Differential Calculus - Rectilinear Motion

RECTILINEAR MOTION

It is a particle's motion along a straight line path. The motion of the particle was perfectly characterized by an equation of the form s=f(t), also known as the equation of motion, where t>=0 is the time and is the displacement of the particle at any given time, measured from a selected fixed point in its path known as the reference point.

The chosen reference point is typically the initial position of the particle at time zero (t = 0). The equations below can be used to calculate the particle's velocity (v) and acceleration (a) at time t.

HORIZONTAL RECTILINEAR MOTION

This is motion of the particle is along a horizontal straight line. Since displacement, velocity and acceleration are all vector quantities, the following are the sign conventions:

• If s > 0 , right beside the reference point the particle is located.
• If s < 0 , left beside the reference point, the particle is located. 
• If v > 0 , the particle is moving in an increasing direction. s (moving to the right away from the reference point) 
• If v < 0 ,the particle is moving in an decreasing direction. s (moving to the left, towards or away from the reference point)
• if v = 0, the particle is at rest at that particular time
• If a > 0 , velocity is increasing
• If a < 0 , velocity is decreasing which means the particle is decelerating 

VERTICAL RECTILINEAR MOTION

A good illustration of a vertical rectilinear motion is free-fall motion. The moving particle, which is referred to as a freely falling body, is only influenced by its weight, and air resistance is considered to be negligible, It's acceleration is due to gravity.

The first two values are rounded-off to the nearest integer for computational convenience. For s and v , the following sign convention applies:

• If s > 0 , the particle is above the reference point
• if s < 0, the particle is below the reference point
• if v < 0, the particle is moving in upward direction
• if v > 0, the particle is moving in downward direction

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Problem Solving of Rectilinear Equation

$\mathbf{T}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{p}\mathbf{a}\mathbf{t}\mathbf{h}\mathbf{ }\mathbf{t}\mathbf{a}\mathbf{k}\mathbf{e}\mathbf{n}\mathbf{ }\mathbf{b}\mathbf{y}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{p}\mathbf{a}\mathbf{r}\mathbf{t}\mathbf{i}\mathbf{c}\mathbf{l}\mathbf{e}\mathbf{ }\mathbf{i}\mathbf{s}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{h}\mathbf{o}\mathbf{r}\mathbf{i}\mathbf{z}\mathbf{o}\mathbf{n}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{ }\mathbf{l}\mathbf{i}\mathbf{n}\mathbf{e}\mathbf{ }\mathbf{a}\mathbf{s}\mathbf{ }\mathbf{i}\mathbf{t}$

$$\begin{gathered} \text{ \\ moves according to the law s=16t4-76t3-7t2+12t+1.} \end{gathered}$$

$$\begin{gathered} \text{ \\ At what time will its acceleration be equal to one?} \end{gathered}$$

$$\begin{gathered} \text{ \\ What its velocity at that time? \\ } \end{gathered}$$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{1}\mathbf{:}\mathbf{ }\mathbf{G}\mathbf{e}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{D}\mathbf{e}\mathbf{r}\mathbf{i}\mathbf{v}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{s}\mathbf{=}\frac{\mathbf{1}}{\mathbf{6}}{\mathbf{t}}^{\mathbf{4}}\mathbf{-}\frac{\mathbf{7}}{\mathbf{6}}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{7}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{t}\mathbf{+}\mathbf{1}$

$$\begin{gathered} \text{ \\ to get the velocity. \\ } \end{gathered}$$

$\mathrm{v}=\frac{2}{3}{\mathrm{t}}^3-\frac{7}{2}{\mathrm{t}}^2-14\mathrm{t}+\frac{1}{2}$


$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{2}\mathbf{:}\mathbf{ }\mathbf{G}\mathbf{e}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{D}\mathbf{e}\mathbf{r}\mathbf{i}\mathbf{v}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{v}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\frac{\mathbf{7}}{\mathbf{2}}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{14}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}$

$$\begin{gathered} \text{ \\ to get the acceleration \\ } \end{gathered}$$

$\mathrm{a}=2{\mathrm{t}}^2-7\mathrm{t}-14$


$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{3}\mathbf{:} \mathbf{E}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{t}\mathbf{e}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{a}\mathbf{c}\mathbf{c}\mathbf{e}\mathbf{l}\mathbf{e}\mathbf{r}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{ }\mathbf{t}\mathbf{o}\mathbf{ }\mathbf{o}\mathbf{n}\mathbf{e}$

$$\begin{gathered} \text{ \\ At what time will its acceleration be equal to one? \\ } \end{gathered}$$

$2{\mathrm{t}}^2-7\mathrm{t}-14=1$

$$\begin{gathered} \text{ \\ 2t2-7t-15=0} \end{gathered}$$

$$\begin{gathered} \text{ \\ (2t+3)(t-5)=0 \\ t=5 \\ } \end{gathered}$$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{4}\mathbf{:}\mathbf{ }\mathbf{W}\mathbf{h}\mathbf{e}\mathbf{n}\mathbf{ }\mathbf{t}\mathbf{=}\mathbf{5}\mathbf{ }\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{v}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\frac{\mathbf{7}}{\mathbf{2}}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{14}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}$

$$\begin{gathered} \text{ \\ What its velocity at that time? \\ } \end{gathered}$$

$\mathrm{a}\left(5\right)=\frac{2}{3}(5{)}^3-\frac{7}{2}(5{)}^2-14\left(5\right)+\frac{1}{2}$

$$\begin{gathered} \text{ \\ a(5)=2503-1752-1392} \end{gathered}$$

$$\begin{gathered} \text{ \\ a(5)=2503-157} \end{gathered}$$

$$\begin{gathered} \text{ \\ a(5)=-2213 \\ } \end{gathered}$$

$\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{a}\mathbf{l}\mathbf{ }\mathbf{A}\mathbf{n}\mathbf{s}\mathbf{w}\mathbf{e}\mathbf{r}\mathbf{:}\mathbf{ }\mathbf{t}\mathbf{=}\mathbf{5}\mathbf{,}\mathbf{ }\mathbf{a}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{221}}{\mathbf{3}}$

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Solving a Rectilinear Motion

$\mathbf{A}\mathbf{ }\mathbf{p}\mathbf{a}\mathbf{r}\mathbf{t}\mathbf{i}\mathbf{c}\mathbf{l}\mathbf{e}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{v}\mathbf{e}\mathbf{s}\mathbf{ }\mathbf{h}\mathbf{o}\mathbf{r}\mathbf{i}\mathbf{z}\mathbf{o}\mathbf{n}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{l}\mathbf{y}\mathbf{ }\mathbf{a}\mathbf{c}\mathbf{c}\mathbf{o}\mathbf{r}\mathbf{d}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{ }\mathbf{t}\mathbf{o}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{l}\mathbf{a}\mathbf{w}\mathbf{ }\mathbf{s}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{12}\mathbf{t}\mathbf{+}\mathbf{1.}\mathbf{ }$

$\mathbf{A}\mathbf{t}\mathbf{ }\mathbf{w}\mathbf{h}\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{v}\mathbf{e}\mathbf{l}\mathbf{o}\mathbf{c}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{ }\mathbf{i}\mathbf{s}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{p}\mathbf{a}\mathbf{r}\mathbf{t}\mathbf{i}\mathbf{c}\mathbf{l}\mathbf{e}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{v}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{ }\mathbf{w}\mathbf{h}\mathbf{e}\mathbf{n}\mathbf{ }\mathbf{t}\mathbf{=}\mathbf{1}\mathbf{?}\mathbf{ }\mathbf{w}\mathbf{h}\mathbf{e}\mathbf{n}\mathbf{ }\mathbf{w}\mathbf{i}\mathbf{l}\mathbf{l}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{p}\mathbf{a}\mathbf{r}\mathbf{t}\mathbf{i}\mathbf{c}\mathbf{l}\mathbf{e}\mathbf{ }$

$\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{e}\mathbf{ }\mathbf{t}\mathbf{o}\mathbf{ }\mathbf{r}\mathbf{e}\mathbf{s}\mathbf{t}\mathbf{ }\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{ }\mathbf{w}\mathbf{h}\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{i}\mathbf{s}\mathbf{ }\mathbf{i}\mathbf{t}\mathbf{s}\mathbf{ }\mathbf{a}\mathbf{c}\mathbf{c}\mathbf{e}\mathbf{l}\mathbf{e}\mathbf{r}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{ }\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{i}\mathbf{m}\mathbf{e}\mathbf{?}$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{1}\mathbf{:}\mathbf{ }\mathbf{G}\mathbf{e}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{d}\mathbf{e}\mathbf{r}\mathbf{i}\mathbf{v}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{s}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{12}\mathbf{t}\mathbf{+}\mathbf{1}\mathbf{ }\mathbf{t}\mathbf{o}\mathbf{ }\mathbf{g}\mathbf{e}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{v}\mathbf{e}\mathbf{l}\mathbf{o}\mathbf{c}\mathbf{i}\mathbf{t}\mathbf{y}$

