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MATH1901 Quizzes

Quiz 11: Chain rules for functions of two variables
Question 1 Questions
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 asz = 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 and 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
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
( Harder ) A function z = f(x,y) is defined implicitly by ln(x + yz) = 1 + xy2z3. Find z y.
Exactly one option must be correct)
a)
(1 y2z3(x + yz)) y 3xy2z2(x + yz)
b)
(y 2xyz3(x + yz)) z 3xy2z2(x + yz)
c)
y 3xy2z2(x + yz) z 2xyz3(x + yz)
d)
(z 2xyz3(x + yz)) y 3xy2z2(x + yz)
e)
z 2xyz3(x + yz) y 3xy2z2(x + yz)

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Choice (e) is incorrect