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:
20,909,909,090,000,000,000,000,000,000,000,000
Answer:
Yes, it's an arithmetic sequence.
Step-by-step explanation:
Each term in this sequence is 4 more than the previous term. Therefore we have an arithmetic sequence.
Answer:
The Temperature in the afternoon was
.
Step-by-step explanation:
Given:
Temperature in the morning = 
Rise in temperature = 
We need to find the temperature in the afternoon.
Solution:
Now we know that;
temperature in the afternoon is equal to Temperature in the morning plus Rise in temperature in afternoon.
framing in equation form we get;
temperature in the afternoon = 
Hence the Temperature in the afternoon was
.
Answer: 40
Step-by-step explanation: