When \(\mathbf{r}\) have continuous first derivative and \(\mathbf{r}'(t)\neq\mathbf{0}~\) for all \(~t~\) in the open interval \((a,b),\)\(\,\)then \(\mathbf{r}\) is said to be a smooth function
In the case of the second derivative, \(\,\)we have
If \(\mathbf{r}\) is a differentiable vector function and \(s=u(t)\) is a differentiable scalar function, then the derivative of \(\mathbf{r}(s)\) with respect to \(t\) is
If \(~\mathbf{r}(t) = f(t)\mathbf{i} +g(t)\mathbf{j} +h(t)\mathbf{k}\,\) is a smooth function, \(\,\)then it can be shown that the length of the smooth curve traced by \(\,\mathbf{r}\,\) is given by
Thus, \(\displaystyle\,\frac{d\mathbf{v}}{dt} \cdot \mathbf{v}=0\)\(\text{ }\)or\(\text{ }\)\(\mathbf{a}(t) \cdot \mathbf{v}(t)=0~\,\) for all \(\,t\)
Centripetal Acceleration
For circular motion in the plane, described by \(\mathbf{r}(t) = r_0 \cos\omega t \,\mathbf{i} +r_0\sin\omega t \,\mathbf{j},\)\(\,\)it is evident that
\[\mathbf{r}''=-\omega^2\mathbf{r}\]
This means that the acceleration vector \(\mathbf{a}(t)=\mathbf{r}''(t)\) points in the direction opposite to that of the position vector \(\mathbf{r}(t)\). \(\,\) We then say \(~\mathbf{a}(t)\) is centripetal acceleration
10.3 Curvature and Components of Acceleration
We know that \(\mathbf{r}'(t)\) is a tangent vector to the curve \(C\), and consequently
\[
\begin{aligned}
\frac{d\mathbf{r}}{dt} &=\frac{d\mathbf{r}}{ds}
\frac{ds}{dt}\;\text{ and so }\\[5pt]
\frac{d\mathbf{r}}{ds} =
\frac{\displaystyle\frac{d\mathbf{r}}{dt}} {\displaystyle\frac{ds}{dt}}
&=\frac{\mathbf{r}'(t)}{\left\| \mathbf{r}'(t) \right\|}
= \mathbf{T}(t)
\end{aligned}\]
Let \(\,\mathbf{r}(t)\,\) be a vector function defining a smooth curve \(C\). \(\,\)If \(s\) is the arc length parameter and \(\mathbf{T}\) is the unit tangent vector, \(\,\)then the curvature of \(C\) at a point is
In other words, \(\,\)curvature is given by \(\,\kappa(t)=\frac{\displaystyle\left\|\mathbf{T}'(t)\right\|}{\displaystyle\left\|\mathbf{r}'(t)\right\|}\)
Tangential and Normal Components of Acceleration
The velocity of the particle on \(C\) is \(~\mathbf{v}(t)=\mathbf{r}'(t)\), \(\,\)whereas its speed is \(\displaystyle\frac{ds}{dt}=v=\| \mathbf{v}(t) \|\). \(\,\) Thus, \[\mathbf{v}(t)=v\mathbf{T}\]
Differentiating this last expression with respect to \(t\) gives acceleration
It follows from the differentiation of \(\mathbf{T}\cdot\mathbf{T}=1\,\) that \(\displaystyle\,\mathbf{T}\cdot \frac{d\mathbf{T}}{dt}=0\). \(\,\)If \(\displaystyle\,\left\|\frac{d\mathbf{T}}{dt}\right\| \neq 0\), \(\,\)the vector
is a unit normal to the curve \(C\). \(\,\)The vector \(\mathbf{N}\) is also called the principal normal
Since curvature is \(\displaystyle\,\kappa = \frac{\left\|\frac{d\mathbf{T}}{dt} \right\|}{v}\), \(\,\) it follows that \(\displaystyle\frac{d\mathbf{T}}{dt}=\kappa v \mathbf{N}\). \(\,\)Thus
If \(z=f(u,v)\) is differentiable and \(u=g(t)\,\) and \(v=h(t)\) are differentiable functions of a single variable \(t\), \(\,\)the ordinary derivative \(\displaystyle\frac{dz}{dt}\) is
