# Section 2 Parametric Differentiation. Theorem Let x = f(t), y = g(t) and dx/dt is nonzero, then dy/dx

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Example 1 Let x = 4sint, y = 3cost. Find: 1. dy/dx and d 2 y/dx 2 2. dy/dx and d 2 y/dx 2 at t = π /4 3. Find the slope & the equation of the tangent at t = π /4

### Text of Section 2 Parametric Differentiation. Theorem Let x = f(t), y = g(t) and dx/dt is nonzero, then...

Section 2

Parametric Differentiation

Theorem

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

and dx/dt is nonzero, then

dy/dx = (dy/dt) / (dx/dt)

; provided the given derivatives exist

Example 1

Let x = 4sint, y = 3cost.

Find:1. dy/dx and d2y/dx2

2. dy/dx and d2y/dx2 at t = /4

3. Find the slope & the equation of the tangent at t = /4

Solution- Part1

dx/dt = 4cost, dy/dt = -3sint

dy/dx = (dy/dt) / (dx/dt) = - 3sint / 4cost

= -(3/4)tant

d(dy/dx)/dt = -(3/4)sec2t

d2y/dx2 = d(dy/dx )/dx= [d(dy/dx)/dt] / [dx/dt]

= [-(3/4)sec2t ] / 4cost = (-3/16) sec3t

(dy/dx)( /4) = -(3/4)tan( /4) = -(3/4)

(d2y/dx2 )( /4) = (-3/16) sec3( /4) = -(3/16)(2) 3 = - 32/8

Part 2

The slope of the tangent at t = /4 is equal to the derivative dy/dy at t = /4, which is

-(3/4).

The Cartesian coordinates of the point

t = /4 are:

x = 4sint /4 = 4(1/2)= 2 2 and

y = 3cos /4= 3(1/2)= (3/2) 2

The equation of the tangent at t = /4 is:

y -(3/2) 2 = -(3/4) ( x - 2 2 )

Example 2

Let x = 4sint, y = 3cost.

Find:

dy/dx and d2y/dx2 at the point (0, -3 )

The equation of the tangent to the curve at that point.

Solution Part 1

. Let x = 4sint, y = 3cost.

First we find any value of t corresponding to the point (0, -3 ).

It is clear that one such value is t = . Why?*

Now, we substitute that in the formulas of dy/dx

and d2y/dx2 , which we have already deduced in the Example(1) (dy/dx)( ) = -(3/4)tan( ) = 0 (d2y/dx2 )( ) = (-3/16) sec3( )

= -(3/16)(-1) 3 = 3/16

We have:

0 = 4sint, -3 = 3cost.

sint = 0 & cost = -1

t = ,-3, -, , 3, 5,..

Notice that the values for any trigonometric

function at any of these numbers (angles) are the

same. Take: t =

Solution Part 2

From the slope of the tangent at (0 ,-3 ) it is clear that the tangent is horizontal, and hence its equation is: y = -3

We could also get that from the straight lines formula:

y (-3)= ( x -0 )

y + 3 =0

y = - 3

But this is not very smart. It is like catching a fly with a hammer!

Book Example (1)

Let x =t2, y = t3 - 3t.

1. Find the equations of all tangents at (3,0)

2. Determine at which point (points), the graph has a horizontal tangent.

3. . Determine at which point (points), the graph has a vertical tangent.

*

4. . Determine when the curve is concave upward / concave downward

Solution-Part 1

We have:

dx/dt = 2t , dy/dt = 3t2 -3

dy/dx = (dy/dt) / (dx/dt) = (3t2 -3) / t

When x =3 & y=0 3=t2 t=3 or t= -3

At t=3, we have dy/dx=(9-3)/(23 )= 3

At t= -3, we have dy/dx=(9-3)/(-23 )= -3

Thus, the equations of the tangents to the

curve at (x,y) = (3,0) are:

y - 0 = 3 (x - 3) & y - 0 = -3 (x - 3)

Thats:

y = 3 (x - 3) & y = -3 (x - 3)

Solution-Part 2

We have: x =t2, y = t3 - 3t

dx/dt = 2t , dy/dt = 3t2 -3

dy/dx = (dy/dt) / (dx/dt) = (3t2 -3) / 2t

The curve has horizontal tangent when dy/dt = 3t2 -3 =0, while dx/dt = 2t 0 t = 1 or t = -1 Why?

