# 11.4 Perpendicular Cross Sectionsap Calculus

11.4 Cross Product The dot product, takes two vectors in Rn and produces a scalar. The cross product takes two vectors in R3 and produces another vector in R3 that is per-pendicular to both input vectors and in a direction determined by the right-hand rule.' Let u = (u1;u2;u3) and v = (v1;v2;v3). Then the cross product given.

- 11.4 Perpendicular Cross Sectionsap Calculus Formulas
- 11.4 Perpendicular Cross Sections Ap Calculus Formulas

Given a smooth vector-valued function (vrttext{,}) we defined in Definition 11.2.16 that any vector parallel to (vrp(t_0)) is *tangent* to the graph of (vrt) at (t=t_0text{.}) It is often useful to consider just the *direction* of (vrp(t)) and not its magnitude. Therefore we are interested in the unit vector in the direction of (vrp(t)text{.}) This leads to a definition.

Defining the Cross Product We willintroduce a new binary operation called the cross product. U The cross product should be a way to take two vectors, and, andfindathird, that is perpendicular to b a a b b oth. For theright hand rule with a b, the first u vector is always your index finger and the second vector is always your middle finger! Calculus Notes x y 2. Find the volume of the solid whose base is bounded by ݕ ൌ ݔ ଷ, ݕ ൌ 0, and ݔ ൌ 2 with cross sections taken perpendicular to the ݕ -axis that form a square. 11.4 Perpendicular Cross Sections Calculus The base of an object is bounded by the lines ൌ െ , ൌ െ , and ൌ . Volumes of Solids of Know Cross Sections AP Calculus l. The base of a solid is the region enclosed by a circle centered a the origin with a radius of 5 inches. Find the volume of the solid if all cross sections perpendicular to the x- xis are s uares. (too 500 — 500 t 000 2. The Cross Product and Its Properties. The dot product is a multiplication of two vectors that results in a scalar. In this section, we introduce a product of two vectors that generates a third vector orthogonal to the first two.

SolutionWe apply Definition 11.4.1 to find (vec T(t)text{.})begin{align*}vec T(t) amp = frac{1}{norm{vrp(t)}}vrp(t)amp =frac{1}{sqrt{big(-3sin(t) big)^2+big(3cos(t) big)^2+ 4^2}}la -3sin(t) ,3cos(t) , 4raamp = la -frac35sin(t) ,frac35cos(t) ,frac45ra.end{align*}

We can now easily compute (vec T(0)) and (vec T(1)text{:})begin{equation*}vec T(0) = la 0,frac35,frac45ra,; vec T(1) = la -frac35sin(1) ,frac35cos(1) ,frac45ra approx la -0.505,0.324,0.8ra.end{equation*}

These are plotted in Figure 11.4.3 with their initial points at (vec r(0)) and (vec r(1)text{,}) respectively. (They look rather “short” since they are only length 1.)

The unit tangent vector (vec T(t)) always has a magnitude of 1, though it is sometimes easy to doubt that is true. We can help solidify this thought in our minds by computing (norm{vec T(1)}text{:})begin{equation*}norm{vec T(1)} approx sqrt{(-0.505)^2+0.324^2+0.8^2} = 1.000001.end{equation*}

We have rounded in our computation of (vec T(1)text{,}) so we don't get 1 exactly. We leave it to the reader to use the exact representation of (vec T(1)) to verify it has length 1.

In many ways, the previous example was “too nice.” It turned out that (vrp(t)) was always of length 5. In the next example the length of (vrp(t)) is variable, leaving us with a formula that is not as clean.

