2.4 Connections F, F' And Trigap Calculus

• 2.4 Connections F F' And Trigap Calculus Solutions
• 2.4 Connections F F' And Trigap Calculus Answers

These connections between the major ideas of calculus are important enough to be called the Fundamental Theorem of Calculus. These connections will also explain why we use the term indefinite integral for the set of all antiderivatives, and why we use such similar notations for antiderivatives and definite integrals. The derivative of a function f(x) is the function whose value at x is f′(x). The graph of a derivative of a function f(x) is related to the graph of f(x). Where (f(x) has a tangent line with. 2.4 Connections f, f' and Trig Notes 2.4 Key. Powered by Create your own unique website with customizable templates.

Whenever lim x → a f (x) is in the domain of F and the composition F (f (x)) makes sense. The reason for which (1.2) happens is in fact a more general (at least formally).

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Section 5-4 : Line Integrals of Vector Fields

In the previous two sections we looked at line integrals of functions. In this section we are going to evaluate line integrals of vector fields. We'll start with the vector field, Papa louie hot doggeria game.

[vec Fleft( {x,y,z} right) = Pleft( {x,y,z} right)vec i + Qleft( {x,y,z} right)vec j + Rleft( {x,y,z} right)vec k]

and the three-dimensional, smooth curve given by

[vec rleft( t right) = xleft( t right)vec i + yleft( t right)vec j + zleft( t right)vec khspace{0.25in}hspace{0.25in}a le t le b]

The line integral of (vec F) along (C) is

[intlimits_{C}{{vec Fcenterdot d,vec r}} = int_{{,a}}^{{,b}}{{vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right),dt}}]

Note the notation in the integral on the left side. That really is a dot product of the vector field and the differential really is a vector. Also, (vec Fleft( {vec rleft( t right)} right)) is a shorthand for,

[vec Fleft( {vec rleft( t right)} right) = vec Fleft( {xleft( t right),yleft( t right),zleft( t right)} right)]

We can also write line integrals of vector fields as a line integral with respect to arc length as follows,

[intlimits_{C}{{vec Fcenterdot d,vec r}} = intlimits_{C}{{vec Fcenterdot vec T,ds}}]

where (vec Tleft( t right)) is the unit tangent vector and is given by,

[vec Tleft( t right) = frac{{vec r'left( t right)}}{{left| {vec r'left( t right)} right|}}]

If we use our knowledge on how to compute line integrals with respect to arc length we can see that this second form is equivalent to the first form given above.

[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}} & = intlimits_{C}{{vec Fcenterdot vec T,ds}} & = int_{{,a}}^{{,b}}{{vec Fleft( {vec rleft( t right)} right)centerdot frac{{vec r'left( t right)}}{{left| {vec r'left( t right)} right|}},left| {vec r'left( t right)} right|,dt}} & = int_{{,a}}^{{,b}}{{vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right),dt}}end{align*}]

In general, we use the first form to compute these line integral as it is usually much easier to use. Let's take a look at a couple of examples.

Example 1 Evaluate ( displaystyle intlimits_{C}{{vec Fcenterdot d,vec r}}) where (vec Fleft( {x,y,z} right) = 8{x^2}y,z,vec i + 5z,vec j - 4x,y,vec k) and (C) is the curve given by (vec rleft( t right) = t,vec i + {t^2},vec j + {t^3},vec k), (0 le t le 1). Show Solution

Okay, we first need the vector field evaluated along the curve.

[vec Fleft( {vec rleft( t right)} right) = 8{t^2}left( {{t^2}} right)left( {{t^3}} right)vec i + 5{t^3},vec j - 4tleft( {{t^2}} right)vec k = 8{t^7},vec i + 5{t^3},vec j - 4{t^3},vec k]

Next, we need the derivative of the parameterization.

[vec r'left( t right) = ,vec i + 2t,vec j + 3{t^2},vec k]

Finally, let's get the dot product taken care of.

[vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right) = 8{t^7} + 10{t^4} - 12{t^5}]

The line integral is then,

[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}} & = int_{{,0}}^{{,1}}{{8{t^7} + 10{t^4} - 12{t^5},dt}} & = left. {left( {{t^8} + 2{t^5} - 2{t^6}} right)} right|_0^1 & = 1end{align*}] Example 2 Evaluate ( displaystyle intlimits_{C}{{vec Fcenterdot d,vec r}}) where (vec Fleft( {x,y,z} right) = x,z,vec i - y,z,vec k) and (C) is the line segment from (left( { - 1,2,0} right)) to (left( {3,0,1} right)). Show Solution

We'll first need the parameterization of the line segment. We saw how to get the parameterization of line segments in the first section on line integrals. We've been using the two dimensional version of this over the last couple of sections. Here is the parameterization for the line.

