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d d x a x f ( u ) d u = f ( x ) .

Therefore

s ( t ) = d d t [ s ( t ) ] = d d t [ 0 t 140 2 + ( −32 v + 2 ) 2 d v ] = 140 2 + ( −32 t + 2 ) 2 = 1024 t 2 128 t + 19604 = 2 256 t 2 32 t + 4901 .

One third of a second after the ball leaves the pitcher’s hand, the distance it travels is equal to

s ( 1 3 ) = ( 1 / 3 2 1 32 ) 1024 ( 1 3 ) 2 128 ( 1 3 ) + 19604 1225 4 ln | ( −32 ( 1 3 ) + 2 ) + 1024 ( 1 3 ) 2 128 ( 1 3 ) + 19604 | + 19604 32 + 1225 4 ln ( 2 + 19604 ) 46.69 feet .

This value is just over three quarters of the way to home plate. The speed of the ball is

s ( 1 3 ) = 2 256 ( 1 3 ) 2 16 ( 1 3 ) + 4901 140.34 ft/s .

This speed translates to approximately 95 mph—a major-league fastball.

Surface area generated by a parametric curve

Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area , we derived a formula for finding the surface area of a volume generated by a function y = f ( x ) from x = a to x = b , revolved around the x -axis:

S = 2 π a b f ( x ) 1 + ( f ( x ) ) 2 d x .

We now consider a volume of revolution generated by revolving a parametrically defined curve x = x ( t ) , y = y ( t ) , a t b around the x -axis as shown in the following figure.

A curve is drawn in the first quadrant with endpoints marked t = a and t = b. On this curve, there is a point marked (x(t), y(t)). There is a circle with an arrow drawn around the x axis that seems to indicate a rotation about the x axis, and there is a shape that accompanies that curve that seems to be what you would obtain if you rotated the curve about the x axis.
A surface of revolution generated by a parametrically defined curve.

The analogous formula for a parametrically defined curve is

S = 2 π a b y ( t ) ( x ( t ) ) 2 + ( y ( t ) ) 2 d t

provided that y ( t ) is not negative on [ a , b ] .

Finding surface area

Find the surface area of a sphere of radius r centered at the origin.

We start with the curve defined by the equations

x ( t ) = r cos t , y ( t ) = r sin t , 0 t π .

This generates an upper semicircle of radius r centered at the origin as shown in the following graph.

A semicircle is drawn with radius r. On the graph there are also written three equations: x(t) = r cos(t), y(t) = r sin(t), and 0 ≤ t ≤ π.
A semicircle generated by parametric equations.

When this curve is revolved around the x -axis, it generates a sphere of radius r . To calculate the surface area of the sphere, we use [link] :

S = 2 π a b y ( t ) ( x ( t ) ) 2 + ( y ( t ) ) 2 d t = 2 π 0 π r sin t ( r sin t ) 2 + ( r cos t ) 2 d t = 2 π 0 π r sin t r 2 sin 2 t + r 2 cos 2 t d t = 2 π 0 π r sin t r 2 ( sin 2 t + cos 2 t ) d t = 2 π 0 π r 2 sin t d t = 2 π r 2 ( cos t | 0 π ) = 2 π r 2 ( cos π + cos 0 ) = 4 π r 2 .

This is, in fact, the formula for the surface area of a sphere.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

Find the surface area generated when the plane curve defined by the equations

x ( t ) = t 3 , y ( t ) = t 2 , 0 t 1

is revolved around the x -axis.

A = π ( 494 13 + 128 ) 1215

Got questions? Get instant answers now!

Key concepts

  • The derivative of the parametrically defined curve x = x ( t ) and y = y ( t ) can be calculated using the formula d y d x = y ( t ) x ( t ) . Using the derivative, we can find the equation of a tangent line to a parametric curve.
  • The area between a parametric curve and the x -axis can be determined by using the formula A = t 1 t 2 y ( t ) x ( t ) d t .
  • The arc length of a parametric curve can be calculated by using the formula s = t 1 t 2 ( d x d t ) 2 + ( d y d t ) 2 d t .
  • The surface area of a volume of revolution revolved around the x -axis is given by S = 2 π a b y ( t ) ( x ( t ) ) 2 + ( y ( t ) ) 2 d t . If the curve is revolved around the y -axis, then the formula is S = 2 π a b x ( t ) ( x ( t ) ) 2 + ( y ( t ) ) 2 d t .

Key equations

  • Derivative of parametric equations
    d y d x = d y / d t d x / d t = y ( t ) x ( t )
  • Second-order derivative of parametric equations
    d 2 y d x 2 = d d x ( d y d x ) = ( d / d t ) ( d y / d x ) d x / d t
  • Area under a parametric curve
    A = a b y ( t ) x ( t ) d t
  • Arc length of a parametric curve
    s = t 1 t 2 ( d x d t ) 2 + ( d y d t ) 2 d t
  • Surface area generated by a parametric curve
    S = 2 π a b y ( t ) ( x ( t ) ) 2 + ( y ( t ) ) 2 d t

For the following exercises, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.

Questions & Answers

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What is lattice structure?
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of graphene you mean?
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Graphene has a hexagonal structure
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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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
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this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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