Use differentials to approximate the change in z for the given change in the ind
ID: 2867956 • Letter: U
Question
Use differentials to approximate the change in z for the given change in the independent variables, z = x2 - 3xy + y when (x,y) changes from (5, 5) to (5.03,4.99) dz = (Type an integer or a decimal.) Find the linear approximation for the following function at the given point. Use part (a) to estimate the given function value. f(x, y) = 6x - 9y + 6xy; (2,3); estimate f(2,2,3.53). Find an equation of the plane tangent to the following surface at the given point. z = 3 - 3x2 - y2; (3,3, - 33) (simplify your answer. Do not factor.). Use differentials to approximate the change in z for the given change in the independent variables. z = In (x6y) when (x,y) changes from (-2, 4) to (-1.98, 3.98). (Type an integer or a decimal).Explanation / Answer
z = x^2 - 3xy + y
zx = partial derivative with x
zx = 2x - 3y
zx = 2(5) - 3(5) = -5
zy = -3x + 1
zy = -3(5) + 1
zy = -14
dz = zx(x - x0) + zy(y - y0)
dz = -5(x - 5) - 14(y - 5)
dz = -5x - 14y + 95
Plug in x = x = 5.03 and y = 4.99 :
dz = -5(5.03) - 14(4.99) + 95
dz = -0.01 ---> ANSWER
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f(x,y) = 6x - 9y + 6xy at (2,3)
fx = 6 + 6y = 6 + 6(3) = 24
fy = -9 + 6x = -9 + 6(2) = 3
When x = 2 , y = 3, we get : z = 12 - 27 + 36 --> z = 21
So, L(x,y) = fx(x-x0) + fy(y-y0) + z0
L(x,y) = 24(x - 2) + 3(y - 3) + 21 ---> ANSWER
L(2.2 , 3.53) = 24(2.2 - 2) + 3(3.53 - 3) + 21 = 27.39 ---> ANSWER
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z = 3 - 3x^2 - y^2
zx = -6x = -6(3) = -18
zy = -2y = -2(3) = -6
z - z0 = zx(x - x0) + zy(y - y0)
z - (-33) = -18(x - 3) - 6(y - 3)
z + 33 = -18x + 54 - 6y + 18
z = -18x - 6y + 39 ---> ANSWER
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z = ln(x^6 * y)
Lets expand the log :
z = ln(x^6) + ln(y)
z = 6ln(x) + ln(y)
zx = 6/x = 6/-2 = -3
zy = 1/y = 1/4
dz = zx(x-x0) + zy(y-y0)
dz = -3(x + 2) + (1/4)(y - 4)
dz = -3x + (1/4)y - 7
Plug in -1.98 and 3.98 :
dz = -3(-1.98) + (1/4)(3.98) - 7
dz = -0.065 ----> ANSWER
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