[begin{align*}vec rleft( t right) & = left( {1 - t} right)leftlangle { - 1,2,0} rightrangle + tleftlangle {3,0,1} rightrangle & = leftlangle {4t - 1,2 - 2t,t} rightrangle ,hspace{1.0in} 0 le t le 1end{align*}]

2.4 Connections F F' And Trigap Calculus Solutions

So, let's get the vector field evaluated along the curve.

[begin{align*}vec Fleft( {vec rleft( t right)} right) & = left( {4t - 1} right)left( t right),vec i - left( {2 - 2t} right)left( t right),vec k & = left( {4{t^2} - t} right)vec i - left( {2t - 2{t^2}} right)vec kend{align*}]

Now we need the derivative of the parameterization.

[vec r'left( t right) = leftlangle {4, - 2,1} rightrangle ]

The dot product is then,

[vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right) = 4left( {4{t^2} - t} right) - left( {2t - 2{t^2}} right) = 18{t^2} - 6t]

The line integral becomes,

[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}}& = int_{{,0}}^{{,1}}{{18{t^2} - 6t,dt}} & = left. {left( {6{t^3} - 3{t^2}} right)} right|_0^1 & = 3end{align*}]

Let's close this section out by doing one of these in general to get a nice relationship between line integrals of vector fields and line integrals with respect to (x), (y), and (z).

Given the vector field (vec Fleft( {x,y,z} right) = P,vec i + Q,vec j + R,vec k) and the curve (C) parameterized by (vec rleft( t right) = xleft( t right)vec i + yleft( t right)vec j + zleft( t right)vec k), (a le t le b) the line integral is,

[vec Fleft( {vec rleft( t right)} right) = 8{t^2}left( {{t^2}} right)left( {{t^3}} right)vec i + 5{t^3},vec j - 4tleft( {{t^2}} right)vec k = 8{t^7},vec i + 5{t^3},vec j - 4{t^3},vec k]

Next, we need the derivative of the parameterization.

[vec r'left( t right) = ,vec i + 2t,vec j + 3{t^2},vec k]

Finally, let's get the dot product taken care of.

[vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right) = 8{t^7} + 10{t^4} - 12{t^5}]

The line integral is then,

[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}} & = int_{{,0}}^{{,1}}{{8{t^7} + 10{t^4} - 12{t^5},dt}} & = left. {left( {{t^8} + 2{t^5} - 2{t^6}} right)} right|_0^1 & = 1end{align*}] Example 2 Evaluate ( displaystyle intlimits_{C}{{vec Fcenterdot d,vec r}}) where (vec Fleft( {x,y,z} right) = x,z,vec i - y,z,vec k) and (C) is the line segment from (left( { - 1,2,0} right)) to (left( {3,0,1} right)). Show Solution

We'll first need the parameterization of the line segment. We saw how to get the parameterization of line segments in the first section on line integrals. We've been using the two dimensional version of this over the last couple of sections. Here is the parameterization for the line.

[begin{align*}vec rleft( t right) & = left( {1 - t} right)leftlangle { - 1,2,0} rightrangle + tleftlangle {3,0,1} rightrangle & = leftlangle {4t - 1,2 - 2t,t} rightrangle ,hspace{1.0in} 0 le t le 1end{align*}]

2.4 Connections F F' And Trigap Calculus Solutions

So, let's get the vector field evaluated along the curve.

[begin{align*}vec Fleft( {vec rleft( t right)} right) & = left( {4t - 1} right)left( t right),vec i - left( {2 - 2t} right)left( t right),vec k & = left( {4{t^2} - t} right)vec i - left( {2t - 2{t^2}} right)vec kend{align*}]

Now we need the derivative of the parameterization.

[vec r'left( t right) = leftlangle {4, - 2,1} rightrangle ]

The dot product is then,

[vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right) = 4left( {4{t^2} - t} right) - left( {2t - 2{t^2}} right) = 18{t^2} - 6t]

The line integral becomes,

[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}}& = int_{{,0}}^{{,1}}{{18{t^2} - 6t,dt}} & = left. {left( {6{t^3} - 3{t^2}} right)} right|_0^1 & = 3end{align*}]

Let's close this section out by doing one of these in general to get a nice relationship between line integrals of vector fields and line integrals with respect to (x), (y), and (z).

