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What is the Jacobian of the transformation given by x = 6e−4r sin(3), y = e4r cos(3)?

a) -24e^(-4r)cos(3), 4e^(4r)sin(3)
b) 24e^(-4r)cos(3), -4e^(4r)sin(3)
c) 24e^(-4r)sin(3), 4e^(4r)cos(3)
d) -24e^(-4r)sin(3), -4e^(4r)cos(3)

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Final answer:

The Jacobian of the transformation is the product of the partial derivatives of x and y with respect to r, which are -24e−4r sin(3) and 4e4r cos(3), respectively. However, the provided options do not match the simplified Jacobian which would be -24e0 cos(3) sin(3), suggesting there might be a typo in the question. None of the given option is correct.

Step-by-step explanation:

The question asks to find the Jacobian of the transformation given by x = 6e−4r sin(3), y = e4r cos(3). The Jacobian of a transformation, given by the variables x and y in terms of r and a constant, is a determinant that represents the rate of change of these variables. To find the Jacobian, we take the partial derivatives of x and y with respect to r.


The derivative of x with respect to r is dx/dr = -24e−4r sin(3) and the derivative of y with respect to r is dy/dr = 4e4r cos(3). Thus, the Jacobian J can be expressed as:


J(r) = dx/dr × dy/dr


This gives us the product of the two derivatives: J(r) = (-24e−4r sin(3))(4e4r cos(3)). Simplifying the expression gives us the Jacobian of the transformation: -24e0 cos(3) sin(3), which simplifies further since e0 = 1. However, as none of the options matches this simplification exactly, there might be a typo in the provided expressions, and we would need the specific and correct transformation equations to provide the accurate Jacobian.

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