Answer:
4
Step-by-step explanation:
Answer:
Equation of tangent plane to given parametric equation is:

Step-by-step explanation:
Given equation
---(1)
Normal vector tangent to plane is:


Normal vector tangent to plane is given by:
![r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]](https://tex.z-dn.net/?f=r_%7Bu%7D%20%5Ctimes%20r_%7Bv%7D%20%3Ddet%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Chat%7Bi%7D%26%5Chat%7Bj%7D%26%5Chat%7Bk%7D%5C%5Ccos%28v%29%26sin%28v%29%260%5C%5C-usin%28v%29%26ucos%28v%29%261%5Cend%7Barray%7D%5Cright%5D)
Expanding with first row

at u=5, v =π/3
---(2)
at u=5, v =π/3 (1) becomes,



From above eq coordinates of r₀ can be found as:

From (2) coordinates of normal vector can be found as
Equation of tangent line can be found as:

Answer:
B and C
Step-by-step explanation:
They're congruent
24 = -v/3 (first you multiply 3 to each side)
<u>×3 ×3</u>
<u />72 = -v (then you divide -1 from each side)
-72 = v<u><u /></u><u />
Answer:
b) square root of 29
Step-by-step explanation:
i just took it on edge