Given the vector field (vec Fleft( {x,y,z} right) = P,vec i + Q,vec j + R,vec k) and the curve (C) parameterized by (vec rleft( t right) = xleft( t right)vec i + yleft( t right)vec j + zleft( t right)vec k), (a le t le b) the line integral is,

[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}} & = int_{{,a}}^{{,b}}{{left( {P,vec i + Q,vec j + R,vec k} right)centerdot left( {x',vec i + y',vec j + z',vec k} right),dt}} & = int_{{,a}}^{{,b}}{{Px' + Qy' + Rz',dt}} & = int_{{,a}}^{{,b}}{{Px',dt}} + int_{{,a}}^{{,b}}{{Qy',dt}} + int_{{,a}}^{{,b}}{{Rz',dt}} & = intlimits_{C}{{P,dx}} + intlimits_{C}{{Q,dy}} + intlimits_{C}{{R,dz}} & = intlimits_{C}{{P,dx}} + Q,dy + R,dzend{align*}]

So, we see that,

Note that this gives us another method for evaluating line integrals of vector fields.

This also allows us to say the following about reversing the direction of the path with line integrals of vector fields.

Fact

This should make some sense given that we know that this is true for line integrals with respect to (x), (y), and/or (z) and that line integrals of vector fields can be defined in terms of line integrals with respect to (x), (y), and (z).

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Section 2-4 : Limit Properties

The time has almost come for us to actually compute some limits. However, before we do that we will need some properties of limits that will make our life somewhat easier. So, let's take a look at those first. The proof of some of these properties can be found in the Proof of Various Limit Properties section of the Extras chapter.

Properties

First, we will assume that (mathop {lim }limits_{x to a} fleft( x right)) and (mathop {lim }limits_{x to a} gleft( x right)) exist and that (c) is any constant. Then,

1. (mathop {lim }limits_{x to a} left[ {cfleft( x right)} right] = cmathop {lim }limits_{x to a} fleft( x right))

   In other words, we can 'factor' a multiplicative constant out of a limit.

2. (mathop {lim }limits_{x to a} left[ {fleft( x right) pm gleft( x right)} right] = mathop {lim }limits_{x to a} fleft( x right) pm mathop {lim }limits_{x to a} gleft( x right))

   So, to take the limit of a sum or difference all we need to do is take the limit of the individual parts and then put them back together with the appropriate sign. This is also not limited to two functions. This fact will work no matter how many functions we've got separated by '+' or '-'.

3. (mathop {lim }limits_{x to a} left[ {fleft( x right)gleft( x right)} right] = mathop {lim }limits_{x to a} fleft( x right),mathop {lim }limits_{x to a} gleft( x right))

   We take the limits of products in the same way that we can take the limit of sums or differences. Just take the limit of the pieces and then put them back together. Also, as with sums or differences, this fact is not limited to just two functions.

4. (displaystyle mathop {lim }limits_{x to a} left[ {frac{{fleft( x right)}}{{gleft( x right)}}} right] = frac{{mathop {lim }limits_{x to a} fleft( x right)}}{{mathop {lim }limits_{x to a} gleft( x right)}}{rm{,}},{rm{provided }},mathop {lim }limits_{x to a} gleft( x right) ne 0)

   As noted in the statement we only need to worry about the limit in the denominator being zero when we do the limit of a quotient. If it were zero we would end up with a division by zero error and we need to avoid that.

5. (mathop {lim }limits_{x to a} {left[ {fleft( x right)} right]^n} = {left[ {mathop {lim }limits_{x to a} fleft( x right)} right]^n},{mbox{where }}n{mbox{ is any real number}})

   In this property (n) can be any real number (positive, negative, integer, fraction, irrational, zero, etc.). In the case that (n) is an integer this rule can be thought of as an extended case of 3.

   For example, consider the case of (n = )2.

   [begin{align*}mathop {lim }limits_{x to a} {left[ {fleft( x right)} right]^2} & = mathop {lim }limits_{x to a} left[ {fleft( x right)fleft( x right)} right] & = mathop {lim }limits_{x to a} fleft( x right)mathop {lim }limits_{x to a} fleft( x right)hspace{0.5in}{mbox{using property 3}} & = {left[ {mathop {lim }limits_{x to a} fleft( x right)} right]^2}end{align*}]

   The same can be done for any integer (n).

6. (mathop {lim }limits_{x to a} left[ {sqrt[n]{{fleft( x right)}}} right] = sqrt[n]{{mathop {lim }limits_{x to a} fleft( x right)}})

   This is just a special case of the previous example.