SolutionWe find (vrp(t) = la 2t-1,2t+1ratext{,}) andbegin{equation*}norm{vrp(t)} = sqrt{(2t-1)^2+(2t+1)^2} = sqrt{8t^2+2}.end{equation*}

Thereforebegin{equation*}vec T(t) = frac{1}{sqrt{8t^2+2}}la 2t-1,2t+1ra = la frac{2t-1}{sqrt{8t^2+2}},frac{2t+1}{sqrt{8t^2+2}}ra.end{equation*}

When (t=0text{,}) we have (vec T(0) = la -1/sqrt{2},1/sqrt{2}ratext{;}) when (t=1text{,}) we have (vec T(1) = la 1/sqrt{10}, 3/sqrt{10}ratext{.}) We leave it to the reader to verify each of these is a unit vector. They are plotted in Figure 11.4.5

Just as knowing the direction tangent to a path is important, knowing a direction orthogonal to a path is important. When dealing with real-valued functions, we defined the normal line at a point to the be the line through the point that was perpendicular to the tangent line at that point. We can do a similar thing with vector-valued functions. Given (vrt) in (mathbb{R}^2text{,}) we have 2 directions perpendicular to the tangent vector, as shown in Figure 11.4.6. It is good to wonder “Is one of these two directions preferable over the other?”

Given (vrt) in (mathbb{R}^3text{,}) there are infinite vectors orthogonal to the tangent vector at a given point. Again, we might wonder “Is one of these infinite choices preferable over the others? Is one of these the `right' choice?”

### 11.4 Perpendicular Cross Sectionsap Calculus Formulas

The answer in both (mathbb{R}^2) and (mathbb{R}^3) is “Yes, there is one vector that is not only preferable, it is the `right' one to choose.” Recall Theorem 11.2.26, which states that if (vrt) has constant length, then (vrt) is orthogonal to (vrp(t)) for all (ttext{.}) We know (vec T(t)text{,}) the unit tangent vector, has constant length. Therefore (vec T(t)) is orthogonal to (vec T,'(t)text{.})

We'll see that (vec T,'(t)) is more than just a convenient choice of vector that is orthogonal to (vrp(t)text{;}) rather, it is the “right” choice. Since all we care about is the direction, we define this newly found vector to be a unit vector.

SolutionIn Example 11.4.2, we found (vec T(t) = la (-3/5)sin(t) ,(3/5)cos(t) ,4/5ratext{.}) Thereforebegin{equation*}vec T,'(t) = la -frac35cos(t) ,-frac35sin(t) ,0ra text{ and } norm{vec T,'(t)} = frac35.end{equation*}

Thusbegin{equation*}vec N(t) = frac{vec T,'(t)}{3/5} = la -cos(t) ,-sin(t) ,0ra.end{equation*}

We compute (vec T(pi/2) = la -3/5,0,4/5ra) and (vec N(pi/2) = la 0,-1,0ratext{.}) These are sketched in Figure 11.4.9.

The previous example was once again “too nice.” In general, the expression for (vec T(t)) contains fractions of square roots, hence the expression of (vec T,'(t)) is very messy. We demonstrate this in the next example.

SolutionIn Example 11.4.4, we foundbegin{equation*}vec T(t) = la frac{2t-1}{sqrt{8t^2+2}},frac{2t+1}{sqrt{8t^2+2}}ra.end{equation*}

Finding (vec T,'(t)) requires two applications of the Quotient Rule:begin{align*}T,'(t) amp = la frac{sqrt{8t^2+2}(2)-(2t-1)left(frac12(8t^2+2)^{-1/2}(16t)right)}{8t^2+2},right.amp left.frac{sqrt{8t^2+2}(2)-(2t+1)left(frac12(8t^2+2)^{-1/2}(16t)right)}{8t^2+2} raamp = la frac{4 (2 t+1)}{left(8 t^2+2right)^{3/2}},frac{4(1-2 t)}{left(8 t^2+2right)^{3/2}}raend{align*}

This is not a unit vector; to find (vec N(t)text{,}) we need to divide (vec T,'(t)) by it's magnitude.begin{align*}norm{vec T,'(t)} amp = sqrt{frac{16(2t+1)^2}{(8t^2+2)^3}+frac{16(1-2t)^2}{(8t^2+2)^3}}amp = sqrt{frac{16(8t^2+2)}{(8t^2+2)^3}}amp = frac{4}{8t^2+2}.end{align*}

Finally,begin{align*}vec N(t) amp = frac1{4/(8t^2+2)}la frac{4 (2 t+1)}{left(8 t^2+2right)^{3/2}},frac{4(1-2 t)}{left(8 t^2+2right)^{3/2}}raamp = la frac{2t+1}{sqrt{8t^2+2}},-frac{2t-1}{sqrt{8t^2+2}}ra.end{align*}

Using this formula for (vec N(t)text{,}) we compute the unit tangent and normal vectors for (t=-1,0) and 1 and sketch them in Figure 11.4.11.

