Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane is measured in meters. A particle is moving clockwise around the circle of radius 1 m centered at the origin at the constant rate of 2 m/sec.How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point?

Answers

Answer 1

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

[tex]P = (\frac{1}{2}, \frac{\sqrt 3}{2})[/tex]

Answer:

[tex]Rate = 0.935042^\circ /cm[/tex]

Step-by-step explanation:

Given

[tex]P = (\frac{1}{2}, \frac{\sqrt 3}{2})[/tex]

[tex]T(x,y) =x\sin2y[/tex]

[tex]r = 1m[/tex]

[tex]v = 2m/s[/tex]

Express the given point P as a unit tangent vector:

[tex]P = (\frac{1}{2}, \frac{\sqrt 3}{2})[/tex]

[tex]u = \frac{\sqrt 3}{2}i - \frac{1}{2}j[/tex]

Next, find the gradient of P and T using: [tex]\triangle T = \nabla T * u[/tex]

Where

[tex]\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})} = (sin \sqrt 3)i + (cos \sqrt 3)j[/tex]

So: the gradient becomes:

[tex]\triangle T = \nabla T * u[/tex]

[tex]\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j][/tex]

By vector multiplication, we have:

[tex]\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}[/tex]

[tex]\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)[/tex]

[tex]\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5[/tex]

[tex]\triangle T = 0.935042[/tex]

Hence, the rate is:

[tex]Rate = \triangle T = 0.935042^\circ /cm[/tex]


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Step-by-step explanation:

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Step-by-step explanation:

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answers is




explaining

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