Asked by Kristian Lou
Find the slope of the tangent line to the curve of intersection of the
surface 36๐ฅยฒ โ 9๐ฆยฒ + 4๐งยฒ + 36 = 0
with the plane ๐ฅ = 1 at the point (1, โ12, โ3) .
surface 36๐ฅยฒ โ 9๐ฆยฒ + 4๐งยฒ + 36 = 0
with the plane ๐ฅ = 1 at the point (1, โ12, โ3) .
Answers
Answered by
oobleck
at x=1, we have
36 - 9y^2 + 4z^2 + 36 = 0
y^2/8 - z^2/18 = 1
If by slope you mean dy/dz, then
y/4 dy/dz - z/9 = 0
โ3/2 dy/dz = -1/3
dy/dz = -2/โ27
36 - 9y^2 + 4z^2 + 36 = 0
y^2/8 - z^2/18 = 1
If by slope you mean dy/dz, then
y/4 dy/dz - z/9 = 0
โ3/2 dy/dz = -1/3
dy/dz = -2/โ27
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