Tangent Lines and Arc Length for Parametric and Polar Curves.

In earlier lectures we looked at parametric curves, that is those that can be written as

( ); ( );x f t y g t a t b= = ≤ ≤

Then it can be shown that

/dy dy dxdx dt dt

=

This allows us to find the slope of tangent lines without eliminating the parameter.

Example. Find the tangent line to the curve

at the point where t = 9π/4.

2sin ; 2 cos ; 0x t t y t t= − = − ≥

We have 1 2cos 1 2dx tdt

= − = − =

1sin2

dy tdt

= =

Thus ( )1 1 1/ / 1 2 1.72 2(1 2) 2 2

dy dy dxdx dt dt

= = − = = ≈−− −

This situation is shown below.

Problem. Find dy/dx and d2y/dx2 at the point on the curve

where t = 2.

1 12 3;2 3

x t y t= =

Solution. dx tdt

= 2/ /dy dy dx t t tdx dt dt

= = =Therefore2dy tdt

=

At the point where t = 2, dy/dx = 2.

Now if we let y′ = dy/dx = t, then2

/ 1/2

d y dy dy dx tdx dt dtdx

′ ′= = =

At the point where t = 2, d2y/dx2 = 1/2.

Problem. Find dy/dx and d2y/dx2 at the point on the curve

where ϕ = 5π/6cos ; 3sinx yϕ ϕ= =

Solution. sindxd

ϕϕ

=−

/ 3cos / sin 3cotdy dy dxdx d d

ϕ ϕ ϕϕ ϕ

= = − =−Therefore

3cosdyd

ϕϕ

=

At the point where ϕ = 5π/6, dy/dx = 3√3.Now if we let y′ = dy/dx = −3cot ϕ , then

2 23csc 3/ 3csc2 sin

d y dy dy dxdx d ddx

ϕ ϕϕ ϕ ϕ

′ ′ == = = =−−

At the point where ϕ = 5π/6, d2y/dx2 = −24.

Problem. Find all values of t at which the parametric curve

has (a) a horizontal tangent line and (b) a vertical tangent line.

3 2 22 15 24 7; 1x t t t y t t= − + + = + +

Solution. 2 26 30 24 6( 5 4) 6( 4)( 1)dx t t t t t tdt

= − + = − + = − −

2 1dy tdt

= + Thus 2 16( 4)( 1)

dy tdx t t

+=− −

We see that the tangent line is horizontal when 2t + 1 = 0, or t = −1/2. The tangent line is vertical when t = 4 or t = 1.

We can apply this procedure to find the tangent line to a polar curve. A polar curve r = f(θ) can be written parametrically by noting that x = rcos(θ) and y = rsin(θ). Thus

( )cosx f θ θ= ( )siny f θ θ=

This means that

( )sin ( )cos sin cosdx drf f rd d

θ θ θ θ θ θθ θ

′=− + =− +

( )cos ( )sin cos sindy drf f rd d

θ θ θ θ θ θθ θ

′= + = +

Therefore at any point we havecos sin

sin cos

drrdy ddrdx rd

θ θθ

θ θθ

+=

− +

Problem. Find the slope of the tangent line to the polar curve

for θ = π/4.1 sinr θ= +

cos sin (1 sin )cos sin cos(1 sin )sin cos cossin cos

drrdy ddrdx rd

θ θ θ θ θ θθθ θ θ θθ θ

θ

+ + += =− + +− +

Solution. dr/dθ = cosθ . Thus

1 1 1 1 1(1 ) 12 2 2 2 2 2 11 1 1 1 1(1 )2 2 2 2 2

+ + += = =− −

− + + −

Problem. Find the slope of the tangent line to the polar curve

for θ = π/6.sec2r a θ=

Solution. dr/dθ = 2asec2θtan2θ . Thus at θ = π/6, we have

dr/dθ = 2asec2θ tan2θ = 4a√3, r = 2a, sinθ = 1/2, cosθ = √3/2.

3 12 4 3cos sin 2 2

1 3sin cos 2 4 32 2

dr a ardy ddrdx r a ad

θ θθ

θ θθ

+ +

= = − + − +

Thus

3 3 3 35 5a

a= =

We can find out some useful information about polar curves that pass through the pole.

Theorem. Suppose that the curve r = f(θ) passes through the pole, or origin, at θ = θ0. If dr/dθ is not 0 at θ = θ0, then the line θ = θ0 is the tangent line to the curve at θ = θ0 at the origin.