$$\begin{gathered} \text{ \\ v=t2-t-12 \\ } \end{gathered}$$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{2}\mathbf{:}\mathbf{ }\mathbf{A}\mathbf{t}\mathbf{ }\mathbf{w}\mathbf{h}\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{v}\mathbf{e}\mathbf{l}\mathbf{o}\mathbf{c}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{ }\mathbf{i}\mathbf{s}\mathbf{ }\mathbf{p}\mathbf{a}\mathbf{r}\mathbf{t}\mathbf{i}\mathbf{c}\mathbf{l}\mathbf{e}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{v}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{ }\mathbf{w}\mathbf{h}\mathbf{e}\mathbf{n}\mathbf{ }\mathbf{t}\mathbf{=}\mathbf{1}\mathbf{?}$

$$\begin{gathered} \text{ \\ v=(1)2-(1)-12} \end{gathered}$$

$$\begin{gathered} \text{ \\ v=-12 \\ } \end{gathered}$$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{3}\mathbf{:}\mathbf{ }\mathbf{G}\mathbf{e}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{f}\mathbf{a}\mathbf{c}\mathbf{t}\mathbf{o}\mathbf{r}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{v}\mathbf{e}\mathbf{l}\mathbf{o}\mathbf{c}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{ }\mathbf{e}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{,}$

$$\begin{gathered} \text{ \\ When will the particle come to rest?} \end{gathered}$$

$$\begin{gathered} \text{ \\ t2-t-12=0} \end{gathered}$$

$$\begin{gathered} \text{ \\ (t-4)(t+3)=0} \end{gathered}$$

$$\begin{gathered} \text{ \\ t=4 \\ } \end{gathered}$$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{4}\mathbf{:} \mathbf{W}\mathbf{h}\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{i}\mathbf{s}\mathbf{ }\mathbf{i}\mathbf{t}\mathbf{s}\mathbf{ }\mathbf{a}\mathbf{c}\mathbf{c}\mathbf{e}\mathbf{l}\mathbf{e}\mathbf{r}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{ }\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{i}\mathbf{m}\mathbf{e}\mathbf{?}$

$$\begin{gathered} \text{ \\ Get the derivative of velocity equation to get the acceleration \\ } \end{gathered}$$

${\mathrm{t}}^2-\mathrm{t}-12$

$$\begin{gathered} \text{ \\ a=2t-1 \\ } \end{gathered}$$

$\mathrm{when} \mathrm{t}=4$

$$\begin{gathered} \text{ \\ a=2(4)-1} \end{gathered}$$

$$\begin{gathered} \text{ \\ a=7 \\ } \end{gathered}$$

$\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{a}\mathbf{l}\mathbf{ }\mathbf{A}\mathbf{n}\mathbf{s}\mathbf{w}\mathbf{e}\mathbf{r}\mathbf{:}\mathbf{ }\mathbf{v}\mathbf{=}\mathbf{-}\mathbf{12}\mathbf{,}\mathbf{ }\mathbf{t}\mathbf{=}\mathbf{4}\mathbf{,}\mathbf{ }\mathbf{a}\mathbf{=}\mathbf{7}$

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Equation of a Normal Line - Algebraic Functions

$\mathit{y}\mathbf{=}\frac{{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}}{\mathit{x}\mathbf{+}\mathbf{1}}\mathbf{ }\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{p}\mathbf{o}\mathbf{i}\mathbf{n}\mathbf{t}\mathbf{ }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{-}\mathbf{4}\mathbf{)}\mathbf{ }$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{1}\mathbf{:}\mathbf{ }\mathbf{G}\mathbf{e}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{D}\mathbf{e}\mathbf{r}\mathbf{i}\mathbf{v}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{ }\mathbf{b}\mathbf{y}\mathbf{ }\mathbf{A}\mathbf{p}\mathbf{p}\mathbf{l}\mathbf{y}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{ }\mathbf{Q}\mathbf{u}\mathbf{o}\mathbf{t}\mathbf{i}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{ }\mathbf{R}\mathbf{u}\mathbf{l}\mathbf{e}$

