Mathematics College

Compute the directional derivatives of the following functions along unit vectors at the indicated points in directions parallel to the given vector. (a) rx, y) = x^y, (Xo, yo) = (e, e), d = 8i + 15j (b) f(x, y, z) = e^x + yz, (Xo, yo, Zo) = (1, 1, 1), d = (1,-4, 4) (c) f(x, y, z) = xyz, (xo, yo, Zo) = (1, 0, 1), d = (1, 0,-1)

Answers

Answer 1

The derivative of given function is D(yz+e^x)d(1,1,1) = (3*((59)^1/2)*e)/59≈1.06167045295814

We have to find the directional derivative of the given function

f(x,y,z) = yz+e^x at (x0,y0,z0)=(1,1,1) in the direction of the vector

d⃗ =(3,−5,5)

Find the gradient of the function and evaluate it at the given point:

To find the gradient of a function (which is a vector), differentiate the function with respect to each variable.

∇f= (∂f/∂x,∂f/∂y,∂f/∂z)

∂f/∂x = ∂( yz+e^x )/∂x

The derivative of a sum/difference is the sum/difference of derivatives:

∂f/∂x = ∂( yz)/∂x +∂(e^x )/∂x since the derivative of y and z with respect of x is zero.

∂f/∂x = e^x The derivative of exponential is d(e^x)/dx=e^x.

Similarly

d(yz+e^x)/dy = z and d(yz+e^x)/dz = y

∇(yz+e^x)(x0,y0,z0)=(e^x,z,y)

Finally plug in the point we get the result

∇(yz+e^x)|(x0,y0,z0)=(1,1,1)=(e,1,1)

∇(yz+ex)|(x0,y0,z0)=(1,1,1)=(e,1,1)

Now we Find the length of the vector: |d⃗ |=((3)^2+(−5)^2+(5)^2)^1/2 =(59)^1/2

To normalize the vector, divide each component by length:

d = ((3*((59)^1/2))/59 ,(−5*(59)^1/2)/59,(5*(59)^1/2)/59)

Finally, the directional derivative is the dot product of the gradient and the normalized vector:

D(yz+e^x)d(1,1,1) = (e,1,1). ((3*((59)^1/2))/59 ,(−5*(59)^1/2)/59,(5*(59)^1/2)/59)

D(yz+e^x)d(1,1,1) = (3*((59)^1/2)*e)/59≈1.06167045295814

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Related Questions

I don’t understand number 2

Answers

Answer:

x=-6.-3.0.3.6

y=6.3.0.-3.-6

Step-by-step explanation:

So in this problem you're given the table at the top for your values of X is going across their corresponding P of X values and their corresponding Q of X values. So in part A we want to find P fq of X. So first off we're gonna start when X is equal to zero. So that would mean we want to find p. of Q. of zero. So remember you start with your inside function. Well Q of zero is equal to zero. So now we need to find P. Of zero which is equal to three. Alright so next we want to find when X is one. So we have P of Q of one. Well Q of one is equal to one and I'm sorry that would leave us with PS one and PS one happens to also be one. So that'll be value for one. Next for one. X. is to so api of Q of two. Well Q two is equal to four and then P. Of four. That would leave us a pr four and pr four is equal to five. So that will be our answer for the next one. It will be five. Alright, next for three. So we have p. of Q. of three. So we have two of three which is equal to two. So now sorry that should be a P. Then we find P. F two which is also to So that'll be the answer to the 3rd 1 Next we need it for four. So we're looking for p of Q of four. Well looking according to our thing. Q4 is equal to five and then we have to find P. Of five which is zero. And lastly for five. So first we'll start with p. of Q. of five. So Q. A. Five we're told is three. So then we have to find pf three which is equal to four. So that's how you would find your first values. Okay, So now part B. We want to find sfx but no, this is the opposite way this time it's Q. A. P. Of X. So when X zero, we're going to be looking for Q. Of P. Of zero. Well no this P. Of zero is equal to three. So then we have to find Q. of three which is equal to two. It'll be our first answer. Next we'll do one. So we're gonna have Q. Of P. Of one. Well p. of one is equal to one. So then we find Q. Of one which is also one. So bring that down. All right, so now for two. First we're gonna start with Q. A. P. F. Two. Well P. F two is two. So that means we're essentially just finding Q. Of two which is four. Alright, next for three. So you're gonna have cute Of p. of 3? Well p of three is equal to four And Q four is equal to five. So I'll be your answer there. I'll do a new color because I think we're running out of room here. So next we're doing for, so I'm gonna go right here. So we're gonna have Q. A. P. Of four. Well if you four is equal to five, so that means we have to find Q. Of five which is three. And lastly for five. So we're gonna have Q. A. P. Of five. So first we'll find PFE which is zero. So that means now we need Q. Of zero which is zero. And now we filled out both of your tables.

(Graphing Proportional Relationships LC)

The table shows a proportional relationship.

x 15 9 21
y 5 3 7

Describe what the graph of the proportional relationship would look like.
A line passes through the point (0, 0) and continues through the point (15, 5).
A line passes through the point (0, 0) and continues through the point (7, 21).
A line passes through the point (0, 0) and continues through the point (5, 15).
A line passes through the point (0, 0) and continues through the point (3, 9).

