Horizontal Tangent Plane In Exercises 37-42, find the point(s) on the surface at which the tangent plane is horizontal. z = 3 x 2 + 2 y 2 − 3 x + 4 y − 5
Solution Summary: The author calculates the tangent plane at the point where the slope of the plane would be zero. They then partial differentiate the function with respect to y.
Finding Curvature In Exercises 23-28, find the curvature of the plane curve at the given value of the parameter.
Curve In Exercise 56, sketch the plane curve and find its length over the given interval.
56. r(t) = t 2i + 2tk, [0, 3]
Check that the point (−1,−1,2) lies on the given surface. Then, viewing the surface as a level surface for a function f(x,y,z) find a vector normal to the surface and an equation for the tangent plane to the surface at (−1,−1,2).
4x^2−y^2+4z^2=19
vector normal = ?
tangent plane: z = ?
Thomas' Calculus: Early Transcendentals (14th Edition)
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