$$\begin{gathered} \text{ \\ ddx(uv)=vdudx-udvdxv2} \end{gathered}$$

$\mathrm{Given}:$

$$\begin{gathered} \text{ \\ u=x2-4} \end{gathered}$$

$$\begin{gathered} \text{ \\ dudx=2x} \end{gathered}$$

$$\begin{gathered} \text{ \\ v=x+1} \end{gathered}$$

$$\begin{gathered} \text{ \\ dvdx=1 \\ } \end{gathered}$$

$\frac{dy}{\mathit{dx}}\mathit{=}\frac{x\mathit{+}\mathit{1}\mathit{\left(}\mathit{2}x\mathit{\right)}\mathit{-}{x}^{\mathit{2}}\mathit{-}\mathit{4}\mathit{\left(}\mathit{1}\mathit{\right)}}{\mathit{(}x\mathit{+}\mathit{1}{\mathit{)}}^{\mathit{2}}}$

$$\begin{gathered} \text{ \\ dydx=2x2+2x-x2-4(x+1)2} \end{gathered}$$

$$\begin{gathered} \text{ \\ dydx=x2+2x+4(x+1)2 \\ } \end{gathered}$$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{2}\mathbf{:}\mathbf{ }\mathbf{G}\mathbf{e}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{S}\mathbf{l}\mathbf{o}\mathbf{p}\mathbf{e}\mathbf{ }\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{p}\mathbf{o}\mathbf{i}\mathbf{n}\mathbf{t}\mathbf{ }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{-}\mathbf{4}\mathbf{)}$

$$\begin{gathered} \text{ \\ m=02+2(0)+4(0+1)2} \end{gathered}$$

$$\begin{gathered} \text{ \\ m=4 \\ } \end{gathered}$$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{3}\mathbf{:}\mathbf{ }\mathbf{A}\mathbf{p}\mathbf{p}\mathbf{l}\mathbf{y}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{e}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{N}\mathbf{o}\mathbf{r}\mathbf{m}\mathbf{a}\mathbf{l}\mathbf{ }\mathbf{L}\mathbf{i}\mathbf{n}\mathbf{e} \mathrm{at} \mathrm{point} (0, -4)$

$$\begin{gathered} \text{ \\ y-y1=-1m(x-x1)} \end{gathered}$$

$$\begin{gathered} \text{ \\ y+4=-14(x-0)-4} \end{gathered}$$

$$\begin{gathered} \text{ \\ -4y-16=x} \end{gathered}$$

$$\begin{gathered} \text{ \\ x+4y+16=0 \\ } \end{gathered}$$

$\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{a}\mathbf{l}\mathbf{ }\mathbf{A}\mathbf{n}\mathbf{s}\mathbf{w}\mathbf{e}\mathbf{r}\mathbf{:}\mathit{ }\mathit{x}\mathit{+}\mathit{4}\mathit{y}\mathit{+}\mathit{16}\mathit{=}\mathit{0}$

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Normal Line Equation to the curves at the given point of tangency

$\mathit{y}\mathbf{=}\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\mathbf{ }\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{p}\mathbf{o}\mathbf{i}\mathbf{n}\mathbf{t}\mathbf{ }\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{ }\mathbf{9}\mathbf{)}\mathbf{ }$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p} 1: \mathbf{G}\mathbf{e}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{D}\mathbf{e}\mathbf{r}\mathbf{i}\mathbf{v}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{ }\mathbf{b}\mathbf{y}\mathbf{ }\mathbf{A}\mathbf{p}\mathbf{p}\mathbf{l}\mathbf{y}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{G}\mathbf{e}\mathbf{n}\mathbf{e}\mathbf{r}\mathbf{a}\mathbf{l} \mathbf{P}\mathbf{o}\mathbf{w}\mathbf{e}\mathbf{r}\mathbf{ }\mathbf{R}\mathbf{u}\mathbf{l}\mathbf{e}$

