tg

tg
tgt

Saturday, 2 August 2014

CHAPTER 5- Differentiation of Parametric and Implicit Functions

(A) PARAMETRIC FUNCTONS
Sometimes, when expressing y as a function of x, one might not use a direction relation between x and y; instead, one might express both x and y as functions of a third variable, say t:
x = f\left( t \right)
y = g\left( t \right)
In that case, how would \dfrac{{dy}}{{dx}} be evaluated?
One option is to eliminate the parameter t and obtain a relation involving only x and y, from which \dfrac{{dy}}{{dx}} may be obtained; however, this could lead to cumbersome expressions.
Another alternative can be taken as follows; we rearrange \dfrac{{dy}}{{dx}} to involve t also:
\dfrac{{dy}}{{dx}} = \,\,\dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}
This relation says that for evaluating the derivative of y w.r.t x, we evaluate the derivative of y and x w.r.t the parameter t, and then take their ratio.
Let us try this on some examples:
(i) x = r\cos \theta \,\,\,\,\,\,\,y = r\sin \theta;\,\,\,\,\,\,\, r \,{\rm{is\, a\, constant}}
This parametric relation represents a circle of radius r. We will follow both the approaches to determine \dfrac{{dy}}{{dx}}:
 \Rightarrow  ELIMINATION:
Square and add the two relations for x and y to obtain:
{x^2} + {y^2} = {r^2}
 \Rightarrow  \,y =  \pm \sqrt {{r^2} - {x^2}}
 \Rightarrow  \, \dfrac{{dy}}{{dx}} =  \pm \dfrac{x}{{\sqrt {{r^2} - {x^2}} }}  \,\,\,\,\left( \begin{array}{l}  {\rm{For\, each }}\,x,\,\,{\rm{we}}\,\,{\rm{obtain}}\,\,{\rm{two}}\,y'\,\\  {\rm{values}}\,{\rm{because}}\,{\rm{the}}\,{\rm{curve}}\,{\rm{is}}\,{\rm{a}}\,{\rm{circle}}  \end{array} \right)
 \Rightarrow  PARAMETRIC DIFFERENTIATION
\dfrac{{dx}}{{d\theta }} =  - r\sin \theta
  \dfrac{{dy}}{{d\theta }} = r\cos \theta
 \Rightarrow \,  \,\dfrac{{dy}}{{dx}} = \dfrac{{dy/d\theta }}{{dx/d\theta }} =  - \cot \theta
(ii) x = a\cos \theta \,\,\,\,\,\,\,y = b\sin \theta;\,\,\,\,\,\,\, a \,{\rm{and}}\, b\, {\rm{are \,constants}}
This parametric relation represents an ellipse with major and minor axis 2a and2b respectively.
 \Rightarrow  ELIMINATION
\theta  can easily be eliminated to obtain:
\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1
 \Rightarrow \,  y =  \pm \dfrac{b}{a}\sqrt {{a^2} - {x^2}}
 \Rightarrow \,  \dfrac{{dy}}{{dx}} =  \pm \,\,\dfrac{b}{a}\,\,\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}
 \Rightarrow  PARAMETRIC DIFFERENTIATION
\dfrac{{dx}}{{d\theta }} =  - a\sin \theta
\dfrac{{dy}}{{d\theta }} = b\cos \theta
 \Rightarrow \,\,  \dfrac{{dy}}{{dx}} = \dfrac{{dy/d\theta }}{{dx/d\theta }} = \dfrac{{ - b}}{a}\,\,\cot \theta
In later examples, we will observe that in many cases, parametric differentiation turns out to be much more convenient than differentiation after elimination.
(B) IMPLICIT FUNCTIONS
Sometimes, the relation between the variables x and y is specified in the form f(x,y) = 0 that is, y is not explicitly specified in terms of x, since this explicit expression is either not possible or not convenient.
In such a case, y is said to be an implicit function of x.
How do we find \dfrac{{dy}}{{dx}} in such a case?
We simply differentiate the relation f(x,y) = 0 with respect to x, using \dfrac{{dy}}{{dx}} for the derivative of the variable y. Then we solve for \dfrac{{dy}}{{dx}}.
This will become clear from some examples:
 \Rightarrow\,\, {x^2} + {y^2} = 1
Differentiating both sides w.r.t x:
2x + 2y\dfrac{{dy}}{{dx}} = 0
 \Rightarrow  \,\, \dfrac{{dy}}{{dx}} = \dfrac{{ - x}}{y}
 \Rightarrow  \,\, {x^3} + {y^3} + 2xy = 2
Differentiating both sides w.r.t x:
3{x^2} + 3{y^2}\dfrac{{dy}}{{dx}} + 2x\dfrac{{dy}}{{dx}} + 2y = 0
 \Rightarrow  \,\, \dfrac{{dy}}{{dx}} = \dfrac{{ - \left( {3{x^2} + 2y} \right)}}{{3{y^2} + 2x}}
 \Rightarrow  \,\, y = \cos \left( {x + y} \right)
Differentiating both sides w.r.t x:
\dfrac{{dy}}{{dx}} =  - \sin \left( {x + y} \right)\left( {1 + \dfrac{{dy}}{{dx}}} \right)
 \Rightarrow  \,\, \dfrac{{dy}}{{dx}} = \dfrac{{ - \sin \left( {x + y} \right)}}{{1 + \sin \left( {x + y} \right)}}
Observe that in case of differentiation of implicit functions, the expression for the derivative \dfrac{{dy}}{{dx}} will generally not be independent of y.
Post a Comment