# Applications of Derivatives Set 8

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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 $$c \in [0,\,\,1]$$

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