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MATH1001 Quizzes

Quiz 8: More on the partial derivative
Question 1 Questions
What is the differential dz of the function given by z = 1 1xyx2+y2 , at the point (1,2) ? Exactly one option must be correct)
a)
dz = 3 5dx dy
b)
dz = 3 25dx + 5 25dy
c)
dz = 3 5dx + 5 5dy
d)
dz = 3 25dx 5 25dy
e)
dz = 3 25dx + 5 25dy

Choice (a) is incorrect
The first partial derivatives are z x = 1+2x (1xyx2+y2)2 and z y = 12y (1xyx2+y2)2 .
Choice (b) is correct!
Choice (c) is incorrect
The first partial derivatives are z x = 1+2x (1xyx2+y2)2 and z y = 12y (1xyx2+y2)2 .
Choice (d) is incorrect
The first partial derivatives are z x = 1+2x (1xyx2+y2)2 and z y = 12y (1xyx2+y2)2 .
Choice (e) is incorrect
The first partial derivatives are z x = 1+2x (1xyx2+y2)2 and z y = 12y (1xyx2+y2)2 .
You are given that z = ln(1 + 1 + x + y). Use differentials to find the approximate change in z when (x,y) changes from (1,7) to (0.9,6.9). Exactly one option must be correct)
a)
An increase of approximately 1 12
b)
A decrease of approximately 1 24
c)
An increase of approximately 1 120
d)
An increase of approximately 1 24
e)
A decrease of approximately 1 120

Choice (a) is incorrect
The first partial derivatives are z x = z y = 1 21+x+y(1+1+x+y).
Choice (b) is incorrect
The first partial derivatives are z x = z y = 1 21+x+y(1+1+x+y).
Choice (c) is incorrect
The first partial derivatives are z x = z y = 1 21+x+y(1+1+x+y).
Choice (d) is incorrect
The first partial derivatives are z x = z y = 1 21+x+y(1+1+x+y).
Choice (e) is correct!
The first partial derivatives are z x = z y = 1 21+x+y(1+1+x+y). They equal 1 24 at (1,7), and so dz = 1 24(dx + dy) = 0.2 24 = 1 120. Thus z has decreased by approximately 1 120.
If z = sinxcosy,x = πt and y = t then dz dt is Exactly one option must be correct)
a)
πcos(πt)cos(t) 1 2tsin(πt)sin(t)
b)
1 2tcos(πt)cos(t) + πsin(πt)sin(t)
c)
πsin(πt)cos(t) 1 2tcos(πt)sin(t)
d)
π 1 2t
e)
πsinxcosy 1 2tcosxsiny

Choice (a) is correct!
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
If z = lny(2x + y),x = sint and y = cost then dz dt is Exactly one option must be correct)
a)
2cost 2x + y + 2sint(x + y) y(2x + y)
b)
2y2 cost 2x + y 2sint(x + y) y(2x + y)
c)
2(y2 x2 xy) y(2x + y)
d)
2cost 2x + y sint 2x + y
e)
None of the above.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
z = lny(2x + y) is easier to differentiate when it is rewritten as z = lny + ln(2x + y) using log laws. Here dx dt = cost = y and dy dt = sint = x which makes it easy to write dz dt in terms of x and y only.
Choice (d) is incorrect
Choice (e) is incorrect
Let z = xy2 + x3y and let x and y be functions of t with x(1) = 1 , y(1) = 2 , x(1) = 3 and y(1) = 4. The value of dz dt when t = 1 is

Correct!
We have
dz dt = z xdx dt + z ydy dt .

Now, z x = y2 + 3x2y and z y = 2xy + x3, so z x = 10 and z y = 5 when t = 1 . Therefore, when t = 0 we have dz dt = 10 3 + 5 4 = 50.