$$\begin{gathered} \text{ \\ ddx(cun)=cnun-1dudx} \end{gathered}$$

$$\begin{gathered} \text{ \\ dydx=2x2-12-12x} \end{gathered}$$

$$\begin{gathered} \text{ \\ dydx=4xx2-1 \\ } \end{gathered}$$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{2}\mathbf{:}\mathbf{ }\mathbf{G}\mathbf{e}\mathbf{t}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{s}\mathbf{l}\mathbf{o}\mathbf{p}\mathbf{e}\mathbf{ }\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{p}\mathbf{o}\mathbf{i}\mathbf{n}\mathbf{t}\mathbf{ }\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{ }\mathbf{9}\mathbf{)}$

$$\begin{gathered} \text{ \\ m=4(-2)-22-1} \end{gathered}$$

$$\begin{gathered} \text{ \\ m=-8(3)} \end{gathered}$$

$$\begin{gathered} \text{ \\ m=-24 \\ } \end{gathered}$$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{3}\mathbf{:}\mathbf{ }\mathbf{A}\mathbf{p}\mathbf{p}\mathbf{l}\mathbf{y}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{E}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{N}\mathbf{o}\mathbf{r}\mathbf{m}\mathbf{a}\mathbf{l}\mathbf{ }\mathbf{L}\mathbf{i}\mathbf{n}\mathbf{e}\mathbf{ }\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{p}\mathbf{o}\mathbf{i}\mathbf{n}\mathbf{t}\mathbf{ }\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{ }\mathbf{9}\mathbf{)}$

$$\begin{gathered} \text{ \\ y-y1=-1m(x-x1)} \end{gathered}$$

$$\begin{gathered} \text{ \\ y-9=--124(x--2)} \end{gathered}$$

$$\begin{gathered} \text{ \\ y-9=124(x+2)24} \end{gathered}$$

$$\begin{gathered} \text{ \\ 24} \end{gathered}$$y-216=x+2

24y-216-x-2=0

$$\begin{gathered} \text{ \\ x-24y+218=0 \\ } \end{gathered}$$

$\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{a}\mathbf{l}\mathbf{ }\mathbf{A}\mathbf{n}\mathbf{s}\mathbf{w}\mathbf{e}\mathbf{r}\mathbf{:}\mathbf{ }\mathit{x}\mathbf{-}\mathbf{24}\mathit{y}\mathbf{+}\mathbf{218}\mathbf{=}\mathbf{0}$

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Tangent Line - Problem Solving

$\mathit{y}\mathbf{=}\sqrt{{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{25}}\mathbf{ }\mathit{a}\mathit{t}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{p}\mathit{o}\mathit{i}\mathit{n}\mathit{t}\mathbf{ }\mathbf{(}\mathbf{13}\mathbf{,}\mathbf{ }\mathbf{12}\mathbf{)}\mathbf{ }$

$\mathit{S}\mathit{t}\mathit{e}\mathit{p}\mathbf{ }\mathbf{1}\mathbf{:}\mathbf{ }\mathit{G}\mathit{e}\mathit{t}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{d}\mathit{e}\mathit{r}\mathit{i}\mathit{v}\mathit{a}\mathit{t}\mathit{i}\mathit{v}\mathit{e}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{y}\mathbf{=}\sqrt{{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{25}}$
$Apply the Derivative of Square Root Rule$

$$\begin{gathered} \text{ \\ ddxu=12ududx} \end{gathered}$$

$$\begin{gathered} \text{ \\ dydx=12x2-25(2x)} \end{gathered}$$

$$\begin{gathered} \text{ \\ dydx=xx2-25 \\ } \end{gathered}$$

$\mathit{S}\mathit{t}\mathit{e}\mathit{p}\mathit{ }\mathit{2}\mathit{:}\mathit{ }\mathit{G}\mathit{e}\mathit{t}\mathit{ }\mathit{t}\mathit{h}\mathit{e}\mathit{ }\mathit{s}\mathit{l}\mathit{o}\mathit{p}\mathit{e}\mathit{ }\mathit{a}\mathit{t}\mathit{ }\mathit{p}\mathit{o}\mathit{i}\mathit{n}\mathit{t}\mathit{ }\mathit{(}\mathit{13}\mathit{,}\mathit{ }\mathit{12}\mathit{)}$

