b)i – ex j – cos ax k
- c)2i – ex j + cos ax k
- d)i – ex j + cos ax k
Correct answer is option 'B'. Can you explain this answer?
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Find the curl of A = (y cos ax)i + (y + ex)ka)2i – ex j – ...
Answer: b
Explanation: Curl A = i(Dy(y + ex)) – j (Dx(y + ex) – Dz(y cos ax)) + k(-Dy(y cos ax))
= 1.i – j(ex) – k cos ax = i – ex j – cos ax k.
Explanation: Curl A = i(Dy(y + ex)) – j (Dx(y + ex) – Dz(y cos ax)) + k(-Dy(y cos ax))
= 1.i – j(ex) – k cos ax = i – ex j – cos ax k.
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Find the curl of A = (y cos ax)i + (y + ex)ka)2i – ex j – ...
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Find the curl of A = (y cos ax)i + (y + ex)ka)2i – ex j – ...
To find the curl of vector A, we can use the formula for the curl of a vector field:
curl(A) = ∇ x A
Where ∇ is the del operator and x represents the cross product.
Let's calculate the curl step by step.
1. Take the determinant of the following matrix:
| i j k |
| ∂/∂x ∂/∂y ∂/∂z |
| ycos(ax) y eax |
2. Expanding the determinant, we have:
i * (∂/∂y * eax - y * ∂/∂z)
- j * (∂/∂x * eax - ycos(ax) * ∂/∂z)
+ k * (∂/∂x * y - ycos(ax) * ∂/∂y)
3. Taking the partial derivatives:
∂/∂x * y = 0 (constant with respect to x)
∂/∂y * eax = 0 (constant with respect to y)
∂/∂z * eax = 0 (constant with respect to z)
∂/∂x * ycos(ax) = -ay * sin(ax)
∂/∂y * ycos(ax) = cos(ax) - ay * x * sin(ax)
∂/∂z * ycos(ax) = 0 (constant with respect to z)
4. Simplifying the expression:
i * (- y * ∂/∂z) = -y * (∂/∂z)
- j * (∂/∂x * eax) = -(0)
+ k * (cos(ax) - ay * x * sin(ax)) = cos(ax) - ay * x * sin(ax)
5. Combining the terms:
curl(A) = -y * (∂/∂z)i + (cos(ax) - ay * x * sin(ax))k
Therefore, the curl of vector A is -y * (∂/∂z)i + (cos(ax) - ay * x * sin(ax))k.
curl(A) = ∇ x A
Where ∇ is the del operator and x represents the cross product.
Let's calculate the curl step by step.
1. Take the determinant of the following matrix:
| i j k |
| ∂/∂x ∂/∂y ∂/∂z |
| ycos(ax) y eax |
2. Expanding the determinant, we have:
i * (∂/∂y * eax - y * ∂/∂z)
- j * (∂/∂x * eax - ycos(ax) * ∂/∂z)
+ k * (∂/∂x * y - ycos(ax) * ∂/∂y)
3. Taking the partial derivatives:
∂/∂x * y = 0 (constant with respect to x)
∂/∂y * eax = 0 (constant with respect to y)
∂/∂z * eax = 0 (constant with respect to z)
∂/∂x * ycos(ax) = -ay * sin(ax)
∂/∂y * ycos(ax) = cos(ax) - ay * x * sin(ax)
∂/∂z * ycos(ax) = 0 (constant with respect to z)
4. Simplifying the expression:
i * (- y * ∂/∂z) = -y * (∂/∂z)
- j * (∂/∂x * eax) = -(0)
+ k * (cos(ax) - ay * x * sin(ax)) = cos(ax) - ay * x * sin(ax)
5. Combining the terms:
curl(A) = -y * (∂/∂z)i + (cos(ax) - ay * x * sin(ax))k
Therefore, the curl of vector A is -y * (∂/∂z)i + (cos(ax) - ay * x * sin(ax))k.
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