The final result for (vec N(t)) in Example 11.4.10 is suspiciously similar to (vec T(t)text{.}) There is a clear reason for this. If (vec u = la u_1,u_2ra) is a unit vector in (mathbb{R}^2text{,}) then the *only* unit vectors orthogonal to (vec u) are (la -u_2,u_1ra) and (la u_2,-u_1ratext{.}) Given (vec T(t)text{,}) we can quickly determine (vec N(t)) if we know which term to multiply by ((-1)text{.})

Consider again Figure 11.4.11, where we have plotted some unit tangent and normal vectors. Note how (vec N(t)) always points “inside” the curve, or to the concave side of the curve. This is not a coincidence; this is true in general. Knowing the direction that (vec r(t)) “turns” allows us to quickly find (vec N(t)text{.})

### 11.4 Perpendicular Cross Sections Ap Calculus Formulas

Let (vrt) be a position function. It is a fact (stated later in Theorem 11.4.13) that acceleration, vat, lies in the plane defined by (vec T) and (vec Ntext{.}) That is, there are scalars (a_{text{T} }) and (a_{text{N} }) such thatbegin{equation*}vat = a_{text{T} }vec T(t) + a_{text{N} }vec N(t).end{equation*}

The scalar (a_{text{T} }) measures “how much” acceleration is in the direction of travel, that is, it measures the component of acceleration that affects the speed. The scalar (a_{text{N} }) measures “how much” acceleration is perpendicular to the direction of travel, that is, it measures the component of acceleration that affects the direction of travel.

We can find (a_{text{T} }) using the orthogonal projection of (vec a(t)) onto (vec T(t)) (review Definition 10.3.17 in Section 10.3 if needed). Recalling that since (vec T(t)) is a unit vector, (vec T(t)cdotvec T(t)=1text{,}) so we havebegin{equation*}proj{a(t)}{T(t)} = frac{vec a(t)cdotvec T(t)}{vec T(t)cdotvec T(t)}vec T(t) = underbrace{big(vec a(t)cdotvec T(t)big)}_{a_{text{T} }}vec T(t).end{equation*}

Thus the amount of vat in the direction of (vec T(t)) is (a_{text{T} }=vatcdotvec T(t)text{.}) The same logic gives (a_{text{N} } = vatcdotvec N(t)text{.})

While this is a fine way of computing (a_{text{T} }text{,}) there are simpler ways of finding (a_{text{N} }) (as finding (vec N) itself can be complicated). The following theorem gives alternate formulas for (a_{text{T} }) and (a_{text{N} }text{.})

Note the second formula for (a_text{T}text{:}) (ds frac{d}{dt}Big(norm{vvt}Big)text{.}) This measures the rate of change of speed, which again is the amount of acceleration in the direction of travel.

SolutionThe previous examples give (vat = la -3cos(t) ,-3sin(t) ,0ra) andbegin{equation*}vec T(t) = la -frac35sin(t) ,frac35cos(t) ,frac45ra text{ and } vec N(t) = la -cos(t) ,-sin(t) ,0ra.end{equation*}

We can find (a_text{T}) and (a_text{N}) directly with dot products:begin{align*}a_text{T} amp = vatcdot vec T(t) = frac95cos(t) sin(t) -frac95cos(t) sin(t) +0 = 0.a_text{N} amp = vatcdot vec N(t) = 3cos^2(t) +3sin^2(t) + 0 = 3.end{align*}

Thus (vat = 0vec T(t) + 3vec N(t) = 3vec N(t)text{,}) which is clearly the case.

What is the practical interpretation of these numbers? (a_text{T} =0) means the object is moving at a constant speed, and hence all acceleration comes in the form of direction change.

