Question

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?

Answer 1

Missing information:

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

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

Answer:

$Rate = 0.935042^\circ /cm$

Step-by-step explanation:

Given

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$T(x,y) =x\sin2y$

$r = 1m$

$v = 2m/s$

Express the given point P as a unit tangent vector:

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$u = \frac{\sqrt 3}{2}i - \frac{1}{2}j$

Next, find the gradient of P and T using: $\triangle T = \nabla T * u$

Where

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

So: the gradient becomes:

$\triangle T = \nabla T * u$

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

By vector multiplication, we have:

$\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}$

$\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)$

$\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5$

$\triangle T = 0.935042$

Hence, the rate is:

$Rate = \triangle T = 0.935042^\circ /cm$