8.3.Smooth Parametric Curves and Their Slopes

8.3.Smooth Parametric Curves and Their Slopes

September 10, 2011

2. solution:

Let x = f (t) = t²− 2t, y = g(t) = t²+ 2t

(a)To find the horizontal tangent to the curve, we need to find t such that

dx

dt = f⁰(t) 6= 0 and ^dy_dt(t) = g⁰(t) = 0. i.e. to solve ^dx_dt = 2t − 2 6= 0 and

dy

dt = 2t + 2 = 0.

For ^dy_dt = g⁰(t) = 2t + 2 = 0, we only get t = −1, and we also have ^dx_dt(−1) = f⁰(−1) = −2 − 2 = −4 6= 0

So, when t = −1, the parametric curve has a horizontal tangent , which is at (x(−1), y(−1)) = (f (−1), g(−1)) = (3, −1).

(b)To find the vertical tangent to the curve, we need to find t such that ^dx_dt = f⁰(t) = 0 and^dy_dt(t) = g⁰(t) 6= 0. i.e. to solve^dx_dt = 2t−2 = 0 and^dy_dt = 2t+2 6= 0.

For^dx_dt = f⁰(t) = 2t − 2 = 0, we only get t = 1, and we also have ^dy_dt(1) = g⁰(1) = 2 + 2 = 4 6= 0

So, when t = 1, the parametric curve has a vertical tangent , which is at (x(1), y(1)) = (f (1), g(1)) = (−1, 3).

4.

solution:

Let x = f (t) = t³− 3t, y = g(t) = 2t³+ 3t²

(a)To find the horizontal tangent to the curve, we need to find t such that

dx

dt = f⁰(t) 6= 0 and ^dy_dt(t) = g⁰(t) = 0. i.e. to solve ^dx_dt = 3t²− 3 6= 0 and

dy

dt = 6t²+ 6t = 0.

For ^dy_dt = g⁰(t) = 6t²+ 6t = 6t(t + 1) = 0, we get t = −1 or 0, and since that we also have ^dx_dt(−1) = f⁰(−1) = 3 − 3 = 0 and^dx_dt(0) = f⁰(0) = 0 − 3 = −3 6= 0, we get that t=-1 is not our choice.

So, when t = 0, the parametric curve has a horizontal tangent , which is at (x(0), y(0)) = (f (0), g(0)) = (0, 0).

(b)To find the vertical tangent to the curve, we need to find t such that ^dx_dt = f⁰(t) = 0 and ^dy_dt(t) = g⁰(t) 6= 0. i.e. to solve ^dx_dt = 3t² − 3 = 0 and

dy

dt = 6t²+ 6t 6= 0.

For ^dx_dt = f⁰(t) = 3t²− 3 = 0, we get t = 1 or t = −1, and since we also have

dy

dt(1) = g⁰(1) = 6 + 6 = 12 6= 0 and ^dy_dt(−1) = g⁰(−1) = 6 − 6 = 0, we get that t = −1 is not our choice.

So, when t = 1, the parametric curve has a vertical tangent , which is at (x(1), y(1)) = (f (1), g(1)) = (−2, 5).

6.

solution:

Let x = f (t) = sin t, y = g(t) = sin t − t cos t

(a)To find the horizontal tangent to the curve, we need to find t such that

dx

dt = f⁰(t) 6= 0 and ^dy_dt(t) = g⁰(t) = 0. i.e. to solve ^dx_dt = cos t 6= 0 and

dy

dt = t sin t = 0.

For ^dy_dt = g⁰(t) = t sin t = 0, we get t = nπ for all n ∈ Z, and since that we also have ^dx_dt(nπ) = f⁰(nπ) = (−1)ⁿ6= 0 for all n ∈ Z .

So, when t = nπ, where n ∈ ,the parametric curve has a horizontal tangent , which is at (x(nπ), y(nπ)) = (f (nπ), g(nπ)) = (0, (−1)ⁿnπ) , n ∈ Z.

(b)To find the vertical tangent to the curve, we need to find t such that ^dx_dt = f⁰(t) = 0 and ^dy_dt(t) = g⁰(t) 6= 0. i.e. to solve ^dx_dt = cos t = 0 and ^dy_dt = t sin t 6= 0.

For ^dx_dt = f⁰(t) = cos t = 0, we get t = (n + ¹₂)π , for all n ∈ Z ,

and since we also have ^dy_dt((n +¹₂)π) = g⁰((n +¹₂)π) = (n +¹₂)π sin[(n +¹₂)π] = (−1)ⁿ(n +¹₂)π 6= 0 , for all n ∈ Z

So, when t = (n + ¹₂)π, where n ∈ Z , the parametric curve has a vertical tangent , which is at (x((n +¹₂)π), y((n +¹₂)π)) = (f ((n +¹₂)π), g((n +¹₂)π)) = ((−1)ⁿ, (−1)ⁿ), where n ∈ Z.

