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You need to remember that a level surface is expressed by the function `f(x,y,z) =0` and you need to find at what point the tangent plane is perpendicular to the line given parametrically:
You should come up with the substitution for the tangency point (a,b,c).
Since the tangency point belongs to level surface `8 - (3*x^2)*y - (2*y^2) - z = 0` , then its coordinates need to check the equation such that:
`f(a,b,c) = 8 - (3*a^2)*b - (2*b^2) -c = 0 `
You need to remember that gradient vector of `f(x,y,z) ` is orthogonal to the level surface such that:
`grad` `f(x,y,z) =` `((del f)/(del x),(del f)/(del y),(del f)/(del z))`
`grad f(x,y,z) = (6xy, -3x^2-4y,-1)`
`grad f(a,b,c) = (6ab, -3a^2-4b,-1)`
You should remember that gradient vector is also orthogonal to the tangent plane.
Notice that the problem provies the information that the tangent plane is orthogonal to the line `x=2-3t,y=7+8t,z=5-` t, hence the line `x=2-3t,y=7+8t,z=5-t` is perpendicular to tangent plane.
You need to remember that two lines perpendicular to a plane are parallel, hence the line `x=2-3t,y=7+8t,z=5-t` is parallel to gradient vector.
You need to express the vector of line `x=2-3t,y=7+8t,z=5-t ` such that `bar d = lt-3,8,-1gt` .
You need to express the condition that the vectors `bar d` is parallel to vector `grad` such that:
`(6ab)/(-3) = (-3a^2-4b)/8 = (-1)/(-1)`
`9a^2 + 12b = 48ab=gt 9a^2= 48ab -12b`
You need to divide by 3 such that:
`3a^2 = 16ab - 4b`
Factoring out b yields:
`3a^2 = b(16a - 4) =gt b = (3a^2)/(16a - 4)`
You need to substitute `-1/(2a)` for b in`-3a^2-4b = 8` such that:
`-3a^2 + 2/a = 8`
You need to bring the terms to a common denominator such that:
`-3a^3 - 8a + 2 = 0 =gt 3a^3+ 8a- 2 = 0`
Notice that graph intercepts x axis at a value between 0 and 1, hence the equation `3a^3 + 8a - 2 = 0` has one real value over `(0,1).`
Hence, evaluating the point where the tangent plane to surface `z=8-(3*x^2)*y-(2*y^2)` is perpendicular to the line given parametrically yields `(a,-1/(2a),1),` where `a in (0,1).`
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