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]