8.

solution:

Let x = f (t) = _1+t^3t₃, y = g(t) = _1+t^3t2₃

(a)To find the horizontal tangent to the curve, we need to find t such that

dx

dt = f⁰(t) 6= 0 and ^dy_dt(t) = g⁰(t) = 0. i.e. to solve ^dx_dt = _(1+t^3−6t3³)² 6= 0 and

dy

dt = _(1+t^6t−3t₃₎⁴₂ = 0.

For ^dy_dt = g⁰(t) = _(1+t^6t−3t3)⁴² = 0, we get t = 0 or √³

2, and since that we also have

dx

dt(0) = f⁰(0) = 3 6= 0 and^dx_dt(√³

2) = f⁰(√³

2) = −1 6= 0, so both of them are of our choice .

So, when t = 0 and√³

2 the parametric curve have horizontal tangents , that are at (x(0), y(0)) = (f (0), g(0)) = (0, 0) and (x(√³

2), y(√³

2)) = (f (√³ 2), g(√³

2)) = (√³

2,√³ 4).

(b)To find the vertical tangent to the curve, we need to find t such that ^dx_dt = f⁰(t) = 0 and ^dy_dt(t) = g⁰(t) 6= 0. i.e. to solve ^dx_dt = _(1+t^3−6t3³)² = 0 and

dy

dt = _(1+t^6t−3t₃₎⁴₂ 6= 0.

For ^dx_dt = f⁰(t) = _(1+t^3−6t3³)² = 0, we get t = √3¹

2, and since we also have ^dy_dt(√3¹

2) = g⁰(√3¹

2) =√³ 4 6= 0 . So, when t = √3¹

2, the parametric curve has a vertical tangent , which is at

(x(√3¹

2), y(√3¹

2)) = (f (√3¹

2), g(√3¹

2)) = (√³ 4,√³

2).

10.

solution:

Let x = f (t) = t⁴− t², y = g(t) = t³+ 2t

Then we have ^dx_dt = f⁰(t) = 4t³− 2t,^dy_dt = g⁰(t) = 3t²+ 2t.

So the slope at t = −1 is _dx^dy|t=−1= ^dy_dtdx^dt|t=−1= ^dydt_dt^|t=−1

dx|t=−1 =_f^g00⁽⁻¹⁾(−1) = −⁵₂ 12.

solution:

Let x = f (t) = e^2t, y = g(t) = te^2t

Then we have ^dx_dt = f⁰(t) = 2e^2t,^dy_dt = g⁰(t) = e^2t+ 2te^2t. So the slope at t = −1 is _dx^dy|t=−2= ^dy_dtdx^dt|t=−2=

dy dt|t=−2 dt

dx|t=−2 =_f^g00⁽⁻²⁾(−2) = −³₂ 14.

solution:

Let x = f (t) = t − cos t, y = g(t) = 1 − sin t

Then we have ^dx_dt = f⁰(t) = 1 + sin t,^dy_dt = g⁰(t) = − cos t.

For t0=^π₄, we have f⁰(^π₄) = 1 +^√1

2, g⁰(t) = −^√1

2 and f (^π₄) = ^π₄−^√1

2, g((^π₄) = 1 −^√1

2

Thus the equation of the tangent line is

( x = ^π₄ −^√1

2 + (1 +^√1

2)(t −^π₄) y = 1 −^√1

2+ (−^√1

2)(t −^π₄) 16.

solution:

Let x = f (t) = sin t, y = g(t) = sin 2t

Then, for the origin of the curve, we need to find t such that f (t) = g(t) = 0 For f (t) = sin t = 0, we have t = nπ, n ∈ Z, but for g(t) = sin 2t = 2 sin t cos t = 0 ,we have t = n^π₂, n ∈ Z, thus the value of t such that f (t) = g(t) = 0 are t = nπ, n ∈ Z.

We also have ^dx_dt = f⁰(t) = cos t,^dy_dt = g⁰(t) = 2 cos 2t.

So the slopes at the origin are _dx^dy|t=nπ = ^dy_dtdx^dt|t=nπ =

dy dt|t=nπ dt

dx|t=nπ = _f^g00^(nπ)(nπ) =

−^{2 cos 2nπ}_{cos nπ} =

 2 if n is even

−2 if n is odd 18.

solution:

Let x = f (t) = (t − 1)⁴, y = g(t) = (t − 1)³

Then we have ^dx_dt = f⁰(t) = 4(t − 1)³,^dy_dt = g⁰(t) = 3(t − 1)².

To find the point that the curve may fail to be smooth ,we need solve ^dx_dt =

dy

dt = 0 ,and we noly get t = 1.

At t = 1, we get x = 0, y = 0

From x = f (t) = (t − 1)⁴, y = g(t) = (t − 1)³, we have y = x³⁴, so ^dy_dx = ³₄x⁻¹⁴ This curve is not smooth at (x, y) = (0, 0), since that ^dy_dx|x=0 does not exist.

20.

solution:

Let x = f (t) = t³, y = g(t) = t − sin t

Then we have ^dx_dt = f⁰(t) = 3t²,^dy_dt = g⁰(t) = 1 − cos t.

To find the point that the curve may fail to be smooth ,we need solve ^dx_dt =

dy

dt = 0 ,and we noly get t = 0 clearly.

At t = 0, we get x = 0, y = 0

To check that the curve is smooth or not at t = 0, we need to check ^dy_dx|x=0

exists or not., i.e, check ^dy_dtdx^dt|t=0=

dy dt|t=0 dt

dx|t=0 exists or not.

We have

dy dt|t=0 dt dx|t=0

= lim

t→0 dy dt dt dx

= lim

t→0

1 − cos t 3t² = lim

t→0

sin t 6t = lim

t→0

cos t 6 = 1

6 ,whic is existed (by L’Hospital’s rule)

Thus there is no points that the curve is not smooth . 22.

solution:

Let x = f (t) = t³, y = g(t) = 3t²− 1

Then we have ^dx_dt = f⁰(t) = 3t²,^dy_dt = g⁰(t) = 6t, and y = 3x²³− 1

First, we need to check whether the curve has nonsmooth points or not. To find the point that the curve may fail to be smooth ,we need solve ^dx_dt = ^dy_dt = 0 ,and we noly get t = 0 clearly. Moreover , when t 6= 0 , we have ^dx_dt 6= 0,^dy_dt 6= 0 At t = 0, we get x = 0, y = 1

To check that the curve is smooth or not at t = 0, we need to check ^dy_dx|x=0

exists or not., i.e, check ^dy_dtdx^dt|t=0=

dy dt|t=0 dt

dx|_t=0 exists or not.

We have

dy dt|t=0 dt

dx|_t=0 = lim

t→0

g⁰(t) f⁰(t) = lim

t→0

6t 3t² = lim

t→0

2 t ,whic does not exist.

Thus the curve is not smooth at (x, y) = (0, −1).

We also have _dx^dy = ^g_f⁰0^(t)(t) = ²_t, which tends to 0 as t → ∞ , and _dx^dy > 0 if t > 0

dy

dx < 0 if t < 0.

For concavity, we calculate ^d_dx^2y₂. Since f⁰⁰(t) = 6t, g⁰⁰(t) = 6, we get d²y

dx² = f⁰(t)g⁰⁰(t) − f⁰⁰(t)g⁰(t)

f⁰(t)³ = (3t²)6 − (6t)(6t) (3t²)³ = − 2

3t⁴

,which is never be zero but fail to be defined at t = 0, that is, the point that is not smooth . Since that ^d_dx^2y₂ = −_3t²₄ < 0, f or all t ∈ R, we get hte curve is concave downward everywhere except at t=0, i.e., (x, y) = (0, −1)

The table is: t -1 0 1 (x, y) (−1, −2) (0, −1) (1, 2)

f⁰(t) 3 0 3

g⁰(t) -6 0 6

dy

dx = ^g_f⁰0^(t)(t) -2 not defined 2

Figure 1: Exercise 8.3.22 24.

solution:

Let x = f (t) = t³− 3t − 2, y = g(t) = t²− t − 2

Then we have ^dx_dt = f⁰(t) = 3t²− 3,^dy_dt = g⁰(t) = 2t − 1.

So, the curve has a horizontal tangent at t =¹₂, that is (x, y) = (−²⁷₈, −⁹₄), and

the curve also have vertial tangents at t = 1, −1, that is , (x, y) = (−4, −2), (0, 0),respectively.

Then, we need to check whether the curve has nonsmooth points or not. To find the point that the curve may fail to be smooth ,we need solve ^dx_dt =^dy_dt = 0 ,and there is no such t clearly. So, the cure is smooth.

At t = 0, we get x = 0, y = 1

We also have ^dy_dx = ^g_f⁰0^(t)(t) = ²_t, which tends to 0 as t → ∞ , and ^dy_dx > 0 if t > 0 ^dy_dx < 0 if t < 0.

For concavity, we calculate ^d_dx^2y₂. Since f⁰⁰(t) = 6t, g⁰⁰(t) = 2, we get d²y

dx² = f⁰(t)g⁰⁰(t) − f⁰⁰(t)g⁰(t)

f⁰(t)³ =(3t²− 3)2 − (6t)(2t − 1)

(3t²− 3)³ = −2(t²− t + 1) 9(t²− 1)³ ,which is never be zero but fail to be defined at t = 1, −1.

Obviously, the curve is concave upward for −1 < t < 1 and concave downward

elsewhere except for t = 1, −1.

The table is:

t -1 −¹₂ 0 ¹₂ 1 2

(x, y) (0,0) (−⁵₈, −⁵₄) (-2,-2) (−²⁷₈, −⁹₄) (-4,-2) (0,0)

f⁰(t) 0 −⁹₄ -3 −⁹₄ 0 9

g⁰(t) -3 -2 -1 0 1 3

dy

dx = ^g_f⁰0^(t)(t) −∞ + ⁸₉ + ¹₃ + 0 - ∞ + ¹₃

Figure 2: Exercise 8.3.24