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using the given R= values we must recognize that this is R=[x1,x2] x [y1,y2]. set up the double integral with these values with respect to x and y. Solving this integral will give you the volume of the area of the rectangle.
I am unsure of how to do D
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Form this problem i recommend starting off drawing the rectangle, this is not necessary though.
Find the x1 and x2 values for your integral with respect to x and do the same for y. use the given function f(x,y) and solve the integral. now since we are looking for the average value of the rectangle, we use the dimensions provided to get LxH. use the reciprocal of this to find the average value
Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=3x−4y subject to the constraint ,x2+3y2=129, if such values exist. maximum = minimum =
The partial derivative of f with respect to a variable = (lambda)(partial derivative of g with respect to a variable)
you know what function is F because it is usually given to us as f(x,y,z).
you know what function is G because it is usually set equal to a constant
Find these values for X, Y, Z
plug these X, Y, Z values into the original function G =constant and solve for lambda
We take the value of Lambda that we have obtained and plug it into our original x=, y= and z= values.
This will give us the values of X, Y, Z
We take the X, Y, Z values an plug it into the function F
this will give us A Max or a Min
Usually if the value is a negative it is a minimum
usually if the value is positive it is a maximum
Let f(x,y)=1/x+2/y+3x+4y in the region R where .x,y>0. Explain why f must have a global minimum at some point in R (note that R is unbounded---how does this influence your explanation?). Then find the global minimum. minimum =
For this qyuestion you first take the aprtia lderivative of x and y, set these =0 and solve for the x= and y= values. next we find the fxx and fyy and fxy/fyx second order partial derivatives. now we plug in our x= and y= values into all of the found partial derivatives and add them up.
(a) In which direction should you move if you want the concentration to increase the fastest?
What should be done for this is taking the partial derivative with respect to x,y,z. set that up in vector form as < fx(xyz), fy(xyz), fz(xyz)> then we plug in the point (-1,1,-1).
6. Find the differential of the function w=x3sin(y5z1) dw=dx+ dy+
For this all we do is take the derivative of the function w with respect to x, y and z. we should also recognize that this is a chain rule for y and z. so the chain rule is d/dx f(g(x)) = f(g(x))(g'(x))
with respect to x: 3x^2 sin(y^5 z) dx
with respect to y: 5(x^3 y^4 z) sin(y^5 z) dy
with respect to z: x^3 y^5 sin(y^5 z) dz
Let z=g(u,v) and .u(r,s),v(r,s). How many terms are there in the expression for ?∂z/∂r? terms Answer. \(2\) 2
For the chain rule we know the formula multiplies the partial derivative with respect to a variable with the derivative of the given variable= values. so, since g(u,v) has only the two values of u and v we can conclude that there are only two expressions for this chain rule.
If ,z=f(x)+yg(x), what can we say about ?zyy? zyy=0 zyy=y zyy=zxx zyy=g(x) We cannot say anything
if we solve this second order derivative with respect to y we get 0 + g(x) for our first derivative because we are to treat it as a constant. when we take the second derivative of this with respect to y we simply get 0. This is because the derivative of any constant is 0.
The largest set on which the function f(x,y)=1/(3−x2−y2) is continuous is All of the xy-plane The interior of the circle x2+y2=3 The exterior of the circle x2+y2=3 The interior of the circle ,x2+y2=3, plus the circle All of the xy-plane except the circle
it is E because the denominator 3-x^2 -y^2 cannot be equal to 0, so x^2 + y^2 cannot be equal to 3.
Find the limit (enter 'DNE' if the limit does not exist) Hint: rationalize the denominato
after the denominator is rationalized, we get sqrt(9x^2 +2y^2+1) +1 which we then get to plug 0,0 in for so we get sqrt(1) + 1 which gives us 2
Suppose the displacement of a particle in motion at time t is given by the parametric equations x(t)=(3t−1)2,y(t)=7,z(t)=54t3−27t2. (a) Find the speed of the particle when .
A) what we havew to do to find the speed first given the particle motion is derive x(t), y(t), z(t). this gives us x'(t) = 18t-6, y'(t) = 0, z'(t) = 162x^2 - 54t. for when the speed is t=3 we plug 3 for t. this gives us 48,0,1296. noe we take the magnitude of this which is Sqrt((48)^2 + (0)^2 + (1298)^2 = 1298.0123 so the speed of this particle at t=3 is 1298.0123
2.
A) We know that acceleration is the derivative of the velocity, so to find the velocity while given acceleration we need to find the integral of each given element. then plug in 0 as it is given that the initial velocity is V(0).
B) the position vector is the integral of the velocity vector . then you also plug in 0 for t as it is given r(0).
The line parametrized by x=7,y=5t,z=6+t is parallel to the x-axis.
A) we can observe the coefficients of x,y,z which gives us the direction vector (0, 5, 1)
Based on what we know is parallel, this direction vector is not parallel to the x-axis.
5. Suppose parametric equations for the line segment between (9,−6) and (−2,5) have the form: x=a+bty=c+dt If the parametric curve starts at (9,−6) when t=0 and ends at (−2,5) at ,t=1, then find ,a, ,b, ,c, and .d. a= , b= , c= , d= .
Using the given equations, our A and C values are simply the respective x,y values for when our curve starts
To find B and D, we take the x,y values of the ending of the curve and subtract the x,y values at the beginning of the curve respectively.
Find a⋅b if |a| = 7, |b| = 7, and the angle between a and b is −π10 radians.
We know the formula a.b=|a||b| cos(theta)
we are given |a|, |b| and cos so this becomes a lot easier.
a.b = (7)(7) cos(-pi/10)
a.b = (49) (0.951)
a.b = 46.602
The angle between the vectors y=⟨1,2,−3⟩ and z=⟨−2,1,1⟩ to the nearest tenth of a degree.
To determine the angle of these two vectors, we use the provided equation 9.3.1
Lets first calculate the length of each of the provided vectors
|y| = sqrt(1^2 + 2^2 + (-3)^2) = sqrt(14), |z| = sqrt((-2)^2 + 1^2 + 1^2) = sqrt(6)
Now let's calculate the dot product of U and V: U.V = 1x-2 + 2x1 + -3x1 = -3
Now we solve for the angle. We have... theta = cos^-1 ((-3)/sqrt(6) x sqrt(14))
We get 109.106 degrees
level curve of
testing