   [begin{align*}mathop {lim }limits_{x to a} left[ {sqrt[n]{{fleft( x right)}}} right] & = mathop {lim }limits_{x to a} {left[ {fleft( x right)} right]^{frac{1}{n}}} & = {left[ {mathop {lim }limits_{x to a} fleft( x right)} right]^{frac{1}{n}}} & = sqrt[n]{{mathop {lim }limits_{x to a} fleft( x right)}}end{align*}]
7. (mathop {lim }limits_{x to a} c = c,c{mbox{ is any real number}})

   In other words, the limit of a constant is just the constant. You should be able to convince yourself of this by drawing the graph of (fleft( x right) = c).

8. (mathop {lim }limits_{x to a} x = a)

   As with the last one you should be able to convince yourself of this by drawing the graph of (fleft( x right) = x).

9. (mathop {lim }limits_{x to a} {x^n} = {a^n})

   This is really just a special case of property 5 using (fleft( x right) = x).

Note that all these properties also hold for the two one-sided limits as well we just didn't write them down with one sided limits to save on space.

Let's compute a limit or two using these properties. The next couple of examples will lead us to some truly useful facts about limits that we will use on a continual basis.

Example 1 Compute the value of the following limit. [mathop {lim }limits_{x to - 2} left( {3{x^2} + 5x - 9} right)] Show Solution

This first time through we will use only the properties above to compute the limit.

First, we will use property 2 to break up the limit into three separate limits. We will then use property 1 to bring the constants out of the first two limits. Doing this gives us,

[begin{align*}mathop {lim }limits_{x to - 2} left( {3{x^2} + 5x - 9} right) & = mathop {lim }limits_{x to - 2} 3{x^2} + mathop {lim }limits_{x to - 2} 5x - mathop {lim }limits_{x to - 2} 9 & = 3mathop {lim }limits_{x to - 2} {x^2} + mathop {5lim }limits_{x to - 2} x - mathop {lim }limits_{x to - 2} 9end{align*}]

We can now use properties 7 through 9 to actually compute the limit.

[begin{align*}mathop {lim }limits_{x to - 2} left( {3{x^2} + 5x - 9} right) & = 3mathop {lim }limits_{x to - 2} {x^2} + mathop {5lim }limits_{x to - 2} x - mathop {lim }limits_{x to - 2} 9 & = 3{left( { - 2} right)^2} + 5left( { - 2} right) - 9 & = - 7end{align*}]

Now, let's notice that if we had defined

[pleft( x right) = 3{x^2} + 5x - 9]

then the proceeding example would have been,

[begin{align*}mathop {lim }limits_{x to - 2} pleft( x right) & = mathop {lim }limits_{x to - 2} left( {3{x^2} + 5x - 9} right) & = 3{left( { - 2} right)^2} + 5left( { - 2} right) - 9 & = - 7 & = pleft( { - 2} right)end{align*}]

In other words, in this case we see that the limit is the same value that we'd get by just evaluating the function at the point in question. This seems to violate one of the main concepts about limits that we've seen to this point.

2.4 Connections F F' And Trigap Calculus Answers

In the previous two sections we made a big deal about the fact that limits do not care about what is happening at the point in question. They only care about what is happening around the point. So how does the previous example fit into this since it appears to violate this main idea about limits?

Despite appearances the limit still doesn't care about what the function is doing at (x = - 2). Your company can reflect your core values. In this case the function that we've got is simply 'nice enough' so that what is happening around the point is exactly the same as what is happening at the point. Eventually we will formalize up just what is meant by 'nice enough'. At this point let's not worry too much about what 'nice enough' is. Let's just take advantage of the fact that some functions will be 'nice enough', whatever that means.

The function in the last example was a polynomial. It turns out that all polynomials are 'nice enough' so that what is happening around the point is exactly the same as what is happening at the point. This leads to the following fact.

Fact

If (p(x)) is a polynomial then,

[mathop {lim }limits_{x to a} pleft( x right) = pleft( a right)]

By the end of this section we will generalize this out considerably to most of the functions that we'll be seeing throughout this course.

Let's take a look at another example.

Example 2 Evaluate the following limit. [mathop {lim }limits_{z to 1} frac{{6 - 3z + 10{z^2}}}{{ - 2{z^4} + 7{z^3} + 1}}] Show Solution

First notice that we can use property 4 to write the limit as,

[mathop {lim }limits_{z to 1} frac{{6 - 3z + 10{z^2}}}{{ - 2{z^4} + 7{z^3} + 1}} = frac{{mathop {lim }limits_{z to 1} 6 - 3z + 10{z^2}}}{{mathop {lim }limits_{z to 1} - 2{z^4} + 7{z^3} + 1}}]

Well, actually we should be a little careful. We can do that provided the limit of the denominator isn't zero. As we will see however, it isn't in this case so we're okay.

