Level curves Consider the paraboloid f ( x, y ) = 16 – x 2 /4 – y ,2 /16 and the point P on the given level curve of f. Compute the slope of the line tangent to the level curve at P and verify that the tangent line is orthogonal to the gradient at that point. 45 f ( x , y ) = 12 ; P ( 4 , 0 )
Level curves Consider the paraboloid f ( x, y ) = 16 – x 2 /4 – y ,2 /16 and the point P on the given level curve of f. Compute the slope of the line tangent to the level curve at P and verify that the tangent line is orthogonal to the gradient at that point. 45 f ( x , y ) = 12 ; P ( 4 , 0 )
Solution Summary: The author calculates the slope of the tangent line to the level curve of f(x,y)=16-x24
Level curvesConsider the paraboloid f(x, y) = 16 – x2/4 – y,2/16 and the point P on the given level curve of f. Compute the slope of the line tangent to the level curve at P and verify that the tangent line is orthogonal to the gradient at that point.
(1 point) If the gradient of ff is ∇f=2z(i) +x (j) +yx (k) and the point P=(−8,−8,−2) lies on the level surface f(x,y,z)=0 find an equation for the tangent plane to the surface at the point P
Let f(x,y)=(2)0.5sin(xy)
a. Find the gradient of f(1/2,pi/2)
b. Use your answer to find the equation of the tangent line (in parametric form) to the level curve f(x,y)=1, at the point (1/2,pi/2)
c. Graph the level curve together with the point and the tangent line you found above.
A function is given by f(x, y, z) = x^2 y^3 + sin za) Find the gradient of f at a general pointb) Use the answer to part a, and a simple substitution, in order to find the gradient of f at a point (x, y, z) = (1, 1, 0)c) Use the answer to part b in order to find the linear approximation of f around the point (x, y, z) = (1, 1, 0)
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (4th Edition)
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