Answer:
a.

b.


Step-by-step explanation:
Remember that for any curve
The tangent vector is given by

And the normal vector is given by

a.
For this case, using the chain rule

And also remember that

Therefore

Similarly, using the quotient rule and the chain rule

And also

Therefore

Notice that
1. 
2. 
b.
Simlarly

and

Therefore

Then

and also

And since

Answer:
The last one, −2.1−(5.9+3.7)
Step-by-step explanation:
−2.1+(−5.9)+(−3.7) = -2.1-5.9-3.7 = -11.7
the only expression that will come out as -11.7 is the last one
This one is for number 26
Answer:
64
Step-by-step explanation:
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