CHAPTER 55 DIFFERENTIATION OF PARAMETRIC EQUATIONS EXERCISE 237 Page 644
1. Given x = 3t – 1 and y = t(t – 1), determine
If x = 3t – 1, then
dx =3 dt
If y = t(t – 1) = t 2 − t , then
Hence,
dy in terms of t. dx
dy = 2t − 1 dt
dy d y d t 2t − 1 1 = ( 2t − 1) = = 3 dx dx 3 dt
2, y 2. A parabola has parametric equations: = x t= 2t . Evaluate
If x = t 2 , then
dx = 2t dt
If y = 2t, then
dy =2 dt
Hence,
dy when t = 0.5 dx
dy d y dt 2 1 = = = d x d x 2t t dt
When t = 0.5,
dy 1 =2 = d x 0.5
3. The parametric equations for an ellipse are x = 4 cos θ, y = sin θ. Determine (a)
(a) If x = 4 cos θ, then If y = sin θ,
then
dy d2 y (b) dx d x2
dx = −4sin θ dθ dy = cos θ dθ
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© 2014, John Bird
dy 1 d y dθ cos θ Hence, = = = − cot θ 4 d x d x −4sin θ dθ
d dy d 1 − cot θ − 1 ( − cosec 2 θ ) d y dθ d x dθ 4 1 1 = 4 (b) = = = − 2 2 dx −4sin θ −4sin θ dx 16 sin θ sin θ dθ 2
1 1 − = 16 sin 3 θ =−
4. Evaluate
1 cosec3 θ 16
π dy at θ = radians for the hyperbola whose parametric equations are x = 3 sec θ, dx 6
y = 6 tan θ
If x = 3 sec θ, then
dx = 3sec θ tan θ dθ
If y = 6 tan θ, then
dy = 6sec 2 θ dθ
Hence,
dy 1 2 d y dθ 6sec θ 2sec θ 2 = = = = 2 cos θ= d x d x 3sec θ tan θ tan θ sin θ sin θ dθ cos θ
π
dy 2 2 , = = =4 6 d x sin π 0.5 6
When θ =
5. The parametric equations for a rectangular hyperbola are x = 2t, y =
2 dy . Evaluate when t dx
t = 0.40
If x = 2t, then If y=
dx =2 dt
2 dy −2 = 2t −1 , then = −2t −2 = t dt t2
Hence,
dy 2 − 2 d y dt 1 = = t = − d x dx 2 t2 dt 962
© 2014, John Bird
When t = 0.04,
dy 1 = − = –6.25 dx (0.4) 2
6. The equation of a tangent drawn to a curve at point ( x1 , y1 ) is given by:
y −= y1
d y1 ( x − x1 ) d x1
Determine the equation of the tangent drawn to the ellipse x = 3 cos θ, y = 2 sin θ at θ =
y −= y1
and
d x1 = −3sin θ dθ
y1 = 2sin θ
and
d y1 = 2 cos θ dθ
d y1 d y1 d θ 2 cos θ 2 = = = − cot θ d x1 d x1 −3sin θ 3 dθ
Hence,
The equation of a tangent is:
π 6
,
2 y – 2 sin θ = − cot θ ( x − 3cos θ ) 3
y – 2 sin
2 π π = − cot x − 3cos 3 6 6 6
π
2 (1.732 )( x − 2.598) 3
i.e.
y–1= −
i.e.
y – 1 = –1.155(x – 2.598)
i.e.
y – 1 = –1.155x + 3
and
y = –1.155x + 4
7. The equation of a tangent drawn to a curve at point ( x1 , y1 ) is given by:
y −= y1
d y1 ( x − x1 ) . d x1
Determine the equation of the tangent drawn to the rectangular hyperbola x = 5t, y =
y −= y1
6
d y1 ( x − x1 ) d x1
At point θ, x1 = 3cos θ
At θ =
π
5 at t = 2 t
d y1 ( x − x1 ) d x1
At point θ, x1 = 5t
and
d x1 =5 dt
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© 2014, John Bird
y=
Hence,
5 = 5t −1 t
and
d y1 5 = −5t −2 = − dt t2
d y1 5 − 2 d y1 1 = dt = t = − t2 d x1 d x1 5 dt
The equation of a tangent is:
y–
5 1 = − ( x − 5t ) t t2
At t = 2 ,
y–
5 1 = − ( x − 10 ) 22 2
i.e.
y–
5 1 5 = − x+ 2 4 2
i.e.