Incorrect. Please try again.
Recall that dz dt = z xx(t) + z yy(t).
Let z = x y, x = set, y = 1 + set. Which of the following alternatives are equal to z t ? (Zero or more options can be correct)
a)
setx y (1 1 y)
b)
x y + x(1 + y) y2
c)
x y x(1 y) y2
d)
set 2s2 (1 + set)2
e)
set + 2s2 (1 + set)2

There is at least one mistake.
For example, choice (a) should be False.
There is at least one mistake.
For example, choice (b) should be False.
There is at least one mistake.
For example, choice (c) should be True.
There is at least one mistake.
For example, choice (d) should be False.
There is at least one mistake.
For example, choice (e) should be True.
Since x t = set = x and y t = set = 1 y it is possible to write z t in terms of x and y only as well as s and t only.
Correct!
  1. False
  2. False
  3. True
  4. False
  5. True Since x t = set = x and y t = set = 1 y it is possible to write z t in terms of x and y only as well as s and t only.
Let z = ex siny and let x and y be functions of s and t with x(0,0) = 0, y(0,0) = 2π, x s = 3 and y s = 4 at s = t = 0. What is the value of z s at s = t = 0.

Correct!
We have that

z s = z xx s + z yy s = ex sinyx s + ex cosyy s.

So when (s,t) = (0,0) we find

z s(s,t)=(0,0) = e0 0 3 + e0 1 4 = 0

Incorrect. Please try again.
Recall that z s = z xx s + z yy s
The point 3,1 lies on the level curve f(x,y) = 10 of the function f(x,y). The partial derivatives at this point are fx(3,1) = 2 and fy(3,1) = 5. What is the slope of the tangent line to the level curve at the point (3,1) ? Exactly one option must be correct)
a)
2
b)
2 5
c)
1 2
d)
5
e)
There is not enough information to be able to decide.

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
A function y = f(x) is defined implicitly by xcosy + ycosx = 1. Find dy dx. Exactly one option must be correct)
a)
cosy ysinx cosx xsiny
b)
ysinx cosy xsiny + cosx
c)
1 cosy + ysinx cosx xsiny
d)
1 ysinx + cosy cosx xsiny
e)
dy dx is not defined because cosx xcosy can be zero.

Choice (a) is incorrect
Choice (b) is correct!
Let g(x,y) = xcosy + ycosx 1 = 0 then

dy dx = gx gy = cosy ysinx cosx xsiny.
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
A function y = f(x) is defined implicitly by ln(x + y) = 1 + xy2. Find dy dx.
Exactly one option must be correct)
a)
dy dx = 1 + xy2 1 2x2y 2xy2
b)
dy dx = 1 xy2 y2 1 + 2x2y 2xy2
c)
dy dx = 1 xy2 2y3 1 + xy2 + 2y3
d)
dy dx = 1 xy2 y3 1 2x2y 2xy2
e)
None of the above

Choice (a) is incorrect
Hint: The partial derivatives of the function g(x,y) = ln(x + y) 1 xy2 are gx(x,y) = 1 x+y y2 and gy(x,y) = 1 x+y 2xy.
Choice (b) is incorrect
Hint: The partial derivatives of the function g(x,y) = ln(x + y) 1 xy2 are gx(x,y) = 1 x+y y2 and gy(x,y) = 1 x+y 2xy.
Choice (c) is incorrect
Hint: The partial derivatives of the function g(x,y) = ln(x + y) 1 xy2 are gx(x,y) = 1 x+y y2 and gy(x,y) = 1 x+y 2xy.
Choice (d) is correct!
Since the partial derivatives of the function g(x,y) = ln(x + y) 1 xy2 are gx(x,y) = 1 x+y y2 and gy(x,y) = 1 x+y 2xy, we have dy dx = gx gy = 1 xy2 y3 1 2x2y 2xy2.
Choice (e) is incorrect
Hint: The partial derivatives of the function g(x,y) = ln(x + y) 1 xy2 are gx(x,y) = 1 x+y y2 and gy(x,y) = 1 x+y 2xy.