Parametric Curves A curve C consisting of points (x,y) where x and y are both functions of a third variable, say t (called a parameter). Thus, we have each point (x,y) determined by the parametric equations x= f( t) and y= gt( ) .
Note: Often, but not always, t represents time and these equations are called equations of motion and the curve C is the trajectory of the moving particle.
Examples: (a) Sketch the curve defined by the parametric equations xt=−+2
yt=52 − We will eliminate the parameter t and sketch. (b) Sketch the curve defined by the parametric equations ( xy,) =( 1,1) +− t( 1,1) This is a compact form of writing the equations separately. Separate this into two individual equations. Eliminate the parameter and sketch the line. Find the coordinates of the point of intersection of the above two curves (c) A spider and a fly crawl so that their positions at time t (in seconds) are: Spider: ( xy ,) =−+−( 2,5) t( 1, 2)
Fly:( xy ,) =( 1,1) +− t( 1,1) Question: Does the spider meet the fly? Explain
As you should have noticed, parametric equations tend to contain more information.
Eliminating the parameter in parametric equations containing trig functions. Note: We often use trig identities, especially cos22tt+= sin 1 Examples: Eliminate the parameter t if:
(a) xt= cos (b) xt = cos+= 2( c) x 2sin t
y=cos( 2 t) yt =sin+= 3 y3cos t
6 Slopes of Tangent lines of Parametric Curves Method: dy dx 1. Find and dt dt dy dy 2. = dt dx dx dt dy Example: Find if x=−+2 ty = 52 − t dx Solution: dy dy dx dy −2 =−= 2 and 1. Hence, ===−dt 2 dt dt dx dx 1 dt
Example: Find an equation of the tangent line to the parametric curve xt=−=+322 yt at the point where t = 1.
dy dy dx dy 1 Solution: =1 and =−== 2t . Hence, dt dt dt dxdx −2 t dt 11 Slope of tangent at t = 1 is m = = − and (x,y) = (2, 3) t −22 1 Therefore, the equation of the tangent line is: yx−=−32( −) 2
In order to draw sketches of these parametric curves we also need to be able to find the second derivative to discuss concavity.
Recall from Calculus 1500: To find the second derivative we differentiated the first derivative. That is exactly what we do with parametric equations. Just be careful!
7 Example: Find the first and second derivatives of the parametric curve xt=−=+322 yt
Solution: As above, the first derivative is dy dy dx dy 1 =1 and =−== 2t . Hence, dt dt dt dxdx −2 t dt 1 That is, y′ = − and the second derivative is 2t dy( ′) dy2 which most people write as . In other words, you find the dx dx2 11 parametric derivative using x=−3 t21 NewY ==−=− y′ t − 22t
1 −2 2 t dy d( NewY ) 1 So = =2 = dx23 dx−−24 t t Be very careful not to confuse this with the derivative of a quotient. It has nothing to do with quotients!
dy d2 y 11 Example: Find and at t= 2 if x = ty23 = t dx dx2 23 dy dy 22d dy t d y dx 1 Solution: =dt = =t and = = dxdx t dx2 dx t dt dy d2 y 1 When t = 2 = 2 and = dx dx2 2
We are now in a position to follow curve sketching steps, very similar to the curve sketching steps you followed in math 1500.
8 Other than plotting some interesting points we will follow these steps: 1. Find the x and y intercepts. 2. Determine the “starting” and “ending” locations of the curve by finding the limits of x and y as t→± ∞. 3. Find all horizontal and vertical tangent lines. 4. Use a first derivative chart to find the direction the curve is following in strategic intervals. 5. Sketch the curve. Using the above information. 6. Check the concavity of your curve using the values of the second derivative in strategic intervals.
Example: Follow the above steps in order to sketch the curve determined by the parametric equations xttytt=−=−3233 3 Solution: Intercepts xy= 00= 03=tt32−=03tt3− 0=−t2 ( t 3) 0=−+tt( 33)( t ) tt= 0,3 =0, 3, − 3 ∴−yxintercepts are: 0,18 −intercepts are: 0,3 3 −9, − 3 3 − 9 Beginning and Ending Positions limx= lim t32 − 3 t = lim tt 2( − 3) = −∞ ( ∴ starts far to the left ) tt→−∞ →−∞ ( ) t→−∞ 32 2 limx= lim( t − 3 t) = lim tt( − 3) =∞∴ ( ends far to the right ) tt→∞ →∞ t→∞ limy= lim t32 − 3 t = lim tt − 3 = ±∞ tt→±∞ →±∞ ( ) t→±∞ ( ) ( ∴ starts far down and ends far up ) Coordinates of points of horizontal and vertical tangent lines. ( These are our horizontal and vertical “min/max” positions. ) dy dx dy Be very careful – we will be using and NOT !!! dt dt dx
9 dy dx =33t22−=36tt− dt dt 03=−+(t 1)( t 1) 03=tt( − 2) dx dy tt=−≠1, 1 and 0 =0,2 and ≠ 0 dt dt ∴∴The curve has horizontal tangents The curve has vertical tangents at (−−2, 2) and ( −4, 2) at (0,0) and (− 4,2)
The Chart We will use the critical values of t to determine the directions of the curve t −1 01 2
dx ++−−+ dt dy +−−++ dt x →→←←→ y ↑↓↓↑↑ curve Use the above information, especially the curve directions in the chart to sketch the curve. dy d2 y Check the concavity [Note: Now we will be using and .] dx dx2
10 dy dy33 t 2 − Note: It is tempting but not advisable =dt = 2 dxdx 36 t− t to reduce this fraction. dt 2 33t − 22 d 63tt( −−− 6 t) ( 6 t 63)( t − 3) 36tt2 − ← [This is the Quotient Rule] 2 2 dy2 (36tt− ) = dt = dx22dx 36t− t dt 18ttttt3− 36 232 − 18 + 18 +− 18 18 = 3 (36tt2 − )
2 −18tt2 +− 18 18 −18(tt −+ 1) = 33= (36tt2 − ) 27tt3 ( − 2)
Therefore, the numbers which appear on the number line are t = 0 and 2. t 02
dy2 − +− dx2 curve ∩ ∪∩
Practice: Follow the steps outlined above to analyze and sketch the curve described by the parametric equations: 1. xt=−=−323 t y 39 t 2. xt=−=−2344 yt t 3. xt=−=−23 t yt3 t 4. xt=+=+222 t yt t 5. xt= 22 yt= − 2 t 6. xt=22 +21 t − yt = +− t 2 7. xttytt=−=−33332 8. xt= 32 yt= − 2 t t 9. xt=−=−3232 t yt t 10. xe=−=t t y4 e2
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