# 1.2 Calculus of parametric curves  (Page 2/6)

 Page 2 / 6

Calculate the derivative $dy\text{/}dx$ for the plane curve defined by the equations

$x\left(t\right)={t}^{2}-4t,\phantom{\rule{1em}{0ex}}y\left(t\right)=2{t}^{3}-6t,\phantom{\rule{1em}{0ex}}-2\le t\le 3$

and locate any critical points on its graph.

${x}^{\prime }\left(t\right)=2t-4$ and ${y}^{\prime }\left(t\right)=6{t}^{2}-6,$ so $\frac{dy}{dx}=\frac{6{t}^{2}-6}{2t-4}=\frac{3{t}^{2}-3}{t-2}.$
This expression is undefined when $t=2$ and equal to zero when $t=±1.$

## Finding a tangent line

Find the equation of the tangent line to the curve defined by the equations

$x\left(t\right)={t}^{2}-3,\phantom{\rule{1em}{0ex}}y\left(t\right)=2t-1,\phantom{\rule{1em}{0ex}}-3\le t\le 4\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}t=2.$

First find the slope of the tangent line using [link] , which means calculating ${x}^{\prime }\left(t\right)$ and ${y}^{\prime }\left(t\right)\text{:}$

$\begin{array}{c}{x}^{\prime }\left(t\right)=2t\hfill \\ {y}^{\prime }\left(t\right)=2.\hfill \end{array}$

Next substitute these into the equation:

$\begin{array}{c}\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt}\hfill \\ \frac{dy}{dx}=\frac{2}{2t}\hfill \\ \frac{dy}{dx}=\frac{1}{t}.\hfill \end{array}$

When $t=2,$ $\frac{dy}{dx}=\frac{1}{2},$ so this is the slope of the tangent line. Calculating $x\left(2\right)$ and $y\left(2\right)$ gives

$x\left(2\right)={\left(2\right)}^{2}-3=1\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y\left(2\right)=2\left(2\right)-1=3,$

which corresponds to the point $\left(1,3\right)$ on the graph ( [link] ). Now use the point-slope form of the equation of a line to find the equation of the tangent line:

$\begin{array}{ccc}\hfill y-{y}_{0}& =\hfill & m\left(x-{x}_{0}\right)\hfill \\ \hfill y-3& =\hfill & \frac{1}{2}\left(x-1\right)\hfill \\ \hfill y-3& =\hfill & \frac{1}{2}x-\frac{1}{2}\hfill \\ \hfill y& =\hfill & \frac{1}{2}x+\frac{5}{2}.\hfill \end{array}$

Find the equation of the tangent line to the curve defined by the equations

$x\left(t\right)={t}^{2}-4t,\phantom{\rule{1em}{0ex}}y\left(t\right)=2{t}^{3}-6t,\phantom{\rule{1em}{0ex}}-2\le t\le 3\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}t=5.$

The equation of the tangent line is $y=24x+100.$

## Second-order derivatives

Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function $y=f\left(x\right)$ is defined to be the derivative of the first derivative; that is,

$\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx}\left[\frac{dy}{dx}\right].$

Since $\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt},$ we can replace the $y$ on both sides of this equation with $\frac{dy}{dx}.$ This gives us

$\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\left(d\text{/}dt\right)\left(dy\text{/}dx\right)}{dx\text{/}dt}.$

If we know $dy\text{/}dx$ as a function of t, then this formula is straightforward to apply.

## Finding a second derivative

Calculate the second derivative ${d}^{2}y\text{/}d{x}^{2}$ for the plane curve defined by the parametric equations $x\left(t\right)={t}^{2}-3,y\left(t\right)=2t-1,-3\le t\le 4.$

From [link] we know that $\frac{dy}{dx}=\frac{2}{2t}=\frac{1}{t}.$ Using [link] , we obtain

$\frac{{d}^{2}y}{d{x}^{2}}=\frac{\left(d\text{/}dt\right)\left(dy\text{/}dx\right)}{dx\text{/}dt}=\frac{\left(d\text{/}dt\right)\left(1\text{/}t\right)}{2t}=\frac{\text{−}{t}^{-2}}{2t}=-\frac{1}{2{t}^{3}}.$

Calculate the second derivative ${d}^{2}y\text{/}d{x}^{2}$ for the plane curve defined by the equations

$x\left(t\right)={t}^{2}-4t,\phantom{\rule{1em}{0ex}}y\left(t\right)=2{t}^{3}-6t,\phantom{\rule{1em}{0ex}}-2\le t\le 3$

and locate any critical points on its graph.

$\frac{{d}^{2}y}{d{x}^{2}}=\frac{3{t}^{2}-12t+3}{2{\left(t-2\right)}^{3}}.$ Critical points $\left(5,4\right),\left(-3,-4\right),\text{and}\phantom{\rule{0.2em}{0ex}}\left(-4,6\right).$

## Integrals involving parametric equations

Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations $x\left(t\right)=t-\text{sin}\phantom{\rule{0.2em}{0ex}}t,\phantom{\rule{1em}{0ex}}y\left(t\right)=1-\text{cos}\phantom{\rule{0.2em}{0ex}}t.$ Suppose we want to find the area of the shaded region in the following graph.

To derive a formula for the area under the curve defined by the functions

$x=x\left(t\right),\phantom{\rule{1em}{0ex}}y=y\left(t\right),\phantom{\rule{1em}{0ex}}a\le t\le b,$

we assume that $x\left(t\right)$ is differentiable and start with an equal partition of the interval $a\le t\le b.$ Suppose ${t}_{0}=a<{t}_{1}<{t}_{2}<\text{⋯}<{t}_{n}=b$ and consider the following graph.

We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is $y\left(x\left({\stackrel{–}{t}}_{i}\right)\right)$ for some value ${\stackrel{–}{t}}_{i}$ in the i th subinterval, and the width can be calculated as $x\left({t}_{i}\right)-x\left({t}_{i-1}\right).$ Thus the area of the i th rectangle is given by

${A}_{i}=y\left(x\left({\stackrel{–}{t}}_{i}\right)\right)\phantom{\rule{0.2em}{0ex}}\left(x\left({t}_{i}\right)-x\left({t}_{i-1}\right)\right).$

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!