Answers

The graph of the given proportional relation shows that the line passes through the point (0, 0) and continues through the point (15, 5). The relation is represented by the equation y = -0.33x. So, option A is correct.

What is the proportional relationship of the given table of proportions?

The given table shows the x and y coordinates of a line.

So, we can find the relationship among them by writing an equation from the given points as

y - y₁ = {(y₂ - y₁)/(x₂ - x₁)} × (x - x₁)

Here we have (x₁, y₁) = (15, 5); (x₂, y₂) = (9, 3)

⇒ y - 5 = (3 - 5)/(9 - 15) × (x - 15)

⇒ y - 5 = (-2)/(-6) × (x - 15)

⇒ y - 5 = 1/3 × (x - 15)

⇒ 3y - 15 = x - 15

⇒ 3y = x + 0

∴ y = 0.33x

Thus, the equation for the given proportions is y = 0.33x.

Since the line is in the form of y = mx, it passes through th point (0, 0) and continues through the point (15, 5).

So, option A is correct.

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take the van der pol equation . (a) set and . convert the above second order differential equation to a two dimensional system of differential equations:

Answers

The van der Pol equation which is a second-order differential equation is given by

d²x/dt² - μ(1 - x²)dx/dt + x = 0

Here we are asked to convert it to a two-dimensional system.

Here we need to set y = dx/dt, where y represents the system's velocity.

Hence we will rewrite it as

d²x/dt² = μ(1 - x²)y + x

Now if we set the derivatives of x and y with respect to time to be equal to different variables we get

dx/dt = y

dy/dt = μ(1 - x²)y + x

Hence we get the above systems of equations as the 2-dimensional representation of the Van der Pol equation where x and y represent the system's position and velocity respectively.

If we set μ = 1 and x = 0, then we get

dx/dt = y

dy/dt = y

Hence we get the system of differential equations representing the behavior of the van der Pol equation with μ = 1 and x = 0 in two dimensions.

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based on his past record, luke, an archer for a college archery team, has a probability of 0.90 of hitting the inner ring of the target with a shot of the arrow. assume that in one practice luke will attempt 5 shots of the arrow and that each shot is independent from the others. let the random variable x represent the number of times he hits the inner ring of the target in 5 attempts. the probability distribution of x is given in the table. P(X) 0 000001 0,00045 0.00810 0.07290 03280S 0.59049
What is the probability that the number of times Luke will hit the inner ring of the target out of the 5 attempts is less than the mean of X?
(A) 0.40951
(B) 0.50000
(C) 0.59049
(D) 0.91854
(E) 0.99144

Answers

The probability that Luke will touch the target's inner ring fewer times than the mean of X out of five attempts is 0.40951.

What is probability ?

Probability refers to possibility.

A random event's occurrence is the subject of this area of mathematics. The range of the value is 0 to 1. Mathematics has incorporated probability to forecast the likelihood of various events.

The degree to which something is likely to happen is basically what probability means.

You will understand the potential outcomes for a random experiment using this fundamental theory of probability, which is also applied to the probability distribution.

Knowing the total number of outcomes is necessary before we can calculate the likelihood that a specific event will occur.

According to our question-

P(target) equals 0.90

Let the random variable X stand in for the number of times the arrow is fired at the target's ring.

The probability distribution of X is

x: 0 1 2 3 4 5

P(x): 0.00001 0.00045 0.00810 0.07290 0.32805 0.59049

Hence, The probability that Luke will touch the target's inner ring fewer times than the mean of X out of five attempts is 0.40951.

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help me solve this please

Answers

Thus, the required trig values corresponding to the cosФ = -3/5 have been shown,

What are trigonometric equations?

These are the equation that contains trigonometric operators such as sin, cos.. etc.. In algebraic operations.

Here,
cosФ = -3/5
cosФ = -3/5 = base / hypotenuse,

perpendicular = √5² - 3² = 4

sinФ = perpendicular / hypotenus
= 4/5

Simultaneously,
cosecФ = 5/4
secФ = -5/4
tanФ = -4/3
cotФ = -3/4

Here,
sin and cosec are positive because this function as a positive positive value between π/2 and π, while else trigonometric function as a negative value for π/2 to π.

Thus, the required trig values are shown above.

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Which of these relations on the set of all people are equivalence relations? Determine the properties of an equivalence relation that the others lack.
a) {(a, b) | a and b are the same age}
b) {(a, b) | a and b have the same parents}
c) {(a, b) | a and b share a common parent}
d) {(a, b) | a and b have met} e) {(a, b) | a and b speak a common language}

Answers

Relations in the options (a), (b) and (e) are the Equivalence relations as, the relations are reflexive, symmetric and transitive all.

Given, five relations as,

a) {(a, b) | a and b are the same age}

b) {(a, b) | a and b have the same parents}

c) {(a, b) | a and b share a common parent}

d) {(a, b) | a and b have met}

e) {(a, b) | a and b speak a common language}

we have to find which of them are equivalence relations

a) as, (a , a) ∈ R ∀ a hence, Reflexive.