x = 1
y = -3/2
Solution: (1, -3/2)
//Hope it helps.
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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:
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: