Notes - Mathematics Olympiads -Application of Derivatives
Category : 12th Class
Application of Derivatives
\[x=a\] then \[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\,\,f(x)=f(a)\]
Generally, a function is said to be continuous at \[x=a\] when the graph of that function can be drawn/sketched without lefting the pencil.
Then function \[y=f(x)\] is differentiable at \[x=c\]
i.e. \[\underset{x\to c}{\mathop{\lim }}\,\frac{f(x+c)-(c)}{c}=f'(c)\]
Solved Example
Let \[y=f(x)=\sin x\] ...(1)
Let \[\delta \]x be the small increment in x then \[\delta \]y be the corresponding increment in y.
\[y+\delta y=\sin (x+\delta x)\] ...(2)
Now, on subtracting equation (1) from (2), we get
\[y+\delta y-y=\sin (x+\delta x)-\sin x\]
\[\delta y=2.\cos \frac{x+\delta x+x}{2}.\sin \left( \frac{x+\delta x+x}{2} \right)\]
Dividing \[\delta \]x on both sides and taking limit \[\delta x\to 0,\]we get
\[\underset{\delta x\to 0}{\mathop{\lim }}\,\frac{\delta y}{\delta x}=\underset{\delta x\to 0}{\mathop{\lim }}\,2.\frac{\cos \left( \frac{2x+\delta x}{2} \right).\sin \left( \frac{\delta x}{2} \right)}{\delta x}\]
\[\frac{dy}{dx}\underset{\delta x\to 0}{\mathop{\lim }}\,2.\cos \left( x+\frac{\delta x}{2} \right).\left( \frac{\sin \frac{\delta x}{2}}{\frac{\delta x}{2}\times 2} \right)\]
On applying limit \[\delta x\to 0,\]we get
\[\frac{dy}{dx}=\cos x\times 1=\cos x\left[ \underset{\theta \to 0}{\mathop{\lim }}\,\frac{\sin \theta }{\theta }=1 \right]\]
Thus,\[\frac{d}{dx}(\sin x)=\cos x\]
\[\frac{dy}{dx}\]is said to be differential coefficient of \[y=f(x).\] It is denoted by\[{{y}_{1}}\]or f \['(x)\]
e.g. \[\frac{d}{dx}(5{{x}^{2}})=5.\frac{d}{dx}({{x}^{2}})=5[2.{{(x)}^{2-1}}]=10{{x}^{1}}=10x\]
e.g. \[\frac{d}{dx}({{e}^{x}}.\sin x)={{e}^{x}}.\frac{d}{dx}(\sin x)+\sin x\frac{d}{dx}({{e}^{x}})={{e}^{x}}\cos x+\sin x.{{e}^{x}}\]
\[={{e}^{x}}(\cos x+\sin x)\]
Geometrical meaning of derivative at point: The derivative of a function \[f(x)\] at a point \[x=a\] is the of the tangent of the curve \[y=f(x)\] at the point \[(a,f(a)).\]
Let us consider a curve \[y=f(x)\] & take a point P\[(x,y)\] on it. We draw the tangent to the curve at \[P(x,y)\]which makes an angle \[\alpha \] with positive direction of x-axis. Then,
\[{{\left. \frac{dy}{dx} \right|}_{\,\,at\,\,P(x,y)\,=\,tan\,\,\alpha \,=\,m(say)}}\]
It is said to be gradient or slope of the tangent to the curve \[y=f(x)\]at\[p(x,y)\].
\[y-{{y}_{1}}=m(x-{{x}_{1}})\]
where m = slope of the tangent
\[y-{{y}_{1}}={{\left. \frac{-dy}{dx} \right|}_{\,\,at\,\,P({{x}_{1}},\,{{y}_{1}})\,\,\times \,\,(x-{{x}_{1}}).}}\]
\[y-{{y}_{1}}=\frac{-1}{m}({{x}_{1}}-{{x}_{1}})\,\,y-{{y}_{1}}=\frac{-1}{{{\left. \frac{dy}{dx} \right|}_{\,\,at\,\,P({{x}_{1}},{{y}_{1}})}}}\times (x-{{x}_{1}})\]
Note: Since tangent & normal to a curve \[y=f(x)\] at point \[({{x}_{1}},{{y}_{1}})\] is perpendicular to each other at that point.
\[\because \,\,\,{{m}_{1}}.{{m}_{2}}=-1\]
Where, \[{{m}_{1}}=\] slope of the tangent to the curve
\[{{m}_{2}}=\] slope of the normal to the curve
Solved Examples
Sol. The given equation of the curve is
\[y=2{{x}^{2}}+3\sin x\] ....(1)
Putting\[x=0\]
then \[y=2\times 0+3\text{ }sin0=0.\]
Here, the point of the contact is (0, 0).
\[\because \] Differentiating (1) w.r.t. x, we get
\[\frac{dy}{dx}=2(2x)+3.\cos x{{\left. \frac{dy}{dx} \right|}_{\,\,at\,\,(0,0)}}=4\times 0+3cos({{0}^{\underline{o}}})=0+3=3\]
i.e. \[{{m}_{1}}=3\]
\[\therefore \] Equation of the tangent is
\[y-{{y}_{1}}=m(x-{{x}_{1}})\]
\[\Rightarrow y-0=3(x-0)\]
\[y=3x\Rightarrow 3x-y=0\]
Now, equation of the normal is
\[y-{{y}_{1}}=\frac{-1}{m}(x-{{x}_{1}})\]
\[\Rightarrow y-0=\frac{-1}{3}(x-0)\]
\[\Rightarrow 3y=-x\Rightarrow x+3y=0\]
\[ax+by+c=0\] is written as
\[m=\frac{-a}{b}=\frac{\text{coefficient of }x\text{ with negative sign}}{\text{coefficient of y}}\]
(a) perpendicular if \[{{m}_{1}}.{{m}_{2}}=-1\]
(b) parallel if \[{{m}_{1}}={{m}_{2}}\]
Let \[{{m}_{1}}=\] slope of the tangent to curve \[y=f(x)\] at \[P(x,y).\]
\[{{m}_{2}}=\] slope of the tangent to curve \[y=g(x)\] at \[P(x,y)\]
(In a triangle, sum of two interior angles is equal to an exterior angle.)
\[\therefore \,\,\,\,\alpha +\theta =\beta \]
So, \[\alpha \]be the angle of the tangent in the positive direction of x-axisa and \[\beta \] be the angle of the tangent to curve \[y=g(x)\] in the positive direction of x-axis.
\[\Rightarrow \theta =\beta -\alpha \Rightarrow \tan \theta =\tan (\beta -\alpha )=\frac{\tan \beta -\tan \alpha }{1+\tan \beta .\tan \alpha }\]
\[\Rightarrow \tan \theta =\frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{2}}.{{m}_{1}}},\,\,\,\theta ={{\tan }^{-1}}\left| \left. \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{2}}.{{m}_{1}}} \right| \right.\]
Let the tangent and normal on the point P(x, y) at the curve meet x-axis at T and N respectively. Let M be the foot of the ordinate at P. Then
In \[\Delta \text{PTM},\,\sin \theta =\frac{y}{\text{PT}}\Rightarrow PT=y\cos ec\,\theta \]
\[=y.\sqrt{1+\frac{1}{{{\left( \frac{dy}{dx} \right)}^{2}}}}=\frac{\sqrt[y]{1+{{\left( \frac{dy}{dx} \right)}^{2}}}}{\left( \frac{dy}{dx} \right)}\]
Similarly,
Length of the normal \[=PN=y.\sec \theta =y.\sqrt{1+{{\tan }^{2}}\theta }=y\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}\]
Length of the sub-tangent \[(TM)=y.cot\theta =\frac{y}{\tan \theta }=\frac{y}{\left( \frac{dy}{dx} \right)}\]
Length of the subnormal \[(MN)=y.\tan \theta =\left| \left. y.\frac{dy}{dx} \right| \right.\]
Solved Examples
Sol. Let AB is a lamp post. At any instant t, A man MN away x m from AB and its shadow is y metre. Given that
\[AB=5m\]
\[MN=2m\] and \[\frac{dx}{dt}=6\]metre / min
\[AM=x\]metre and \[MC=y\] metre.
Since, \[\Delta ABC\] and \[\Delta MNC\] be similar
So, \[\frac{MN}{AB}=\frac{MC}{AC}\] \[\Rightarrow \,\,\,\,\frac{AB}{MN}=\frac{AC}{MC}\]
\[\Rightarrow \,\,\,\,\frac{5}{2}=\frac{x+y}{y}\] \[\Rightarrow \,\,\,\,\frac{5}{2}=1+\frac{x}{y}\]
\[\Rightarrow \,\,\,\,\frac{5}{2}-1=\frac{x}{y}\] \[\Rightarrow \,\,\,\,\frac{3}{2}=\frac{x}{y}\,\,\,\,\Rightarrow 3y=2x\]
Differentiating w.r.t. to t, we get
\[3.\frac{dy}{dt}=2\left( \frac{dx}{dt} \right)\] \[\Rightarrow \,\,\,\,\frac{dy}{dt}=\frac{2}{3}\times 6=4\]m/sec
Sol. Let \[y=f(x)=\sqrt{x}\]
Let \[x=0.040\] and \[x+\delta x=0.037\]
\[\delta x=x+\delta x-x=0.037-0.040=-0.003\]
For
\[\because y=f(x)=\sqrt{0.04}=0.2\]
\[\therefore dy=\left( \frac{\delta y}{\delta x} \right).dx\]
\[\therefore \frac{dy}{dx}=\frac{\delta }{\delta x}(\sqrt{x})=\frac{1}{2\sqrt{x}}\le \]
\[{{\left. \frac{dy}{dx} \right|}_{\,\,at\,\,x=0.04}}=\frac{1}{2\sqrt{x}}=\frac{1}{2\times 0.2}=\frac{1}{0.4}\]
So, \[\delta y={{\left( \frac{\delta y}{\delta x} \right)}_{at\,\,x=0.04}}\times \delta x=\frac{1}{0.4}\times (-0.003)=-\frac{3}{400}\]
\[\therefore \] Approximate value of \[y=f(x)=y+\delta y=0.2+\left( \frac{-3}{400} \right)=\frac{80-3}{400}=\frac{77}{400}\]
(i) A function \[y=f(x)\] is said to be increasing on an interval \[[a,b]\]if\[{{x}_{1}}<{{x}_{2}}\]in\[(a,b)\]\[\Rightarrow f({{x}_{1}})\le f({{x}_{2}})\]\[\forall \,{{x}_{1}},{{x}_{2}}\in (a,b)\]
(ii) A function \[y=f(x)\] is said to be strictly increasing on\[(a,b)\]if\[{{x}_{1}}<{{x}_{2}}\]in\[(a,b)\]\[\Rightarrow f({{x}_{1}})<f({{x}_{2}})\] for all \[{{x}_{1}},{{x}_{2}}\in (a,b)\]
(i) A function \[y=f(x)\] is said to be decreasing on the interval\[(a,b)\] if \[{{x}_{1}}>{{x}_{2}}\]in\[(a,b)\]\[\Rightarrow f({{x}_{1}})\ge f({{x}_{2}})\]\[\forall \,{{x}_{1}},{{x}_{2}}\in (a,b)\]
(ii) A f unction\[y=f(x)\] is said to be strictly decreasing on \[(a,b)\] if\[{{x}_{1}}>{{x}_{2}}\]in\[(a,b)\]\[\Rightarrow f({{x}_{1}})>f({{x}_{2}})\]\[\forall \,{{x}_{1}},{{x}_{2}}\in (a,b)\]
(i) \[f(x)\] is increasing in \[[a,b]\] if \[f'(x)\ge 0\]\[\forall x\in [a,b]\]
(ii) \[f(x)\] is strictly increasing in \[[a,b]\]if \[f'(x)>0\]\[\forall x\in [a,b]\]
(iii) \[f(x)\] is decreasing in \[[a,b]\] if \[f'(x)\le 0\]\[\forall x\in [a,b]\]
(iv) \[f(x)\] is strictly decreasing in \[[a,b]\]if \[f'(x)<0\]\[\forall x\in [a,b]\]
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