
The chain rule
applications 
Derivative of
parametric functions, parametric derivatives 
Derivative of
parametric functions example

Derivative
of vectorvalued functions 






Derivative of
parametric functions or parametric derivatives 
When
Cartesian coordinates of a curve is represented as
functions of the same variable (usually written t),
they are
called the parametric equations. 

Thus,
parametric equations in the xyplane 
x
= x (t)
and y
= y (t)
or x
= f (t)
and y
= g (t), 
denote
the x
and y
coordinate of the graph of a curve in the plane. 
Assume
that f
and g
are differentiable and f
'(t) is not
0 then, given parametric curve can be
expressed as y
= y (x) and this
function is differentiable at x,
that is 

x
= f (t)
or
t = f ^{1}(x), 
by plugging
into y
= g (t)
obtained is y
= g [ f ^{1}(x)]. 
Therefore,
we use the chain rule and the derivative of the inverse function to find
the derivative of the parametric
functions, 



Derivative of
parametric functions example

Example:
Write
equation of the line tangent to the curve x
= t
+
1 and
y
=

t^{2}
+
4
at the point t
= 1.

Solution:
The equation x
=
t + 1
solve for t
and plug into y
=
 t^{2} + 4,
thus 
t
=
x
 1
=> y
=

t^{2} + 4,
y
=
 (x
 1)^{2} + 4 
i.e., y

4
=
 (x
 1)^{2}
or y
=
 x^{2} + 2x + 3 translated parabola with the vertex V(x_{0},
y_{0}),
so V(1,
4). 
When
plotting points of a parametric curve by increasing t, the
graph of the function is traced
out in the direction of motion. 

The
derivative of the given parametric equations at t
= 1 is the slope of the tangent
line, 

since
t
=
x
 1
then m
=
y' (x)
=

2(x
 1)
= 
2(2
 1)
=

2,
m
=

2.

Therefore,
the equation of the line tangent to the given parametric curve at t
= 1 or the point P_{1}(2,
3) is 
y
 y_{1}
= m(x
 x_{1}),
y
 3
= 
2(x
 2) =>
y
= 
2x
+ 7. 

Derivative
of vectorvalued functions 
If
the radius vector r
of a point in a plane depends on a parameter t,
say t
represents time, so that its magnitude
and direction change continuously while t
changes, then its arrow sweeping out a curve. 
Let
r (t)
denotes its value at the moment t
and r (t
+ h)
represents its value at t
+ h,
and P
and P_{1}
are the corresponding
points of the curve, as is shown in the figure below. 
The
increment Dr
= r (t +
h)

r (t)
is the vector that falls in the direction of the secant line
PP_{1
}
and points from P
to P_{1}.
The difference quotient 

obtained
by division with the scalar
h,
is the vector of the 
same
direction but of different length. 
The
limit of the difference quotient as
h
®
0 

is
the
derivative vector of the
vectorvalued
function that
falls 
in
the direction of the line tangent to the curve at P. 



If
x (t)
and y (t)
are the scalar components of the vector r
(t)
then, according to rules of
vector algebra, 
Dr
= [x (t +
h)

x (t)] i
+ [y (t +
h)

y (t)] j 
and
by use of the definition of the derivative 

is
the derivative vector,
where x' (t)
and
y' (t)
are its scalar components, and where 

is
its length or magnitude. 









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