SolutionThe previous examples give (vat = la 2,2ra) andbegin{equation*}vec T(t) = la frac{2t-1}{sqrt{8t^2+2}},frac{2t+1}{sqrt{8t^2+2}}ra text{ and } vec N(t) = la frac{2t+1}{sqrt{8t^2+2}},-frac{2t-1}{sqrt{8t^2+2}}ra.end{equation*}

While we can compute (a_text{N}) using (vec N(t)text{,}) we instead demonstrate using another formula from Theorem 11.4.13.begin{align*}a_text{T} amp = vatcdotvec T(t) = frac{4t-2}{sqrt{8t^2+2}}+frac{4t+2}{sqrt{8t^2+2}} = frac{8t}{sqrt{8t^2+2}}.a_text{N} amp = sqrt{norm{vat}^2-a_text{T} ^2} = sqrt{8-left(frac{8t}{sqrt{8t^2+2}}right)^2}=frac{4}{sqrt{8t^2+2}}.end{align*}

When (t=2text{,}) (ds a_text{T} = frac{16}{sqrt{34}}approx 2.74) and (ds a_text{N} = frac{4}{sqrt{34}} approx 0.69text{.}) We interpret this to mean that at (t=2text{,}) the particle is acclerating mostly by increasing speed, not by changing direction. As the path near (t=2) is relatively straight, this should make intuitive sense. Figure 11.4.16 gives a graph of the path for reference.

Contrast this with (t=0text{,}) where (a_text{T} = 0) and (a_text{N} = 4/sqrt{2}approx 2.82text{.}) Here the particle's speed is not changing and all acceleration is in the form of direction change.

SolutionUsing Key Idea 11.3.15 of Section 11.3 we form the position function of the ball:begin{equation*}vrt = la big(64cos(30^circ) big)t, -16t^2+big(64sin(30^circ) big)t+240ra,end{equation*}which we plot in Figure 11.4.18.

From this we find (vvt = la 64cos(30^circ) , -32t+64sin(30^circ) ra) and (vat = la 0,-32ratext{.}) Computing (vec T(t)) is not difficult, and with some simplification we findbegin{equation*}vec T(t) = la frac{sqrt{3}}{sqrt{t^2-2t+4}}, frac{1-t}{sqrt{t^2-2t+4}}ra.end{equation*}

With (vat) as simple as it is, finding (a_text{T}) is also simple:begin{equation*}a_text{T} = vatcdot vec T(t) = frac{32t-32}{sqrt{t^2-2t+4}}.end{equation*}

We choose to not find (vec N(t)) and find (a_text{N}) through the formula (a_text{N} = sqrt{norm{vat}^2-a_text{T} ^2,}) :begin{equation*}a_text{N} = sqrt{32^2-left(frac{32t-32}{sqrt{t^2-2t+4}}right)^2} = frac{32sqrt{3}}{sqrt{t^2-2t+4}}.end{equation*}

Figure 11.4.19 gives a table of values of (a_text{T}) and (a_text{N}text{.}) When (t=0text{,}) we see the ball's speed is decreasing; when (t=1) the speed of the ball is unchanged. This corresponds to the fact that at (t=1) the ball reaches its highest point.

After (t=1) we see that (a_text{N}) is decreasing in value. This is because as the ball falls, it's path becomes straighter and most of the acceleration is in the form of speeding up the ball, and not in changing its direction.

Our understanding of the unit tangent and normal vectors is aiding our understanding of motion. The work in Example 11.4.17 gave quantitative analysis of what we intuitively knew.

The next section provides two more important steps towards this analysis. We currently describe position only in terms of time. In everyday life, though, we often describe position in terms of distance (“The gas station is about 2 miles ahead, on the left.”). The *arc length parameter* allows us to reference position in terms of distance traveled.

We also intuitively know that some paths are straighter than others — and some are curvier than others, but we lack a measurement of “curviness.” The arc length parameter provides a way for us to compute *curvature*, a quantitative measurement of how curvy a curve is.