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Quiz 11: Chain rules for functions of two variables

Last unanswered question  Question  Next unanswered question
 

Question 1

 
 
If z = sinxcosy, x = πt and y = √t then dz dt is
a) πcos(πt)cos(√t)- -1√--sin(πt)sin(√t) 2 t   b) - - - -1√--cos(πt)cos(√ t) + πsin(πt)sin(√ t) 2 t
c) √ - 1 √- π sin(πt)cos( t)- -√-cos(πt) sin( t) 2 t   d) 1 π - 2√t-
e) -1-- π sinx cosy- 2√t-cosxsiny

 

Your answer is correct.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 2

 
 
If z = lny(2x + y), x = sint and y = cost then dz dt is
a) -2cost 2sint(x+-y)- -2x + y + y(2x+ y)   b) - 2y2cost- 2-sint(x+-y) 2x+ y y(2x + y)
c) 2(y2 --x2---xy) y(2x+ y)   d) -2cost -sint-- 2x + y - 2x+ y
e) None of the above.

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
z = ln y(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 = - sint = - x dt which makes it easy to write dz -- dt in terms of x and y only.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 3

 
 
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

 

Your answer is correct
We have
dz -∂zdx- ∂z-dy dt = ∂x dt + ∂y dt.

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

Not correct. You may try again.
Recall that dz ∂z- ′ ∂z-′ dt = ∂x x(t)+ ∂yy (t) .
 

Question 4

 
 
Let z = x- y , x = set and y = 1 +se-t . Which of the following alternatives are equal to ∂z- ∂t?
a) t se-x(1- 1) y y   b) x-+ x(1-+2y) y y
c) x-- x(1-2y) y y   d) set - 2s2 -------t-2 (1 +se )
e) set + 2s2 -------t2- (1 + se )

 

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- = -se-t = 1 -y it is possible to write ∂z ∂t- in terms of x and y only as well as s and t only.
Your answers are correct
  1. False.
  2. False.
  3. True.
  4. False.
  5. True. Since ∂x -∂t = set = x and ∂y ∂t- = -se-t = 1 -y it is possible to write ∂z ∂t- in terms of x and y only as well as s and t only.
 

Question 5

 
 
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-= 3 ∂s and ∂y-= 4 ∂s at s = t = 0. What is the value of ∂z- ∂s at s = t = 0.

 

Your answer is correct
We have that

∂z-= -∂z∂x-+ ∂z-∂y-= exsiny∂x-+ excosy∂y-. ∂s ∂x ∂s ∂y ∂s ∂s ∂s

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

∂z|| 0 0 ∂s|(s,t)=(0,0) = e ⋅0⋅3 + e ⋅1⋅4 = 0

Not correct. You may try again.
Recall that ∂z ∂z ∂x ∂z ∂y ∂s-= ∂x--∂s + ∂y-∂s
 

Question 6

 
 
A function y = f(x) is defined implicitly by xcosy + y cosx = 1. Find dy- dx .
a) -cosy---ysin-x cosx - xsin y   b) -y-sinx---cos-y - xsiny+ cosx
c) 1 - cosy+ ysinx --cosx--x-siny--   d) 1---ysin-x+-cosy cosx- x siny
e) -dy dx is not defined because cosx - xcosy can be zero.

 

Not correct. Choice (a) is false.
Your answer is correct.
Let g(x,y) = xcosy+ y cosx - 1 = 0 then

dy- gx cosy--y-sinx- dx = - gy = - cosx - x siny .
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 7

 
 
( Harder ) A function z = f(x,y) is defined implicitly by ln(x + yz) = 1 + xy2z3. Find ∂z- ∂y.
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)

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
Not correct. Choice (e) is false.