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

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

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

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.

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).

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

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 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.

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.

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

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

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).

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.

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.

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

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

testing