Derivative Worksheet and Answers - PDF

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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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