Graded Assignment 5: Modules 12 and 13
Geometric applications and some Maple plotting
FYI: technology heavy. Most numbers are horrible – don’t expect pretty answers. You should
strategically employ Maple/Wolfram for evaluating and checking things.
Problem 1: Consider the function f ( x, y )  2 x3 y 2  7 xy 3 at the point
( a, b, f ( a, b ))  (2, 1, f (2, 1)) .
(a) Find f x ( x, y ) , f x (a, b) , f y ( x, y ) , f y (a, b) , and f (a, b) .
(b) Give a vector tangent to the surface at the point, and in the direction of the y axis. Then, write
the equation of the tangent line that has that vector as its direction vector.
(c) Give a vector tangent to the surface at the point, and in the direction of the x axis. Then, write
the equation of the tangent line that has that vector as its direction vector.
(d) Give a vector normal to the surface at the point. Then, write the equation of the tangent plane
that has that vector as its normal.
(e) Produce a Maple plot of the surface, the two tangent lines, and the tangent plane. Recommend
setting x = 0..4, y = -3..1, and t = -2..2. Then set
. That should
produce a decent picture.
Problem 2: Working with the above function f ( x, y )  2 x3 y 2  7 xy 3 at the point
( a, b, f ( a, b ))  (2, 1, f (2, 1)) .
(a) Write L( x, y ) , the linearization of f at the given point.
(b) Approximate f (2.03, 0.98) by computing L(2.03, 0.98)
(c) And compare the approximate value to the exact value (what's the absolute value of the
difference?).
3 2
3
Problem 3: Still working with the function f ( x, y )  2 x y  7 xy at the point
( a, b, f ( a, b ))  (2, 1, f (2, 1)) .
(a) Write the vectors f ( x, y ) and f (a, b) .
(b) Find the derivative of f at (a, b, f (a, b)) in the direction of the vector v  3, 2  .
(c) Give a vector tangent to the surface at the point, and in the direction of v . Then, write the
equation of the tangent line that has that vector as its direction vector.
(d) Maple plot the surface and this tangent line.
Problem 4: Yep, still working with the function f ( x, y )  2 x3 y 2  7 xy 3 at the point
( a, b, f ( a, b ))  (2, 1, f (2, 1)) .
(a) What is the direction of maximum decrease in the surface from that point?
(b) What is the value of that decrease (the slope of the tangent in that direction)?
(c) Give a vector tangent to the surface at the point, and in the direction you found above. Then,
write the equation of the tangent line that has that vector as its direction vector.
(d) Maple plot the surface and this tangent line.
Problem 5: Considering the function that won't die, f ( x, y )  2 x3 y 2  7 xy 3 .
(a) Write expressions for the level curves at k  0 and k  2 . Use Maple to sketch (there’s an
example of this posted with the assignment and a template in BBoard).
(b) Let z  0 , y  1 , and solve for x (these ( x, y ) points will lie on the level curve k  0 ). Give
the values of f at these points (approximate x solutions to two decimals and gradient values
to one decimal for convenience). Produce a plot that displays this level curve, and adds the
gradient vectors to the curve. Since you’re checking for orthogonality, square off the view at 10..10 in each direction (you’ll clip off some of the vectors, but that’s not a big deal).
(c) Let z  2 , y  1 , and solve for x (these ( x, y ) points will lie on the level curve k  2 ). This
equation won’t be hand solvable, by the way – it’s a cubic – so I’m expecting a Maple “fsolve.”
Give the values of f at these points (approximate x solutions to two decimals and gradient
values to one decimal for convenience). Produce a plot that displays this level curve, and adds
the gradient vectors to the curve.
Problem 6: For the function f ( x, y )  e xy ( 4 x  y ) .
(a) Find f x and f y and solve for the critical points of f . OK to use Wolfram (recommended over
Maple, although you can interpret Maple’s somewhat cryptic response for exact values if you
know what you’re looking at) to solve simultaneous system - it’s doable by hand, but tedious
algebra. DO show setup and differentiation and clearly indicate WHAT system you are asking it
to solve. You should get two critical points.
(b) Compute the expressions for the second partials f xx , f xy  f yx and f yy .
(c) For each critical point, evaluate the second partials, evaluate the quantity D 
f xx
f xy
f yx
f yy
, and
perform the second partials test to determine whether the critical point is a local maximum,
minimum, or saddle point.
(d) Give all maximum and minimum values, and coordinates of saddle points (evaluate f at the
appropriate locations).
(e) Maple plot the surface and points to verify your answer makes sense. Recommend view = [3..3,-3..3,-10..10].
Problem 7: Use the method of Lagrange multipliers to find the minimum value of f ( x, y )  x 2  y 2
subject to the constraint xy 2  54 . Don’t forget the method described in the notes does include testing
to verify that you in fact have a minimum (don’t assume).
Note that what you’re doing there is minimizing the value of the square of the distance function
d 2  x 2  y 2 , and minimizing d 2 effectively minimizes d , and locates the points on the constraint
curve that are closest to the origin. Plot (however you want- Maple, Winplot, MVT) the level curve for
f at the solution value(s) and the constraint equation and verify that they are tangential, and see if you
really have found what appear to be the points closest to the origin.