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 = π/4
3. Find the slope & the equation of the
tangent at t = π/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)( π/4) = -(3/4)tan( π/4) = -(3/4)
(d2y/dx2 )( π/4) = (-3/16) sec3( π/4) = -(3/16)(√2) 3 = - 3√2/8
Part 2
The slope of the tangent at t = π/4 is equal
to the derivative dy/dy at t = π/4, which is
-(3/4).
The Cartesian coordinates of the point
t = π/4 are:
x = 4sint π/4 = 4(1/√2)= 2 √2 and
y = 3cos π/4= 3(1/√2)= (3/2) √2
The equation of the tangent at t = π/4 is:
y -(3/2) √2 = -(3/4) ( x - 2 √2 )
Example 2
Let x = 4sint, y = 3cost.
Find:
1. dy/dx and d2y/dx2 at the point (0, -3 )
2. 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 = π. 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)( π) = -(3/4)tan( π) = 0
(d2y/dx2 )( π) = (-3/16) sec3( π)
= -(3/16)(-1) 3 = 3/16
*Answering “Why?”
We have:
0 = 4sint, -3 = 3cost.
→ sint = 0 & cost = -1
→ t = ………,-3π, -π, π, 3π, 5π,…..
Notice that the values for any trigonometric
function at any of these numbers (angles) are the
same. Take: t = π
Solution – Part 2
From the slope of the tangent at (0 ,-3 ) it is clear
that the tangent is horizontal, and hence it’s
equation is: y = -3
We could also get that from the straight line’s
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)/(2√3 )= √3
At t= -√3, we have dy/dx=(9-3)/(-2√3 )= -√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)
That’s:
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)
What’s 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)
What’s 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 = π/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)( π/3) = sin( π/3) / [1 - cos( π/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 = nπ and t ≠ 2nπ
Why?
→ t=(2n-1) π ; n is an integer.
→x =r[(2n-1) π - 0)] = r(2n-1) π , y =r(1 – (-1) = 2r,
Thus the curve has horizontal tangent at the points:
(r(2n-1) π ,2r)
Examples: ….,(-3rπ ,2r), (-rπ ,2r), (rπ ,2r), (3rπ ,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 = 2nπ ; n is an integer Why?
→x =r[(2nπ - 0)] = 2rnπ , y =r(1 – (1) = 0
Checking: show that dy/dx → + ∞ as t → 2nπ from the
right*
Thus the curve has vertical tangent at the points:
(2nrπ , 0)
Examples:…, (-4rπ , 0), (-2rπ , 0), (0, 0), (2rπ , 0), (4π , 0),
….
Checking: show that dy/dx → + ∞ as t →2nπ
from the right
lim
t  2 n
dy

dx
 lim
t  2 n
 lim
t  2 n
 lim
t  2 n

sin t

1  cos t
cos t


sin t
cot( t )