$$\begin{gathered} \text{ \\ m=13132-25} \end{gathered}$$

$$\begin{gathered} \text{ \\ m=1312 \\ } \end{gathered}$$

$\mathit{S}\mathit{t}\mathit{e}\mathit{p}\mathbf{ }\mathbf{3}\mathbf{:}\mathbf{ }\mathit{U}\mathbf{s}\mathbf{i}\mathbf{n}\mathit{g}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{p}\mathit{o}\mathit{i}\mathit{n}\mathit{t}\mathbf{-}\mathit{s}\mathit{l}\mathit{o}\mathit{p}\mathit{e}\mathbf{ }\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathbf{ }\mathit{a}\mathit{t}\mathbf{ }\mathit{p}\mathit{o}\mathit{i}\mathit{n}\mathit{t}\mathbf{ }\mathbf{(}\mathbf{13}\mathbf{,}\mathbf{ }\mathbf{12}\mathbf{)}$

$$\begin{gathered} \text{ \\ y-12=1312(x-13)} \end{gathered}$$

$$\begin{gathered} \text{ \\ y-12=1312(x-13)12 \\ 12} \end{gathered}$$

$$\begin{gathered} \text{ \\ y-144=13(x-13) \\ 12} \end{gathered}$$

$$\begin{gathered} \text{ \\ y-144-13x+169=0} \end{gathered}$$

$$\begin{gathered} \text{ \\ 12y-13x+25=0} \end{gathered}$$

$$\begin{gathered} \text{ \\ 13x-12y-25=0 \\ } \end{gathered}$$

$\mathit{F}\mathit{i}\mathit{n}\mathit{a}\mathit{l}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathit{w}\mathit{e}\mathit{r}\mathbf{:}\mathbf{ }\mathbf{13}\mathit{x}\mathbf{-}\mathbf{12}\mathit{y}\mathbf{-}\mathbf{25}\mathbf{=}\mathbf{0}$

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Tangent line to the given curves at the given point of tangency

$\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{3}}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{4}}\mathbf{ }\mathit{a}\mathit{t}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{p}\mathit{o}\mathit{i}\mathit{n}\mathit{t}\mathbf{ }\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{ }\mathbf{8}\mathbf{)}$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{1}\mathbf{:}\mathbf{ }\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{d}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{D}\mathbf{e}\mathbf{r}\mathbf{i}\mathbf{v}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{ }\mathbf{b}\mathbf{y}\mathbf{ }\mathbf{a}\mathbf{p}\mathbf{p}\mathbf{l}\mathbf{y}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{P}\mathbf{r}\mathbf{o}\mathbf{d}\mathbf{u}\mathbf{c}\mathbf{t}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{T}\mathbf{w}\mathbf{o}\mathbf{ }\mathbf{F}\mathbf{a}\mathbf{c}\mathbf{t}\mathbf{o}\mathbf{r}\mathbf{s}\mathbf{ }\mathbf{R}\mathbf{u}\mathbf{l}\mathbf{e}$

$$\begin{gathered} \text{ \\ ddx(uv)=udvdx+vdudx} \end{gathered}$$

$$\begin{gathered} \text{ \\ dudx=3x2} \end{gathered}$$

$$\begin{gathered} \text{ \\ dvdx=4(x-1)3} \end{gathered}$$

$$\begin{gathered} \text{ \\ ddxuv=4x3x-13+3x2x-14 \\ } \end{gathered}$$

$\mathbf{S}\mathbf{t}\mathbf{e}\mathbf{p}\mathbf{ }\mathbf{2}\mathbf{:}\mathbf{ }\mathbf{U}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{p}\mathbf{o}\mathbf{i}\mathbf{n}\mathbf{t}\mathbf{s}\mathbf{l}\mathbf{o}\mathbf{p}\mathbf{e}\mathbf{ }\mathbf{f}\mathbf{o}\mathbf{r}\mathbf{m}\mathbf{ }\mathbf{a}\mathbf{t}\mathbf{ }\mathbf{p}\mathbf{o}\mathbf{i}\mathbf{n}\mathbf{t}\mathbf{ }\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{ }\mathbf{8}\mathbf{)}$

$$\begin{gathered} \text{ \\ y-8=44(x-2)} \end{gathered}$$

$$\begin{gathered} \text{ \\ y-8=44x-88} \end{gathered}$$

$$\begin{gathered} \text{ \\ y-44x-8+88=0} \end{gathered}$$

$$\begin{gathered} \text{ \\ y-44x+80=0} \end{gathered}$$

$$\begin{gathered} \text{ \\ 44x-y-80=0 \\ } \end{gathered}$$

$\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{a}\mathbf{l}\mathbf{ }\mathbf{A}\mathbf{n}\mathbf{s}\mathbf{w}\mathbf{e}\mathbf{r}\mathbf{:}\mathbf{ }\mathbf{44}\mathit{x}\mathbf{-}\mathit{y}\mathbf{-}\mathbf{80}\mathbf{=}\mathbf{0}$

