3.

You see that each step above is just a simple row operation. So the solution is

  1. We’re going to use a substitution to solve this integral. Let’s let , and

No we substitute and take the integral:

Now substitute back in:

5.

The primary rule here is the chain rule. When you take a derivative of an expression that’s a function of another expression, you treat it as a single variable, take it’s derivative and then multiply by the derivative of the stuff inside of it. For example, let’s first find :

V is a function of x and of U (which is also a function of x), so we’ll have to use both the product rule and the chain rule.

The product rule says that if you have the derivative of the product of two factors that you take the derivative as follows. The derivative of the product is equal to

the derivative of the first term x the second + derivative of the second x the first.

So, for example, if we wanted the derivative of the product, , then we see that the first factor is and the second factor is . And using the product we get:

.

The chain rule tells us how to handle something complex, like . We can’t handle that by itself, but we know how to take the derivative of and of . That’s really all we need, if we have the chain rule. The rule says that you treat, in this case, the stuff under the radical as one thing. Then you take the derivative of the outermost function. Then you multiply that by the derivative of the stuff that was on the inside. So, for example, to take the derivate of , just treat the as something else. Let’s call it u. So, we get .

But the chain rule says that since u is a function of x, we need to multiply this expression by the derivative of u with respect to x, so we have:

In that last step, I substituted back in for u.

So you see, that’s all we need for this problem:

We do the same in the next case:

And to find the last one, we just need to take the second partial derivative of what we found in the first part of this:

To find the next part, let’s just try substituting from the results that we’ve just found:

Which is what we wanted to show.

6.So a stationary point just means a minimum, maximum, or an inflection point. With a function of one variable, you just take the first derivative to find out where it is equal to 0, and then the sign of the second derivative tells you if it’s a max, min, or saddle point.

In the case of a function of 2 variables, we need the place where both partial derivatives are equal to 0. So first, we find the partial derivatives:

and

When you take the partial derivative with respect to x, say, you just treat the y as if it were a constant. And the same for x when you take the partial derivative when respect to y.

Now, we have two equations with two unknowns. We set them both equal to 0 and solve for x and y.

Now substituting back into the second equation, we get:

But keep in mind that because it’s in the denominator in the initial function. So we have , and we substitute back into the first equation to get:

So we see that there is one stationary point at .

Now we find the second partial derivatives (there are only three because ).

, , and

So plugging in our point , we get:

, , and

Now, since the is negative while the determinate of the Hessian matrix is positive, we see that we have a maximum at this point.

See the following webpage for a clear and constructive discussion of determining if the point is a minimum or a maximum or a saddle point.

  1. I suggest that we use a simple substitution for this problem. Let’s let and let’s substitute into our function:

Now, we just take the derivative and find the critical points. We’ll use the product rule in this case:

So, . Now we need to determine which of these points is a maximum. We do this with the second derivative test:

Plugging in, we find that the second derivative is negative only for the point , so that’s a maximum.

So we know that , so the point that maximizes this function according to the constraints is .

8a.

So, let’s just set the equations that are supposed to be equal as being equal to one another. Hence, we get:

So, we convert that to the following matrix, and put into row echelon form:

So the solution is

8b.

The key to the first one is getting the trig functions figured out.

Remember that and that . Once we know these, we use the following substitution:

Note that we need to rewrite the bounds of the integral in terms of u as well:

So now we rewrite the integral:

For the second one we’ve just got to integrate by parts. That means we need to find a v and a u in our integral:

.

The integration by parts formula says that . The trick is to pick them in a smart way so that eventually, the integral part gets simpler. Since we know that , it seems smart to make and .

So, integrating by parts, we get:

.

Now we just multiply the stuff inside the remaining integral out and we’ve got it!

9c.The first order partial derivatives are straightforward to find. You just take the derivatives with respect to each variable and treat the others as if they were constants. We will also use the product rule, described above:

The hint really does help here. Let’s do what they say and set these two equal to each other and add them together:

We get:

Now we factor:

So we see that one solution to this system of equations is , which really does simplify matters.

Substituting, then, back into the two partials, we find that:

So the solutions that we’re looking for are (i.e., the stationary points)

11a. All we have to do here is take some partial derivatives. We’ll need the chain rule and we work from the outside in. We’ll treat everything under the largest radical first, then the stuff in it, then the stuff under the smallest radical. We get the following:

and

And now, we just plug in to show that the expression is true: