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MATH 2374, SECTION 13/14, WORKSHEET 3 ANSWERS 1. Consider the following function: x2 −y 2 +x3 x2 +y 2 , (x, y) = (0, 0) f (x, y) = 0, (x, y) = (0, 0) ∂f (a) Find ∂x (1,2) −y +x 2 2 3 Answer: We need to diﬀerentiate x x2 +y2 with respect to x (that is treating y as a constant). Then, we will plug in x = 1 and y = 2 into the expression that we got. Here goes: ∂f (x2 + y 2 )(2x + 3x2 ) − (x2 − y 2 + x3 )(2x) = ∂x (x2 + y 2 )2 2x3 + 3x4 + 2xy 2 + 3x2 y 2 − 2x3 + 2xy 2 − 2x4 = (x2 + y 2 )2 x4 + 3x2 y 2 + 4xy 2 = (x2 + y 2 )2 So: ∂f 14 + 3(12 )(22 ) + 4(1)(22 ) = ∂x (1,2) (12 + 22 )2 29 = 25 ∂f (b) Find ∂x (0,0) Answer: We would prefer to just plug x = 0 and y = 0 into the partial derivative formula we obtained in the previous problem. However, when we do that we get 0 . So, we need to evaluate 0 this partial derivative using the limit deﬁnition. ∂f f (0 + h, 0) − f (0, 0) = lim ∂x (0,0) h→0 h h2 −0+h3 h −0 = lim h→0 h 1+h = lim h→0 h = D.N.E. (Well, I intended for this limit to exist, but I messed up so it doesn’t. Normally when you get a problem like this you should expect an actual limit.) 2. Find the equation of the tangent plane of z = x2 − y at the point (2, −1, 5). Answer: The equation of the tangent plane is given by : ∂z ∂z z= (x − 2) + (y + 1) + z|(2,−1) . ∂x (2,−1) ∂y (2,−1) So, we do the following calculations: ∂z ∂z • ∂x = 2x, so ∂x (2,−1) = 2(2) = 4. 2 MATH 2374, SECTION 13/14, WORKSHEET 3 ANSWERS ∂z ∂z • ∂y = −1, so ∂x (2,−1) = −1. • z|(2,−1) = (22 ) − (−1) = 5. Thus, the tangent plane is z = 4(x − 2) − (y + 1) + 5 or z = 4x − y − 4. 3. Find the equation of the tangent plane to the surface z = f (x, y) = x − y 2 at the point (1, 2, −3). Answer: The equation of the tangent plane is given by : ∂f ∂f z= (x − 1) + (y − 2) + f (1, 2). ∂x (1,2) ∂y (1,2) So, we do the following calculations: ∂f ∂f • ∂x = 1, so ∂x (1,2) = 1. ∂f ∂f • ∂y = −2y, so ∂x (1,2) = −2(2) = −4. • f (1, 2) = 1 − 22 = −3 (or we could have just read oﬀ the z-coordinate of the point given in the problem. Thus, the tangent plane is z = (x − 1) − 4(y − 2) − 3 or z = x − 4y + 4. 4. Use your answer to (3.) to approximate f (0.98, 2.01) where f (x, y) = x − y 2 . Answer: We will approximate f (x, y) using the tangent plane at (1, 2, −3) as a linear ap- proximation. All this really means is that we plug in x = 0.98 and y = 2.01 into the equation of the tangent plane. The resulting z value should be approximately f (0.98, 2.01). So, we get f (0.98, 2.01) ≈ 0.98 − 4(2.01) + 4 = −3.06. Thus, our approximation to f (0.98, 2.01) is −3.06. For the curious, f (0.98, 2.01) = −3.0601 (just plug in x = 0.98 and y = 2.01 into the formula for f ).

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posted: | 1/18/2011 |

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Derivative Worksheet and Answers document sample

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