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

Answer:

Step-by-step explanation:
Given




Express the given point P as a unit tangent vector:

Next, find the gradient of P and T using: 
Where

So: the gradient becomes:

![\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]](https://tex.z-dn.net/?f=%5Ctriangle%20T%20%3D%20%5B%28sin%20%5Csqrt%203%29i%20%2B%20%28cos%20%5Csqrt%203%29j%5D%20%2A%20%20%5B%5Cfrac%7B%5Csqrt%203%7D%7B2%7Di%20-%20%5Cfrac%7B1%7D%7B2%7Dj%5D)
By vector multiplication, we have:




Hence, the rate is:
50% is equal to 5/10 so 5/10 were party cakes 1/5x2=2/10 so 2/10 were fruit cakes 5/10+2/10=7/10 so 10/10 which is 100% of all cakes 10/10-7/10=3/10 so 3/10 of the cakes were sponge cakes.
If something is to the power of 1/2, it is the same as the square root of that number. So, what's the square root of 49?
Radius^2=13, so eqn is x^2+y^2=13