The gradient vector \(\nabla f\) points in the direction in which \(f\) increases most rapidly, \(\,\)whereas \(-\nabla f\) points in the direction of the most decrease of \(\,f\)
\(~\)
10.6 Tangent Planes and Normal Lines
\(\nabla f\) is orthogonal to the level curve at \(P\)
The derivative of \(\,f\left(x(t), \,y(t)\right)=c\,\) with respect to \(\,t\,\) is
Stated another way, \(\,\)if \(C_1\) and \(C_2\) are two different paths between the same points \(A\) and \(B\), \(\,\)then we expect that \[\int_{C_1} \mathbf{F}\cdot \,d\mathbf{r} \neq \int_{C_2} \mathbf{F}\cdot \,d\mathbf{r}\]
Path Independence
A line integral \(\displaystyle\int_C \mathbf{f}\cdot d\mathbf{r}~\) is independence of the path if
for any two paths \(C_1\) and \(C_2\) between \(A\) and \(B\)
A vector function \(\mathbf{f}\) in 2- or 3-space is said to be conservative if \(\,\mathbf{f}\) can be written as the gradient of a scalar function \(\phi\). \(\,\)The function \(\phi\) is called a potential function for \(\,\mathbf{f}\)
In other words, \(\,\mathbf{f}\) is conservative if there exists a function \(\phi\) such that \(\,\mathbf{f}=\nabla \phi\). \(\,\)A conservative vector field is also called a gradient vector field
Fundamental Theorem
If \(~\mathbf{f}(x,y,z)~\) is a conservative vector field in \(R\) and \(\phi\) is a potential function for \(\,\mathbf{f}\), \(\,\)then
is a conservative vector field in an open region of 3-space, and that \(P\), \(Q\), and \(R\) are continuous and have continuous first partial derivatives in that region
Then \(\,\mathbf{f}=\nabla \phi\,\) and \(\,\nabla\times\mathbf{f}=\nabla\times\nabla\phi=\mathbf{0}\), \(\,\)that is
Suppose that \(C\) is a piecewise-smooth simple closed curve bounding a simply connected region \(R\)
If \(\,P\), \(Q\), \(\displaystyle\frac{\partial P}{\partial y}\), and \(\displaystyle\frac{\partial Q}{\partial x}\) are continuous on \(R\), \(\,\)then
\[
\begin{aligned}
\iint_R \left( \frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y} \right)\, dA &=\iint_{R_1} \left( \frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y} \right)\, dA
+\iint_{R_2} \left( \frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y} \right)\, dA\\
&=\oint_{C_1} P \,dx +Q \,dy +\oint_{C_2} P \,dx +Q \,dy =\oint_C P \,dx +Q \,dy
\end{aligned}\]
\(~\)
Example\(\,\) Evaluate \(\displaystyle\oint_{C} \frac{-y}{x^2 +y^2}\, dx + \frac{x}{x^2 + y^2}\,dy\), \(\,\) where \(C=C_1 \cup C_2\) is the boundary of the shaded region \(R\)
\[
\begin{aligned}
\mathbf{u} &= \Delta x_k \mathbf{i} +f_x(x_k,y_k)\Delta x_k \mathbf{k}\\
\mathbf{v} &= \Delta y_k \mathbf{j} +f_y(x_k,y_k)\Delta y_k \mathbf{k}\\
& \Downarrow \\
\mathbf{u} \times \mathbf{v} &={\scriptsize\left[-f_x(x_k,y_k)\mathbf{i} -f_y(x_k,y_k)\mathbf{j}+\mathbf{k}\right] \Delta x_k \Delta y_k}\\
& \Downarrow \\
\Delta S_k &= {\sqrt{1 +\left[\,f_x(x_k,y_k)\right]^2 +\left[\,f_y(x_k,y_k)\right]^2}\Delta A_k }
\end{aligned}\] Then the area of the surface over \(R\) is given by \[ { S=\iint_R \sqrt{1+\left[\,f_x(x,y)\right]^2 +\left[\,f_y(x,y)\right]^2}\,dA} \]
Surface Integral
