MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z.
NAME: Wenhao Li
RUID:201002676
EMAIL:wl545@rutgers.edu
BELOW WRITE THE LIST OF THE ANSWERS
Answer[ 1 ]=z'=-1/(1+2*x*z)
Answer[ 2 ]=f is decreasing at the direction <2,1,-1>
Answer[ 3 ]=The requested directional derivative is 66/sqrt11
Answer[ 4 ]=It has one saddle point at (1,2)
Answer[ 5 ]=the absolute minimum is 1
Answer[ 6 ]=the LIMIT of f(x,y) is 2
Answer[ 7 ]=the curvature of the curve is 2/(sqrt1+4t^2)^3
Answer[ 8 ]=it is locate at [2sinPi,cosPi] at t=Pi
Answer[ 9 ]=the function is changing 1 right now
Answer[ 10 ]= the point of intersection of the three planes is 6
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Instructions: Download this file with its original name, mt1.txt, then rename it, in your computer
mt1FirstLast.txt
Edit it with your answers and solutions
USING COMPUTEREZE: e.g.: x times y IS x*y, x to the power y is x^y
and Email DrZcalc3@gmail.com, 80 minutes (or sooner) after starting (for most people 10:00am, Oct. 15)
Subject: mt1
with an attachment. YOU MUST NAME IT EXACTLY
mt1FirstLast.txt
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For each of the questions you MUST first figure, YOUR version, with the following convention
For i=1,2,3,4,5,6,7,8,9 , a[i]:= The i-th digit of your RUID, BUT of it is zero make it 1
Example: RUID=413200125;
a[1] = 4, a[2] = 1, a[3] = 3, a[4] = 2, a[5] = 1, a[6] = 1, a[7] = 1, a[8] = 2, a[9] = 5
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HERE WRITE THE ACTUAL a[i]
a[1]= 2 , a[2]=1 , a[3]= 1, a[4]=1 , a[5]=1 , a[6]=2 , a[7]=6 , a[8]=7 , a[9]= 6
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Problem 1:
Find dz/dy at the point (1,1,1) if z(x,y) is given implicitly by the equation
x^a[1]+y^a[2]+z^a[3]+a[5]*x*y*z^2 = 3+a[5]
With my RUID data the question is
x^2+y^1+z^1+1*x*y*z^2=3+1
Here is how I do it (Explain everything)
1+1*z'+x*2z*z'=0
1+z'+2*x*z*z'=0
z'+2*x*z*z'=-1
z'*(1+2*x*z)=-1
z'=-1/(1+2*x*z)
Ans.:z'=-y/(1+2*x*y*z)
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Problem 2:
Suppose that grad(f)(P)=. Is f increasing or decreasing at the direction ?
With my RUID data the question is
Suppose that grad(f)(P)=<2,-1,9>. Is f increasing or decreasing at the direction <2,1,-1>?
Here is how I do it (Explain everything)
sqrt2^2+1^2+(-1)^2=sqrt6
u=1/sqrt6<2,1,-1>=<2/sqrt6,1/sqrt6,-1/sqrt6>
<2,-1,9>.<2/sqrt6,1/sqrt6,-1/sqrt6>=-6/sqrt6<0
f is decreasing at the direction <2,1,-1>
Ans.:f is decreasing at the direction <2,1,-1>
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Problem 3:
Find the directional derivative of the function f(x,y,z)
x^3*a[6]+y^3*a[3]+z^3*a[8]
At the point P=(1,-1,1) in the direction pointing to Q=(1,-1,3)
With my RUID data the question is
x^3*2+y^3*1+z^3*7
Here is how I do it (Explain everything)
fx=6x^2 fy=3y^2 fz=21z^2
grad(f)=<6x^2,3y^2,21z^2>
|<1,-1,3>|=sqrt11
u=<1/sqrt11,-1/sqrt11,3/sqrt11>
grad(f)(1,-1,1)=<6,3,21>
<6,3,21>.<1/sqrt11,-1/sqrt11,3/sqrt11>=66/sqrt11
Ans.:The requested directional derivative is 66/sqrt11
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Problem 4:
Find a saddle point of the function f(x,y)=
exp(x-a[4])-(x-a[4])*exp(y-a[6])
If there is no saddle point, write in the Answers: "Does Not exist". Explain what you are doing
With my RUID data the question is
exp(x-1)-(x-1)*exp(y-2)
Here is how I do it (Explain everything)
fx=exp(x-1£©-exp(y-2)
fy=-(x-1)*exp(y-2)
fxx=exp(x-1£©
fxy=-exp(y-2)
fyy=-(x-1)*exp(y-2)
exp(x-1£©-exp(y-2)=0 -(x-1)*exp(y-2)=0
y=2 x=1
fxx£¨1,2)=1
fxy(1,2)=-1
fyy(1,2)=0
D=1*0-1^2=-1 is negative, this is a saddle point.
