MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z.
NAME: Sai Embar
RUID: 193003278
EMAIL: sai.embar@rutgers.edu
BELOW WRITE THE LIST OF THE ANSWERS
Answer[ 1 ]= -2
Answer[ 2 ]= decreasing
Answer[ 3 ]= 60/sqrt(8) (WRONG ANS., RIGHT WAY, CARELESS ERROR, -3 POINTS)
Answer[ 4 ]= (1,3)
Answer[ 5 ]= 14
Answer[ 6 ]= 2
Answer[ 7 ]= 54/729
Answer[ 8 ]= (0,-9)
Answer[ 9 ]= 17
Answer[ 10 ]= (1,2,3)
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SCORE: 97 POINTS (out of 100) GOOD JOB!
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]= 1 , a[2]= 9 , a[3]= 3 , a[4]= 1 , a[5]= 1 , a[6]= 3 , a[7]= 2 , a[8]= 7 , a[9]= 8
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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^1+y^9+z^3+1*x*y*z^2 = 3+1
Here is how I do it (Explain everything)
The shortcut formula for implicit differentiation for dz/dy = -(F_y)/(F_z)
F_y (partial derivative) = 9y^8 + x*z^2
F_z = 3*z^2 + 2*x*y*z
dz/dy = -(9y^8 + xz^2)/(3z^2+2xyz)
To find dz/dy at a specific point, plug in the point into dz/dy
dz/dy at (1,1,1) = (9*1^8 + 1*1^2)/(3*1^2 + 2*1*1*1)
= -10/5
= -2
Ans.: -2
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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)=<1,-1,4>. Is f increasing or decreasing at the direction <1,3,-1>?
Here is how I do it (Explain everything)
First we can label the direction vector v = <1,3,-1>
Next, we take the magnitude to find the unit vector of v. ||v|| = sqrt(1^2 + 3^2 + (-1)^2) = sqrt(11)
u_v (unit vector) = <1/sqrt(11), 3/sqrt(11), -1/sqrt(11)>
Next, we find the directional derivative D_u(f)(P) = grad(f)(P) *(dot product) u_v
= <1,-1,4> * (dot product) <1/sqrt(11), 3/sqrt(11), -1/sqrt(11)>
= -6/sqrt(11) -----> Negative directional derivative means f is decreasing at the direction <1,3,-1>
Ans.: decreasing
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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*3+y^3*3+z^3*7
Here is how I do it (Explain everything)
First find F_x, F_y, and F_z
F_x = 9x^2
F_y = 9y^2
F_z = 21z^2
next find grad(f) = <9x^2, 9y^2, 21z^2>
next, plug in the point P to find grad(f)(P) = <9, -9, 21>
next, find vector v = PQ = <0,-2,2>
find magnitude of v = ||v|| = sqrt(0^2 + (-2)^2 + 2^2) = sqrt(8)
unit vector of v = u_v = <0, -2/sqrt(8), 2/sqrt(8)>
Next, we find the directional derivative D_u(f)(P) = grad(f)(P) *(dot product) u_q
= <1,-9,21> * (dot product) <0, -2/sqrt(8), 2/sqrt(8)>
= 60/sqrt(8)
Ans.: 60/sqrt(8)
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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
Find a saddle point of the function f(x,y)=
exp(x-1)-(x-1)*exp(y-3)
If there is no saddle point, write in the Answers: "Does Not exist". Explain what you are doing
Here is how I do it (Explain everything)
First find F_x and F_y
F_x = exp(x-1)-(x-1)*exp(y-3)
F_y = -exp(y-3)*(x-1)
Set both equations equal to 0. We can then solve for x in the F_y equation and we find that x = 1. Plug x back into the F_x equation to solve for y, which gives y = 3.
The critical point is (1,3).
Find F_xx, F_xy, and F_yy
F_xx = exp(x-1)
F_xy = -exp(y-3)
F_yy = -exp(y-3)*(x-1)
Next, find the discriminant, D = F_xx*F_yy - (F_xy)^2
= exp(x-1)*-exp(y-3)*(x-1)-(-exp(y-3))^2
Calculate it at (1,3)
= -1
Since the discriminant D < 0, (1,3) is a saddle point
Ans.: (1,3)
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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
Let f(x,y) be the function
1*x + 2*y + 9
Find the ABSOLUTE MINIMUM VALUE of f(x,y) INSIDE the TRIANGLE whose VERTICES ARE
A = [1, 9], B = [3, 1], C = [1, 3]
Here is how I do it (Explain everything)
Since it is a linear function, the finalists are automatically the vertices
so f(1,9) = 1+2(9) + 9 = 28
f(3,1) = 3 + 2*1 + 9 = 14
f(1,3) = 1 + 2(3) + 9 = 16
The absolute minimum is 14.
Ans.: 14
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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)
Find the LIMIT of f(x,y) as (x,y) goes to the point [1,1], or show that it does not exist
Here is how I do it (Explain everything)
limit of (x^2 - y^2)/(x-y) as (x,y) goes to (1,1)
we can factor the numerator.
limit of ((x-y)(x+y))/(x-y) as (x,y) goes to (1,1)
we can cancel the (x-y) from the numerator and denominator
limit of x + y as (x,y) goes to (1,1)
Now we can just plug in the point to get 1 + 1 = 2.
Ans: 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) = [1, 9*t, 3*t^2]
At the point (1,0,0)
Here is how I do it (Explain everything)
first find r'(t) and r''(t)
r'(t) = <0,9,6t>
r''(t) = <0,0,6>
do cross product of r'(t) x r''(t)
After doing it you get r'(t) x r''(t) = 54 i.
Find magnitude of r'(t) = 9 at (1,0,0)
Find mangitude of r'(t) x r''(t) = 54
Plug into curvature formula K(t) = ||r'(t) x r''(t)||/(||r'(t)||^3)
= 54/729
Ans: 54/729
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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
[-1*sin(t), -9*cos(t)]
At time t=0 its position is , [0, 9]
and its velocity is , [1, 0]
Where is it located at time , t = Pi
Here is how I do it (Explain everything)
first find velocity r'(t) by integrating acceleration function
integrating r''(t) = r'(t) =
Next, integrate r'(t) to find r(t), the position vector
integrating r'(t) = r(t) =
plug in time = Pi
r(Pi) =
=<0,-9> = (0,-9)
Ans: (0,-9)
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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
A certain function depends on variables x and y
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, 2
Both x and y depend on time
Right now the rate of change of x with respect to time is, 1
and the rate of change of y with respect to time is, 8
How fast is the function changing right now?
Here is how I do it (Explain everything)
df/dx = 1, df/dy = 2, dx/dt = 1, dy/dt = 8
Formula to find df/dt = (df/dx)*(dx/dt) + (df/dy)*(dy/dt)
df/dt = 1*1 + 2*8 = 1 + 16
df/dt = 17
Ans: 17
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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
Find the point of intersection of the three planes
x = 1, y = 2, z = 3
Here is how I do it (Explain everything)
The traditional way of solving for the point of intersection between three planes is by setting up a system of equations
x - 1 = 0
y - 2 = 0
z - 3 = 0
Usually, this is followed up with solving for the three variables x,y, and z using the three equations,
We can solve these three equations which gives us, x = 1, y = 2, and z = 3.
In this case, the values of x,y, and z are already given to us
This means that the point of intersection is simply (1,2,3) because the values are already solved and there is no need for a system of equations to set up.
Ans: (1,2,3)