Applications of Derivatives Set 8

Example - 16

Find the condition that the line \(x\cos \alpha  + y\sin \alpha  = p\) may touch the curve  \(\begin{align}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\)

Solution: The approach we can follow for our purpose here is to first write down the equation of a general tangent to the ellipse at a general variable point \(\left( {a\cos \theta ,b\sin \theta } \right)\) and then make the equation of this tangent identical to the equation of the given line. We can then eliminate \(\theta \) to get some condition on p and \(\alpha\).

We wrote the equation of the required general tangent on an ellipse in example 9.

\[bx\cos \theta  + ay\sin \theta  = ab\]

This is identical to \(x\cos \alpha  + y\sin \alpha  = p\)  for some \(\theta \). Therefore:

\[\begin{align}&\quad\quad\frac{{b\cos \theta }}{{\cos \alpha }} = \frac{{a\sin \theta }}{{\sin \alpha }} = \frac{{ab}}{p}\\& \Rightarrow \quad   \cos \theta  = \frac{{a\cos \alpha }}{p}\,\,\,\,{\rm{and}}\,\,\,\sin \theta  = \frac{{b\sin \alpha }}{p}\\&\Rightarrow  \quad {\cos ^2}\theta  + {\sin ^2}\theta  = \frac{{{a^2}{{\cos }^2}\alpha }}{{{p^2}}} + \frac{{{b^2}{{\sin }^2}\alpha }}{{{p^2}}} = 1\end{align}\]

Therefore, the required condition is:

\[{a^2}{\cos ^2}\alpha  + {b^2}{\sin ^2}\alpha  = {p^2}\]

 

Example - 17

If \(f'\left( \sin x \right)<0\,\,\text{and}\,\,f''\left( \sin x \right)>\,0\,\,\,\,\,\forall x\in \mathbb{R},\) then find the intervals of monotonicity of the function \(g\left( x \right) = f\left( {\sin x} \right) + f\left( {\cos x} \right),\,\,x \in [0,\pi /2].\)

Solution: To obtain the intervals of monotonicity of g(x), we need to analyses g'(x).

\[g'\left( x \right) = f'\left( {\sin x} \right)\cos x - f'\left( {\cos x} \right)\sin x\]

This is 0 when

\[\begin{align}&\qquad\;\; f'\left( {\sin x} \right)\cos x = f'\left( {\cos x} \right)\sin x\\\\& \Rightarrow \quad  f'\left( {\sin x} \right)\cos x = f'\left( {\sin \left( {\frac{\pi }{2} - x} \right)} \right)\cos \left( {\frac{\pi }{2} - x} \right)\end{align}\]

One of the possible roots of this equation is given by:

\[\begin{align}&\qquad\;\; x = \frac{\pi }{2} - x\\\\ &\Rightarrow \quad  x = \frac{\pi }{4}\end{align}\]

Now, we analyze g''(x):

\[g''\left( x \right)=f''\left( \sin x \right){{\cos }^{2}}x+f''\left( \cos x \right){{\sin }^{2}}x-\left\{ f'\left( \sin x \right)\sin x+f'\left( \cos x \right)\cos x \right\}..............(i)\]

It is given that                                \(f'\left( {\sin x} \right) < 0\,\,\,\,\,\forall x \in \mathbb{R}\)

\(\begin{align}\Rightarrow \quad \text{ }\!\!~\!\!\text{ }{f}'\left( \cos x \right)<0\,\,\,\,\, \forall x\in\mathbb{R}\qquad\qquad\quad\left\{ \because \cos x=\sin \left( \frac{\pi }{2}-x \right) \right\}\end{align}\)

Similarly,

 \(\begin{align} &\qquad\;\;\;  f''\left( \sin x \right)>0\,\,\,\,\,\,\forall x\in \mathbb{R} \\\\ & \Rightarrow\quad  f''\left( \cos x \right)>0\,\,\,\,\,\,\forall x\in \mathbb{R}\,\,\,\,\,\,\,\,\,\,\,\{\text{same reason as above}\} \\ \end{align}\)

From (i), observe carefully that these two conditions above imply:

\[g''\left( x \right)>0\,\,\forall x\in \mathbb{R}\]

Since g''(x) is always positive, the graph for g(x) is a concave upwards curve, so that g(x) has only one extremum point (a minimum) which we deduced as  \(\begin{align}x={}^{\pi }\!\!\diagup\!\!{}_{4}\;\end{align}\)

Therefore:

\(\begin{align}& g(x)\,\text{decreases on}\,\,\left( 0,{}^{\pi }\!\!\diagup\!\!{}_{4}\; \right) \\\\ & g(x)\,in\text{creases on}\,\,\left( {}^{\pi }\!\!\diagup\!\!{}_{4}\;,\,\,{}^{\pi }\!\!\diagup\!\!{}_{2}\; \right) \\\\ & x={}^{\pi }\!\!\diagup\!\!{}_{4}\;\,\text{is minimum for}\,\,g(x) \\ \end{align}\)

 

Example - 18

If f(x) and g(x) are differentiable functions for \(0 \le x \le 1\) such that\(f\left( 0 \right) = 2,g\left( 0 \right) = 0,\,\,f\left( 1 \right) = 6\)   and \(g\left( 1 \right) = 2,\) then show that there exists \(c \in \left( {0,1} \right)\) such that\(f'\left( c \right) = 2g'\left( c \right)\).

Solution: The nature of the proof required hints that we have to use one of the Mean Value Theorems. We construct a new function for this purpose:

\[h\left( x \right) = f\left( x \right) - 2g\left( x \right)\]

Now,

\[h\left( 0 \right) = f\left( 0 \right) - 2g\left( 0 \right) = 2\]

and

\[h\left( 1 \right) = f\left( 1 \right) - 2g\left( 1 \right) = 2\]

Also, since f(x) and g(x) are differentiable on [0, 1], h (x) must also be differentiable on [0, 1]. Therefore, Rolle’s theorem can be applied on h(x) for the interval [0, 1]:

There exists \(c \in [0,\,1]\) such that

                 \(h'\left( c \right) = 0\)

    \( \Rightarrow  f'\left( c \right) - 2g'\left( c \right) = 0\) for some \(c \in [0,\,\,1]\)

    \( \Rightarrow  f'\left( c \right) = 2g'\left( c \right)\)  for some \(c \in [0,\,\,1]\)