Ans.:It has one saddle point at (1,2)
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Problem 5:
Let f(x,y) be the function
a[4]*x + a[7]*y + a[2]
Find the ABSOLUTE MINIMUM VALUE of f(x,y) INSIDE the TRIANGLE whose VERTICES ARE
A = [a[1], a[2]], B = [a[3], a[4]], C = [a[5], a[6]]
With my RUID data the question is
f(x,y)=1*x + 6*y + 1
A = [2, 1], B = [1, 1], C = [1, 2]
Here is how I do it (Explain everything)
fx=1 fy=6 1<=x<=2 1<=y<=2
f(1,y)=x+6y+1=Fy
F'y=6
f(2,y)=2x+6y+1=Fy
F'y=6
f(x,1)=x+7=Fx
F'x=1
f(x,2)=x+13=Fx
F'x=1 the absolute minimum is 1
Ans.:
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Problem 6:
Let f(x,y) be the function
(x^2*a[4]^2-y^2*a[5]^2)/(x*a[4]-y*a[5])
Find the LIMIT of f(x,y) as (x,y) goes to the point [a[5],a[4]], or show that it does not exist
With my RUID data the question is
(x^2*1^2-y^2*1^2)/(x*1-y*1) the point [1,1]
Here is how I do it (Explain everything)
(x^2-y^2)/(x-y)
((x+y)-(x-y)£©/(x-y)
=x+y
plug in the point [1,1]
1+1=2
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Problem 7:
Find the curvature of the curve
r(t) = [a[1], a[2]*t, a[3]*t^2]
At the point (a[1],0,0)
With my RUID data the question is
r(t) = [2, 1*t, 1*t^2] At the point (2,0,0)
Here is how I do it (Explain everything)
r'(t)=<0,1,2t>
r''(t)=<0,0,2>
r'(t)*r''(t)=2i
|r'(t)*r''(t)|=2
|r'(t)|=sqrt1+4t^2
k(t)=2/(sqrt1+4t^2)^3
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Problem 8:
A particle is moving in the plane with ACCELERATION given by
[-a[1]*sin(t), -a[2]*cos(t)]
At time t=0 its position is , [0, a[2]]
and its velocity is , [a[1], 0]
Where is it located at time , t = Pi
With my RUID data the question is
[-2*sin(t), -1*cos(t)] At time t=0 its position is , [0, 1 ] its velocity is , [2, 0]
Here is how I do it (Explain everything)
a=[-2*sin(t), -1*cos(t)] v=[2cost,-sint] s=[2sint,cost]
t=Pi s=[2sinPi,cosPi]
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Problem 9:
A certain function depends on variables x and y
Right now the rate of change of the function with respect to x is, a[5]
and the rate of change of the function with respect to y is, a[7]
Both x and y depend on time
Right now the rate of change of x with respect to time is, a[1]
and the rate of change of y with respect to time is, a[9]
How fast is the function changing right now?
With my RUID data the question is
Right now the rate of change of the function with respect to x is, 1
and the rate of change of the function with respect to y is, 6
Both x and y depend on time
Right now the rate of change of x with respect to time is, 2
and the rate of change of y with respect to time is, 6
How fast is the function changing right now?
Here is how I do it (Explain everything)
<1,6> <2,6>
<2,6>-<1,6>=<1,0>
|<1,0>|=1
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Problem 10:
Find the point of intersection of the three planes
x = a[5], y = a[7], z = a[3]
With my RUID data the question is
x = 1, y = 6, z = 1
Here is how I do it (Explain everything)
1*6*1=6