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Local Linearity and the Differential Quiz

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This quiz tests the work covered in lecture on local linearity and the differential and corresponds to Section 14.3 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).
There is a useful applet at http://www.slu.edu/classes/maymk/banchoff/TangentPlane.html - take some time to read the instructions and add your own functions.
There are more web quizzes at Wiley, select Section 3. This quiz has 10 questions.

Question 1

Suppose f (3,2) = 4, fx(3,2) = - 2 and fy(3,2) = 3 for some surface z = f(x,y).
Which of the following is the tangent plane to the surface at (3,2,4)?

a)
4z = - 2(x- 3)+ 3(y- 2)
  b)
z = 4- 2(x- 3)+ 3(y- 2)
c)
z + 4 = 2(x+ 3)+ 3(y+ 2)
  d)
4z = 3(x + 3)- 2(y+ 2)

 

Not correct. Choice (a) is false.
Try again, check the formula for the tangent plane.
Your answer is correct.
The tangent at the point (a,b) on the surface is z = f(a,b)+ fx(a,b)(x - a)+ f(a,b)(y - b) so the above equation is correct.
Not correct. Choice (c) is false.
Try again, check the formula for the tangent plane.
Not correct. Choice (d) is false.
Try again, check the formula for the tangent plane.

Question 2

Is the plane z = 12+ 8(x- 1)+ 7(y- 2) the tangent plane to the surface, f(x,y) = x2 + 3xy+ y2 - 1 at (1,2)?

a)
Yes.
   b)
No

 

Not correct. Choice (a) is false.
f(1,2) = 10 and z = 12 at (1,2) so the plane does not touch the surface.
Your answer is correct.
fx(x,y) = 2x+ 3y so fx(1,2) = 2+ 6 = 8
fy(x,y) = 3x +2y so fy(1,2) = 3+ 4 = 7
f(1,2) = 10 so the tangent plane is z = 10+ 8(x- 1)+ 7(y- 2).

Question 3

Which of the following is the tangent plane to the surface f(x,y) = x2 - 2xy - 3y2 at the point (- 2,1,5)?

a)
z + 6x + 2y+ 15 = 0
  b)
z - 6x- 2y+ 5 = 0
c)
z + 6x+ 2y +5 = 0.
  d)
None of the above, since (- 2,1,5) is not on the surface.

 

Not correct. Choice (a) is false.
Try again, look carefully at the signs of the constant terms.
Not correct. Choice (b) is false.
Try again, carefully rearrange your equation.
Your answer is correct.
fx(x,y) = 2x- 2y so fx(- 2,1) = - 4- 2 = - 6
fy(x,y) = - 2x - 6y so fy(- 2,1) = 4- 6 = - 2
f(- 2,1) = 5 so the tangent plane is z = 5- 6(x+ 2)- 2(y- 1) ⇒ z + 6x + 2y+ 5 = 0 as required.
Not correct. Choice (d) is false.
Try again, the point is on the surface.

Question 4

Which of the following is the differential of xy f(x,y) = sin xye ?

a)
xy 2 2 df = cosxye (y dx+ x dy)
  b)
xy df = e (cos xy+ sin xy)(xdx + ydy)
c)
xy df = e (- cosxy+ sin xy)(ydx + xdy)
  d)
xy df = e (cosxy + sinxy)(ydx+ xdy)

 

Not correct. Choice (a) is false.
Try again, you must use the product rule to differentiate f(x,y).
Not correct. Choice (b) is false.
Try again, you have not differentiated f(x,y) correctly.
Not correct. Choice (c) is false.
Try again, you have not differentiated sin xy correctly.
Your answer is correct.
fx(x,y) = ycosxyexy + sinxy(yexy) = yexy(cosxy + sinxy) using the product rule, and

fy(x,y) = xcosxyexy +sinxy(xexy) = xexy(cosxy + sinxy) using the product rule, so

df = yexy(cosxy+ sinxy)dx+ xexy(cos xy+ sin xy)dy = exy(cosxy + sinxy)(ydx+ xdy).
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