Now, both the numerator and denominator are polynomials so we can use the fact above to compute the limits of the numerator and the denominator and hence the limit itself.

[begin{align*}mathop {lim }limits_{z to 1} frac{{6 - 3z + 10{z^2}}}{{ - 2{z^4} + 7{z^3} + 1}} & = frac{{6 - 3left( 1 right) + 10{{left( 1 right)}^2}}}{{ - 2{{left( 1 right)}^4} + 7{{left( 1 right)}^3} + 1}} & = frac{{13}}{6}end{align*}]

Notice that the limit of the denominator wasn't zero and so our use of property 4 was legitimate.

In the previous example, as with polynomials, all we really did was evaluate the function at the point in question. So, it appears that there is a fairly large class of functions for which this can be done. Let's generalize the fact from above a little.

Fact

Provided (f(x)) is 'nice enough' we have,

[mathop {lim }limits_{x to a} fleft( x right) = fleft( a right)hspace{0.5in}mathop {lim }limits_{x to {a^ - }} fleft( x right) = fleft( a right)hspace{0.5in}mathop {lim }limits_{x to {a^ + }} fleft( x right) = fleft( a right)]

Again, we will formalize up just what we mean by 'nice enough' eventually. At this point all we want to do is worry about which functions are 'nice enough'. Some functions are 'nice enough' for all (x) while others will only be 'nice enough' for certain values of (x). It will all depend on the function.

As noted in the statement, this fact also holds for the two one-sided limits as well as the normal limit.

Here is a list of some of the more common functions that are 'nice enough'.

• Polynomials are nice enough for all (x)'s.
• If (displaystyle fleft( x right) = frac{{pleft( x right)}}{{qleft( x right)}}) then (f(x)) will be nice enough provided both (p(x)) and (q(x)) are nice enough and if we don't get division by zero at the point we're evaluating at.
• (cos left( x right),sin left( x right)) are nice enough for all (x)'s
• (sec left( x right),tan left( x right)) are nice enough provided (x ne ldots , - frac{{5pi }}{2}, - frac{{3pi }}{2},frac{pi }{2},frac{{3pi }}{2},frac{{5pi }}{2}, ldots ) In other words secant and tangent are nice enough everywhere cosine isn't zero. To see why recall that these are both really rational functions and that cosine is in the denominator of both then go back up and look at the second bullet above.
• (csc left( x right),cot left( x right)) are nice enough provided (x ne ldots , - 2pi , - pi ,0,pi ,2pi , ldots ) In other words cosecant and cotangent are nice enough everywhere sine isn't zero.
• (sqrt[n]{x}) is nice enough for all (x) if (n) is odd.
• (sqrt[n]{x}) is nice enough for (x ge 0) if (n) is even. Here we require (x ge 0) to avoid having to deal with complex values.
• ({a^x},{{bf{e}}^x}) are nice enough for all (x)'s.
• ({log _b}x,ln x) are nice enough for (x > 0). Remember we can only plug positive numbers into logarithms and not zero or negative numbers.
• Any sum, difference or product of the above functions will also be nice enough. Quotients will be nice enough provided we don't get division by zero upon evaluating the limit.

The last bullet is important. This means that for any combination of these functions all we need to do is evaluate the function at the point in question, making sure that none of the restrictions are violated. This means that we can now do a large number of limits.

Example 3 Evaluate the following limit. [mathop {lim }limits_{x to 3} left( { - sqrt[5]{x} + frac{{{{bf{e}}^x}}}{{1 + ln left( x right)}} + sin left( x right)cos left( x right)} right)] Show Solution

This is a combination of several of the functions listed above and none of the restrictions are violated so all we need to do is plug in (x = 3) into the function to get the limit.

[begin{align*}mathop {lim }limits_{x to 3} left( { - sqrt[5]{x} + frac{{{{bf{e}}^x}}}{{1 + ln left( x right)}} + sin left( x right)cos left( x right)} right) & = - sqrt[5]{3} + frac{{{{bf{e}}^3}}}{{1 + ln left( 3 right)}} + sin left( 3 right)cos left( 3 right) & = {rm{8}}{rm{.185427271}}end{align*}]

Not a very pretty answer, but we can now do the limit.