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Find the equation of the tangent line and normal line

$\mathit{F}\mathit{i}\mathit{n}\mathit{d}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{e}\mathit{q}\mathit{u}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathbf{t}\mathbf{a}\mathbf{n}\mathit{g}\mathit{e}\mathit{n}\mathit{t}\mathbf{ }\mathit{l}\mathit{i}\mathit{n}\mathit{e}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{n}\mathit{o}\mathit{r}\mathit{m}\mathit{a}\mathit{l}\mathbf{ }\mathit{l}\mathit{i}\mathit{n}\mathit{e}\mathbf{ }$

$$\begin{gathered} \text{ \\ to the graph of y=4x+1 at 1, 2} \end{gathered}$$

$$\begin{gathered} \text{ \\ } \end{gathered}$$

$\mathit{S}\mathit{t}\mathit{e}\mathit{p}\mathbf{ }\mathbf{1}\mathbf{:}\mathbf{ }\mathit{U}\mathit{s}\mathit{e}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{d}\mathit{i}\mathit{f}\mathit{f}\mathit{e}\mathit{r}\mathit{e}\mathit{n}\mathit{t}\mathit{i}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}\mathbf{ }$

$$\begin{gathered} \text{ \\ ddxcu=-cu2ddxu } \end{gathered}$$

$$\begin{gathered} \text{ \\ y'=-4x+12 (1) \\ } \end{gathered}$$

$\mathit{:}\mathit{ }\mathit{A}\mathit{t}\mathit{ }\mathit{p}\mathit{o}\mathit{i}\mathit{n}\mathit{t}\mathit{ }\mathit{(}\mathit{1}\mathit{,}\mathit{ }\mathit{2}\mathit{)}$

$$\begin{gathered} \text{ \\ dydx=-41+12 =-1 \\ } \end{gathered}$$

$\mathrm{Hence}, \mathrm{slope} \mathrm{of} \mathrm{the} \mathrm{tangent} \mathrm{line} \mathrm{at} (1, 2) \mathrm{equals} -1$

$\mathrm{while} \mathrm{slope} \mathrm{of} \mathrm{the} \mathrm{normal} \mathrm{line} \mathrm{at} \mathrm{the} \mathrm{point} \mathrm{is} 1.$

$\mathit{Step\; 2:\; U}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathit{ }\mathit{t}\mathit{h}\mathit{e}\mathit{ }\mathit{p}\mathit{o}\mathit{i}\mathit{n}\mathit{t-}\mathit{s}\mathit{l}\mathit{o}\mathit{p}\mathit{e}\mathit{ }\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{ }\mathit{o}\mathit{f}\mathit{ }\mathit{t}\mathit{h}\mathit{e}\mathit{ }\mathit{e}\mathit{q}\mathit{u}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{ }\mathit{o}\mathit{f}\mathit{ }\mathit{t}\mathit{h}\mathit{e}\mathit{ }\mathit{l}\mathit{i}\mathit{n}\mathit{e}\mathit{.}$

$$\begin{gathered} \text{ \\ y-y1=m(x-x1) \\ } \end{gathered}$$

$\mathbf{E}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{g}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{ }\mathbf{l}\mathbf{i}\mathbf{n}\mathbf{e}\mathbf{:}$

$$\begin{gathered} \text{ \\ y-2=-1(x-1)} \end{gathered}$$

$$\begin{gathered} \text{ \\ y-2=-x+1} \end{gathered}$$

$$\begin{gathered} \text{ \\ x+y-3=0 \\ } \end{gathered}$$

$\mathbf{E}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{n}\mathbf{o}\mathbf{r}\mathbf{m}\mathbf{a}\mathbf{l}\mathbf{ }\mathbf{l}\mathbf{i}\mathbf{n}\mathbf{e}\mathbf{:}$

$$\begin{gathered} \text{ \\ y-2=1(x-1)} \end{gathered}$$

$$\begin{gathered} \text{ \\ y-x-1=0} \end{gathered}$$

$$\begin{gathered} \text{ \\ x-y+1=0 \\ } \end{gathered}$$

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