Let \(\,G\,\) be a function of three variables defined over a region of space containing the surface \(S\). \(\,\)Then the surface integral of \(G\) over \(S\) is given by
Let \(S\) be a piecewise-smooth orientable surface bounded by a piecewise-smooth simple closed curve \(C\)
Let \(~\mathbf{f}(x,y,z)=P(x,y,z)\mathbf{i} +Q(x,y,z)\mathbf{j} +R(x,y,z)\mathbf{k} \,\) be a vector field for which \(P\), \(Q\), and \(R\) are continuous and have continuous first partial derivatives in a region of 3-space containing \(S\). \(~\)If \(C\) is traversed in the positive direction, then
where \(\displaystyle\,\mathbf{n}=\frac{dy}{ds}\mathbf{i} -\frac{dx}{ds}\mathbf{j}\)
Divergence Theorem
Let \(D\,\) be a closed and bounded region in 3-space with a piecewise-smooth boundary \(S\) that is oriented outward
Let \(\;\mathbf{f}(x,y,z)=P(x,y,z)\,\mathbf{i}+Q(x,y,z)\,\mathbf{j} +R(x,y,z)\,\mathbf{k}~\) be a vector field for which \(P\), \(Q,\) and \(R\) are continuous first partial derivatives in a region of 3-space containing \(D\). \(\,\) Then
2.\(~\) Find an equation of the plane containing \((1, 7, -1)\) that is perpendicular to the line of intersection of \(-x +y -8z=4~\) and \(~3x -y +2z=0\)
Solution
Step 1:\(~\) Find direction vector of the line of intersection
The line of intersection of two planes lies in both planes, and its direction vector is perpendicular to both normals of the planes
So, compute the cross product of the normal vectors of the two planes
Normal to \(\pi_1\): \(~\mathbf{n}_1 = \langle -1, 1, -8 \rangle\)
Normal to \(\pi_2\): \(~\mathbf{n}_2 = \langle 3, -1, 2 \rangle\)
4.\(~\) Find the work done by the given force \(~\mathbf{f}=(x-y)\mathbf{i} +(x+y)\mathbf{j}~\) around the closed curve in the following
Solution
The image shows a region bounded between two quarter circles:
Outer curve: \(x^2 + y^2 = 4\) (radius 2)
Inner curve: \(x^2 + y^2 = 1\) (radius 1)
The region is the quarter annulus in the first quadrant, swept from the positive \(x\)-axis to the positive \(y\)-axis, and the curve is oriented counterclockwise, forming a positively oriented closed curve \(C\)
So \(v = 1\) is the upper limit (since \(v > 0\)), and the region \(S\) is bounded by:
\(0 \le v \le 1\)
\(v \le u \le \frac{2}{1 + v}\)
Step 4:\(~\) Change of variables in the integral
We rewrite the integral in terms of \(u\), \(v\):
\[\iint_S \frac{1}{\sqrt{(u + v +1)^2}} \cdot |J| \, du\, dv\]
So the integrand becomes simply 1. Then the integral becomes the area of the region \(S\):
\[\iint_S 1 \, du\, dv = \int_{v=0}^{1} \int_{u=v}^{\frac{2}{1 + v}} 1 \, du\, dv = \int_0^1 \left( \frac{2}{1 + v} - v \right) dv =2 \ln 2 - \frac{1}{2}\]
\(~\)
6.\(~\) Evaluate the integral
\[ \iint_R (x^2 +y^2) \sqrt[3]{x^2 -y^2} \,dA \]
where \(R\) is the region bounded by the graphs of \(x=0\), \(x=1\), \(y=0\), and \(y=1\) by means of the change of variables \(u=2xy\), \(v=x^2 -y^2\)
Solution
Step 1:\(~\) Compute the Jacobian
We need the Jacobian determinant \(J = \left| \frac{\partial(x, y)}{\partial(u, v)} \right|\), but we are given \(u = 2xy\), \(v = x^2 - y^2\), so we compute \(\frac{\partial(u, v)}{\partial(x, y)}\) and then take the reciprocal for the change of variables