At t =1 x=(1) 2=1 & y = (1)3 3(1) = 1 - 3 = -2

At t =-1 x=(-1) 2=1 & y = (-1)3 3(-1) = -1 + 3 = 2

Thus the curve has horizontal tangent at the points:

(1,2) & (1,-2)

Whats the equations of these tangents?

Solution-Part 3

We have: x =t2, y = t3 - 3t

dx/dt = 2t , dy/dt = 3t2 -3

dy/dx = (dy/dt) / (dx/dt) = (3t2 -3) / 2t

The curve has vertical tangent when dy/dt = 3t2 -3 0, while dx/dt = 2t = 0 t = 0 Why?

At t =0 x=(0) 2=0 & y = (0)3 3(0) = 0

Thus the curve has vertical tangent at the point (0,0)

Whats the equations of this tangent?

Solution-4**

We have:

dx/dt = 2t , dy/dt = 3t2 -3

dy/dx = (dy/dt) / (dx/dt) = (3t2 -3) / 2t = (3/2)t (3/2)t-1

d2y/dx2 = d(dy/dx )/dx= [d(dy/dx)/dt] / [dx/dt]

= [(3/2) +(3/2)t-2 ] / 2t = (3/4) t-1 + (3/4)t-3

= [3t2 + 3] / 4t3

d2y/dx2 > 0 if t > 0 & d2y/dx2 < 0 if t < 0

Thus the curve is concave upward if t > 0 and downward if t < 0

We had: x =t2, y = t3 - 3t (t > 0 on the first quadrant. Why? and t < 0 on the fourth quadrant. Why?

,

Book Example (2)

Let x =r(t - sint), y =r(1 - cost), Where r is a constant

1. Find the slope of the tangent at t = /3

2. Determine at which point (points), the graph has a horizontal tangent.

3. . Determine at which point (points), the graph has a vertical tangent

Solution-Part1

We have:

x =r(t - sint), y =r(1 - cost),

dx/dt = r(1 - cost), , dy/dt = rsint

dy/dx = (dy/dt) / (dx/dt) = rsint / r(1 - cost)

= sint / (1 - cost)

(dy/dx)( /3) = sin( /3) / [1 - cos( /3)] = (3/2)/[1-(1/2)]

= 3

Solution-Part 2

We have:

x =r(t - sint), y =r(1 - cost),

dx/dt = r(1 - cost), , dy/dt = rsint

dy/dx = (dy/dt) / (dx/dt) = rsint / r(1 - cost)

= sint / (1 - cost)

The curve has horizontal tangent when dy/dt = rsint

=0, while dx/dt = r(1 - cost), 0 t = n and t 2n Why?

t=(2n-1) ; n is an integer.

x =r[(2n-1) - 0)] = r(2n-1) , y =r(1 (-1) = 2r,

Thus the curve has horizontal tangent at the points:

(r(2n-1) ,2r)

Examples: .,(-3r ,2r), (-r ,2r), (r ,2r), (3r ,2r),

Solution-Part 3

We have:

x =r(t - sint), y =r(1 - cost),

dx/dt = r(1 - cost), , dy/dt = rsint

dy/dx = (dy/dt) / (dx/dt) = rsint / r(1 - cost)

= sint / (1 - cost)

r(1 - cost), = 0 t = 2n ; n is an integer Why?

x =r[(2n - 0)] = 2rn , y =r(1 (1) = 0

Checking: show that dy/dx + as t 2n from the

right*

Thus the curve has vertical tangent at the points:

(2nr , 0)

Examples:, (-4r , 0), (-2r , 0), (0, 0), (2r , 0), (4 , 0), .

Checking: show that dy/dx + as t 2n from the right

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