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?

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]

-3 + 8n = -5
what is n?

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ati
n
100
Sc
Geography English Spelling Science
Mean of all 5th graders in District (150 students) 75
125 90
SD of these 5th graders
12
10 12 20
Mean of the Norm Group
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100 120 90
SD of the Nom Group
10
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ww
sti
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ker
ndo
ted
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Geography only
lccr
Sa
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