Proof.cos sin sin

tansin cos cos

dr drrdy d ddr drdx rd d

θ θ θθ θ θ

θ θ θθ θ

+= = =

− +

at the origin, under the conditions given.

Problem. Sketch the polar curve

and find the polar equations of the tangent line to the curve atthe pole.

4cosr θ=

Solution. The curve can be written as 2 4 cosr r θ=

or x2 + y2 = 4x. On completing the square, this becomes

( )2 22 4.x y− + = The graph is:

Since at θ = π/2, r = 0 and dr/dθ = −4sinθ = −4, the previous theorem tells us that the tangent line at the pole is θ = π/2.

Problem. Sketch the polar curve

and find the polar equations of the tangent line to the curve atthe pole.

sin2r θ=

Solution. The curve is a 4 leaf rose whose graph is:

We have r = 0 when θ = 0, and θ = π/2. At these points, the derivative is 2cos 2θ = 1 or −1. Thus the tangent lines at the pole have equations θ = 0 and θ = π/2, by the previous theorem.

Arc Length

Recall from earlier lectures that the differential element of arc length, ds can be represented by the diagram

ds

dx

dy

From which we see that 2 2ds dx dy= +

Then arc length can be found by integration,

2 2s ds dx dy= = +∫ ∫

dxdx dtdt

= dydy dtdt

=

so that2 2dx dy

s dtdt dt

= + ∫

We previously investigated various forms of this equation, corresponding to y = f(x), x = g(y), and the parametric curve x = x(t), y = y(t). The parametric form follows from the differential substitutions

We apply this formula to find a formula for arc length of a polar curve. We know that a parametric form of the polar curve r = f(θ) is given by

( )cos( ); ( )sin( )x f y fθ θ θ θ= =Then

By squaring these expressions, we get

( )sin( ) ( )cos( ) sin( ) cos( )dx drf f rd d

θ θ θ θ θ θθ θ

′=− + =− +

( )cos( ) ( )sin( ) cos( ) sin( )dy drf f rd d

θ θ θ θ θ θθ θ

′= + = +

We add these equations to get:

2 2 22 2 2cos ( ) sin ( )

22

dx dy drr

d d d

drr

d

θ θθ θ θ

θ

+ = + +

= +

2 22 2 2sin ( ) 2 sin( )cos( ) cos ( )dx dr drr r

d d dθ θ θ θ

θ θ θ = − +

2 22 2 2cos ( ) 2 cos( )sin( ) sin ( )dy dr drr r

d d dθ θ θ θ

θ θ θ = + +

2 2

1 1

2 2 22dx dy dr

s d r dd d d

θ θθ θ

θ θ θθ θ

= + = + ∫ ∫

Thus we finally arrive at the formula for polar arc length.

Problem. Find the length of the spiral r = eθ between θ = 0 and θ = π/2.

This curve is as shown below.

Using the arc length formula, we have

( ) ( )2 2 22 2 22 20 0 0

drs r d e e d e d

d

π π πθ θ θθ θ θ

θ = + = + = ∫ ∫ ∫

22( 1) 31.3eπ

= − ≈

Problem. Find the length of the entire circle r = 2acos(θ).

.

2 2

2 2

22 2 2

drs r d ad a

d

π π

π πθ θ π

θ− −

= + = = ∫ ∫

22 22 2 22 sin( ) so 4 cos ( ) sin ( ) 4drdr a r a a

ddθ θ θ

θθ =− + = + =

Problem. Find the length of the polar curve

from θ = 0 to θ = π .

2sin ( /2)r θ=

0sin( ) 2cos( ) 22 2 0

s dπ π

θ θθ = =− =∫

2 2

2 2 2 2

sin( )cos( ) so

22 4 2 2 2sin ( ) sin ( )cos ( ) sin ( )

drd

drrd

θ θ

θ θ θ θ

θ

θ

=

+ = + =

Problem. Find the length of the polar curve

from θ = 0 to θ = π .

3sin ( /3)r θ=

/ 2/ 2

0

/ 21 1 32 2 2sin ( ) 1 cos( ) sin( )3 3 32 2 2 00s d d

ππ πθ θ θθ θ θ

= = − = − ∫∫

3 3

3 3 3 3

2sin ( )cos( ) so

22 6 4 2 4sin ( ) sin ( )cos ( ) sin ( )

drd

drrd

θ θ

θ θ θ θ

θ

θ

=

+ = + =

31 3 3sin2 2 32 4 8π π π = − = −