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:
Could you attach an image of the slope? Because i dunno what lesson you're on....
:/
The simplified expression should be 3/7
Set y=ax+b pass (7,-9) (5,6)
=》-9=7a+b
6=5a+b
=》2a=-15
a=-15/2
b=6-5a=6+37.5=43.5
=》y= -15/2(x)+43.5
2y+15x=87