ha (lvh262) – Homework 14.4 – karakurt – (56295) 1This print-out should have 8 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsFind the linearization of z = f(x, y) atP (3, −1) whenf(3, −1) = 1andfx(3, −1) = −2 , fy(3, −1) = 1 .1. L(x, y) = 1 − 2x + y2. L(x, y) = 1 − 2x − y3. L(x, y) = 8z − 2x + y4. L(x, y) = 8z + 2x − y5. L(x, y) = 1 + 2 x − y6. L(x, y) = 8 − 2 x + y correctExplanation:The linearization of z = f (x, y) at P (a, b)is given byL(x, y) = f (a, b)+ fx(a, b)(x − a) + fy(a, b)(y − b) ,and so at P (3, −1),L(x, y) = f (3, −1)+ fx(3, −1)(x − 3) + fy(3, −1)(y + 1) .Consequently, the linearization of f at P isL(x, y) = 1 − 2(x − 3) + (y + 1) ,which after rearrangement becomesL(x, y) = 8 − 2x + y .keywords: linearization, partial derivative,radical function, square root function,002 10.0 pointsFind an equation for the tangent planeat the point P (2, 3, 8) on the graph ofz = f(x, y) whenfx(2, 3) = 3 , fy(2, 3) = 1 .1. x − 3y + z − 1 = 02. x − 3y + z + 8 = 03. 3 x + y −z + 8 = 04. 3 x + y −z − 1 = 0 correct5. x − 3y −z + 8 = 06. 3 x + y + z − 1 = 0Explanation:The eq uation of the tangent plane to thegraph of z = f(x, y) at the pointP (a, b, f (a, b) )is given byz = f (a, b)+ fx(a, b)(x − a) + fy(a, b)(y −b) .But whenP (a, b, f (a, b)) = (2, 3, 8) ,we see that(a, b) = (2, 3) , f (a, b) = 8 ,whilefx(a, b) = 3 , fy(a, b) = 1 .So at P the tangent plane has equationz = 8 + 3(x − 2) + (y −3) ,which after rearrangement becomes3x + y − z −1 = 0 .ha (lvh262) – Homework 14.4 – karakurt – (56295) 2keywords: tangent plane, partial derivative003 10.0 pointsUse Linear Approximati on to estimate thevalue of f(0.9, −1.9) whenf(1, −2) = 3andfx(1, −2) = −1 , fy(1, −2) = 4 .1. f (0.9, −1.9 ) ≈ 3.5 correct2. f (0.9, −1.9 ) ≈ 3.83. f (0.9, −1.9 ) ≈ 3.74. f (0.9, −1.9 ) ≈ 3.65. f (0.9, −1.9 ) ≈ 3.9Explanation:Linear Approximation uses the Lineariza-tionL(x, y) = f (a, b)+ fx(a, b)(x − a) + fy(a, b)(y − b) ,of z = f(x, y) at P (a, b) to writef(x, y) ≈ L(x, y) .The value of L(x, y) is usually much easier tocalculate, and for (x, y) near (a, b) it gives agood estimate of f (x, y).Now at P (1, −2),L(x, y) = f (1, −2)+ fx(1, −2)(x − 1) + fy(1, −2)(y + 2) .Consequently, the linearization of f at P isL(x, y) = 3 − ( x − 1) + 4(y + 2) .But thenf(0.9, −1.9) ≈ L(0. 9, −1.9 ) = 3.5 .keywords: linearization, partial derivative,Linear Approximation,004 10.0 pointsUse Linear Approximati on to estimate thechange∆f = f (a + ∆x, b + ∆y) − f(a, b)in f when∆x = −0.3 , ∆y = −0.1 ,andfx(a, b) = 2 , fy(a, b) = −1 .1. ∆f ≈ −0.62. ∆f ≈ −0.43. ∆f ≈ −0.34. ∆f ≈ −0.5 correct5. ∆f ≈ −0.2Explanation:By Linear Approximation of f at (a, b),f(a + ∆x, b + ∆y) ≈ L(a + ∆x, b + ∆y)= f (a, b) + fx(a, b)∆x + fy(a, b)∆y .Thus∆f = f (a + ∆x, b + ∆y) − f(a, b)≈ fx(a, b)∆x + fy(a, b)∆y .Consequently, when∆x = −0.3 , ∆y = −0.1 ,andfx(a, b) = 2 , fy(a, b) = −1 ,Linear Approximation gives the estimate∆f ≈ (2)(−0.3) + (−1)( −0.1) = −0.5 .ha (lvh262) – Homework 14.4 – karakurt – (56295) 3keywords: linearization, partial derivative,Linear Approximation,005 10.0 pointsFind the linearization, L(x, y), off(x, y) =p7 − x2− 2 y2at the point P (2, −1).1. L(x, y) = −5 − 2x − 2y2. L(x, y) = 7 − 2 x + 2y correct3. L(x, y) = 7 − 2 x − 2y4. L(x, y) = −5 − x − 2y5. L(x, y) = −5 − 2x + 2y6. L(x, y) = 7 − x + 2yExplanation:The linearization of z = f(x, y) at the pointP (a, b) is given byL(x, y) = f (a, b)+∂f∂x(a, b)(x − a) +∂f∂y(a, b)(y − b) .Now whenf(x, y) =p7 − x2− 2y2,we see that∂f∂x= −xp7 − x2− 2y2,while∂f∂y= −2yp7 − x2− 2y2.Thus at P ,f(2, −1) = 1 ,while∂f∂x(2, −1)= −2 ,∂f∂y(2, −1)= 2 .So at P the linearization of f isL(x, y) = 1 − 2(x − 2) + 2(y + 1) ,which after rearrangement becomesL(x, y) = 7 − 2x + 2y .keywords: linearization, partial derivative,radical function, square root function,006 10.0 pointsFind the linearizat ion, L(x, y), off(x, y) = y√xat the point (9, −1).1. L(x, y) =32−16x + 3y correct2. L(x, y) = −32+ 3x +16y3. L(x, y) = −3 +32x +16y4. L(x, y) = −3 +16x −32y5. L(x, y) =32+ 3 x −16y6. L(x, y) = −32+16x + 3yExplanation:The linearization of f = f (x, y) at a point(a, b) is given byL(x, y) = f(a, b)+(x−a)∂f∂x(a,b)+(y−b)∂f∂y(a,b).But when f(x, y) = y√x,∂f∂x=y2√x,∂f∂y=√x ;ha (lvh262) – Homework 14.4 – karakurt – (56295) 4thus when (a, b) = (9, −1),∂f∂x(a,b)= −16,∂f∂y(a,b)= 3 ,while f(a, b) = −3. Consequently,L(x, y) =32−16x + 3y.keywords:007 10.0 pointsFind the linearization, L(x, y), of the func-tionf(x, y) = tan−1(x + 8y)at the point (1, 0) .1. L(x, y) = x + 4y +π4−122. L(x, y) =12x + 4y −123. L(x, y) =12x + 4y +π4−12correct4. L(x, y) = x + 4y +π4+125. L(x, y) = x + 4y −12Explanation:The linearization of f = f (x, y) at a point(a, b) is given byL(x, y) = f (a, b)+ (x − a)∂f∂x(a,b)+(y −b)∂f∂y(a,b).But the partial derivatives off(x, y) = tan−1(x + 8y)arefx=11 + (x + 8y)2, fy=81 + (x + 8y)2.It follows that∂f∂x(1, 0)=12and∂f∂y(1, 0)= 4 .Consequently,L(x, y) =12x + 4y +π4−12.keywords:008 10.0 pointsFind an equation for the plane passingthrough the origin that is parallel to the tan-gent plane to the graph ofz = f(x, y) = 3x2− 2y2− 3 x + 3yat the point (1, 1, f (1, 1)).1. z + 3x − y − 3 = 02. z + 3x + y − 5 = 03. z − 3x − y + 3 = 04. z + 3x + y = 05. z − 3x − y = 06. z − 3x + y = 0 correctExplanation:Parallel planes have parallel normals. Onthe other hand, the tangent plane to the graphof z = f ( x , y) at the point (a, b, f (a, b)) hasnormaln = h−fx(a, b), −fy(a, b), 1 i.But whenf(x, y) = 3x2− 2 y2− 3x + 3ywe see thatfx= 6x − 3 , fy= …