4y – 10 = –x + 10
i.e.
4y = –x + 20
and
1 y = − x+5 4
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© 2014, John Bird
EXERCISE 238 Page 646 1. A cycloid has parametric equations x = 2(θ – sin θ), y = 2(1 – cos θ). Evaluate, at θ = 0.62 rad, correct to 4 significant figures, (a)
(a) If x = 2(θ – sin θ),
then
If y = 2(1 – cos θ), then
dy d2 y (b) dx d x2
dx = 2 − 2 cos θ dθ dy = 2sin θ dθ
dy dy 2sin θ sin θ θ Hence, = d= = d x d x 2(1 − cos θ ) 1 − cos θ dθ When θ = 0.62 rad,
dy sin 0.62 = 3.122, correct to 4 significant figures = d x 1 − cos 0.62
d dy d sin θ (1 − cos θ )(cos θ ) − (sin θ )(sin θ ) cos θ − cos 2 θ − sin 2 θ d 2 y d θ d x d θ 1 − cos θ (1 − cos θ ) 2 (1 − cos θ ) 2 (b) = = = = dx d x2 2(1 − cos θ ) 2(1 − cos θ ) 2(1 − cos θ ) dθ =
cos θ − (cos 2 θ + sin 2 θ ) cos θ − 1 = 2(1 − cos θ )3 2(1 − cos θ )3
d2 y (cos 0.62) − 1 −0.1861215 = = When θ = 0.62 rad, = –14.43, correct to 4 significant d x 2 2(1 − cos 0.62)3 2(0.00644748) figures 2. The equation of the normal drawn to a curve at point ( x1 , y1 ) is given by: y − y1 = −
1 ( x − x1 ) d y1 d x1
Determine the equation of the normal drawn to the parabola = x
1 If x1 = t 2 , 4
d x1 1 = t dt 2
1 If y1 = t , 2
d y1 1 = dt 2
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1 2 1 = t , y t at t = 2 4 2
© 2014, John Bird
d y1 1 d y1 t 2 1 = d= = d x1 d x1 1 t t dt 2
Hence,
y − y1 = −
Equation of a normal is:
1 ( x − x1 ) d y1 d x1
1 1 1 y− t = − x − t2 1 2 4 t
i.e.
At t = 2, equation of normal is:
y – 1 = –2(x – 1)
i.e.
y – 1 = –2x + 2
or
y = –2x + 3
3. The equation of the normal drawn to a curve at point ( x1 , y1 ) is given by:
y − y1 = −
1 ( x − x1 ) d y1 d x1
Find the equation of the normal drawn to the cycloid x = 2(θ – sin θ), y = 2(1 – cos θ) at
θ=
π 2
rad .
If x 1 = 2(θ – sin θ) = 2θ – 2 sin θ, If y1 = 2(1 − cos θ ) = 2 − 2 cos θ ,
d y1 = 2sin θ dθ
d y1 d y1 2sin θ dt = = d x1 d x1 2 − 2 cos θ dt
Hence,
Equation of a normal is:
y − y1 = −
1 ( x − x1 ) d y1 d x1
1 ( x − 2(θ − sin θ ) ) 2sin θ 2 − 2 cos θ
y − 2(1 − cos θ ) = −
i.e.
