vyA Section 11.3 Quiz x G The wind chill index is a x WA Be5fc1b2c7ca7df7ca4b7 x

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vyA Section 11.3 Quiz x G The wind chill index is a x WA Be5fc1b2c7ca7df7ca4b7 x y VA Section 11.4 Quiz x C https://www.webassign.net/web/Student/Assignment-Responses/last?dep 14379054 Find an equation of the tangent plane to the given surface at the specified point. z 20x 4 (y 3 3 (2, 1, 21 2. -1 points SEssCalcET2 013 My Notes Ask Your Teacher Explain why the function is differentiable at the given point. fx, y Xvcos y, (IT, 0 so 1, 0) fo( The partial derivatives are f(x, y) and f (x, y and f (n, 0) Both f and f are continuous functions, so fis differentiable at (n, 0 Find the linearization L y) of fx, y) at (n, 0) L(x, y 3. -1 points SEssCalcET2 4.02B My Notes Ask Your Teacher Use differentials to estimate the amount of metal in a closed cylindrical can that is 30 cm high and 10 cm in diameter if the metal in the top and the bottom is 0.4 cm thick and the metal in the sides is 0.05 cm thick. (Round your answer to two decimal places. Cm 10:28 PM O Ask me anything 10/26/2016

Explanation / Answer

An equation of the tangent line to the surface z = f(x , y) at point P( x0, y0, z0) is

z - z0 = fx ( x0 , y0 )(x - x0) + fy( x0, y0) ( y - y0) .------(1)

Therefore , (x0, y0, z0) = ( 2 , -1 , 21) .

fx( x , y) = 4(x -1)   and    fy = 8( y + 3 ) ,

fx ( 2 , -1 ) = 4 and f y ( 2 , -1 )   = 16.

SO the equation (1) becomes ( z - 21 ) = 4( x - 2) + 16( y - (-1)).

Simplify it then we get the euqtion as z = 4x + 16 y + 29 .

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