At θ =
d x1 = 2 − 2 cos θ dθ
1 π − rad , equation of normal is: y − 2(1 − cos ) = π 2 2 2sin 2
π π x − 2( − sin ) 2 2
π
2 − 2 cos 966
π 2 © 2014, John Bird
y − 2 =−
i.e.
y – 2 = –x + π – 2
i.e.
4. Determine the value of
π 1 x − 2( − 1) 2 2 2
or
y=–x+π
d2 y π , correct to 4 significant figures, at θ = rad for the cardioid 2 6 dx
x = 5(2θ – cos 2 θ), y = 5(2 sin θ – sin 2 θ).
If x = 5(2θ – cos 2θ), then
dx = 10 + 10sin 2θ = 10 (1 + sin 2θ ) dθ
If y = 5(2 sin θ – sin 2θ), then
dy =10 cos θ − 10 cos 2θ =10(cos θ − cos 2θ ) dθ
dy d y d θ 10(cos θ − cos 2θ ) cos θ − cos 2θ = = = dx dx 10(1 + sin 2θ ) 1 + sin 2θ dθ d dy d cos θ − cos 2θ (1 + sin 2θ )(− sin θ + 2sin 2θ ) − (cos θ − cos 2θ )(2 cos 2θ ) 2 1 + sin 2θ ) ( d 2 y d θ dx d θ 1 + sin 2θ = = = dx d x2 10(1 + sin 2θ ) 10(1 + sin 2θ ) dθ
When θ =
π 6
rad,
2π 2π 2π 2π π π 1 + sin − sin + 2sin − cos − cos 2 cos 6 6 6 6 6 6 2 2π 1 sin + d2 y 6 = 2π d x2 10 1 + sin 6
(1.86603)(1.23205) − (0.366025)(1) 3.48205 = = 0.02975, correct to 4 significant 18.660254 figures
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© 2014, John Bird
5. The radius of curvature, ρ, of part of a surface when determining the surface tension of a liquid d y 2 1 + d x ρ= d2 y d x2
is given by:
3/2
Find the radius of curvature (correct to 4 significant figures) of the part of the surface having parametric equations (a) x = 3t, y =
3 1 at the point t = t 2
(b) x 4= = cos3 t , y 4sin 3 t at t =
(a) x = 3t ,
y=
hence
π 6
rad .
dx =3 dt
dy 3 3 = −3t −2 = − = 3t −1 , hence t dt t2
dy 3 − 2 1 d y dt 1 = = t = − and at t = , 2 2 dx dx 3 t dt
dy 1 = − = −4 2 dx 1 2
d dy d 1 d − −t −2 ) 1 d2 y 2 2 16 d y d t d x d t t 2 d t ( 2t −3 2 and at t = , = = = = = = = = 3 2 3 2 3 d x 2 dx 3t 3 dx 3 3 3 3t 1 3 dt 2 2
3/2
d y 2 1 + d x Hence, radius of = curvature, ρ = d2 y d x2 (b) x = 4 cos3 t , hence
y = 4sin 3 t , hence
3
1 + ( −4 )2 = 16 3
173 = 13.14 16 3
dx 12 cos 2 t (− sin t ) = = −12 cos 2 t sin t dt dy = 12sin 2 t cos t dt
dy d y dt 12sin 2 t cos t sin t == = − = − tan t d x d x −12 cos 2 t sin t cos t dt
and at t =
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π 6
rad ,
dy π = − tan = −0.57735 dx 6
© 2014, John Bird
d dy d ( − tan t ) d y d t d x − sec 2 t 1 dt = = = = dx d x2 −12 cos 2 t sin t −12 cos 2 t sin t 12 cos 4 t sin t dt 2
π d2 y and at t rad , = 6 dx 2
1 0.29630 = 4 π π 12 cos sin 6 6 3/2
d y 2 1 + d x Hence, radius of= curvature, ρ = d2 y d x2
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3
1 + ( −0.57735 )2 = 0.29630
1.3333333 = 5.196 0.29630
© 2014, John Bird