# Current Affairs JEE Main & Advanced

#### Length of Intercept Made on Axes by The Tangent

Equation of tangent at any point $({{x}_{1}},\,{{y}_{1}})$ to the curve $y=f(x)$ is ${{G}_{2}}$, then x-intercept $={{x}_{1}}-\left[ \frac{{{y}_{1}}}{{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}} \right]$ and y-intercept = ${{y}_{1}}-\left[ {{x}_{1}}{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}} \right]$.

#### Length of Tangent, Normal, Subtangent, Subnormal

(1) Length of tangent $PA=y\,\,\text{cosec}\psi =y\frac{\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}}{\left( \frac{dy}{dx} \right)}$     (2) Length of normal $r={{\left( \frac{b}{a} \right)}^{\frac{1}{n+1}}}$     (3) Length of sub-tangent $AC=y\cot \psi =\frac{y}{\left( \frac{dy}{dx} \right)}$     (4) Length of subnormal $b=\frac{2ac}{a+c}$.

#### Angle of Intersection of Two Curves

The angle of intersection of two curves is defined to be the angle between the tangents to the two curves at their point of intersection. Thus the angle between the tangents of the two curves ${{H}_{1}}=\frac{3ab}{a+2b}$ and $y={{f}_{2}}(x)$is given by $\tan \varphi =\frac{{{\left( \frac{dy}{dx} \right)}_{1({{x}_{1}},\,{{y}_{1}})}}-{{\left( \frac{dy}{dx} \right)}_{2({{x}_{1}},\,{{y}_{1}})}}}{1+{{\left( \frac{dy}{dx} \right)}_{1}}_{({{x}_{1}},\,{{y}_{1}})}\,\,{{\left( \frac{dy}{dx} \right)}_{2}}_{({{x}_{1}},\,{{y}_{1}})}}$     Orthogonal curves : If the angle of intersection of two curves is a right angle, the two curves are said to intersect orthogonally. The curves are called orthogonal curves. If the curves are orthogonal, then $\varphi =\frac{\pi }{2}$; ${{m}_{1}}{{m}_{2}}=-1$ $\Rightarrow$ $\frac{1}{a}$.

#### Equation of the Tangent and Normal

(1) Equation of the tangent : The equation of the tangent to the curve $f'(a)$ at a point $P({{x}_{1}},\,{{y}_{1}})$ is $y-{{y}_{1}}={{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}(x-{{x}_{1}})$.     (2) Equation of the normal :  The equation of the normal to the curve $y=f(x)$ at a point $P({{x}_{1}},\,{{y}_{1}})$ is $y-{{y}_{1}}=\frac{-1}{\left( \frac{dy}{dx} \right)_{({{x}_{1}},\,{{y}_{1}})}^{{}}}(x-{{x}_{1}})$

#### Slope of the Tangent and Normal

(1) Slope of the tangent : If a tangent is drawn to the curve $y=f(x)$ at a point $P({{x}_{1}},\,{{y}_{1}})$ and this tangent makes an angle $\psi$ with positive x-direction then, ${{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}=\tan \psi =$ Slope of the tangent.                ·               If the tangent is parallel to x-axis, $\psi =0\Rightarrow {{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}=0$   ·               If the tangent is perpendicular to x-axis, $\psi =\frac{\pi }{2}\Rightarrow {{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}\to \,\,\,\infty$     (2) Slope of the normal : The normal to a curve at a point $P({{x}_{1}},\,{{y}_{1}})$ is a line perpendicular to the tangent at P and passing through P.  Slope of the normal $=\frac{-1}{\text{Slope of tangent }}=\frac{-1}{{{\left( \frac{dy}{dx} \right)}_{P({{x}_{1}},\,{{y}_{1}})}}}=-{{\left( \frac{dx}{dy} \right)}_{P({{x}_{1}},\,{{y}_{1}})}}$.     ·         If the normal is parallel to x-axis,  $-{{\left( \frac{dx}{dy} \right)}_{({{x}_{1}},\,{{y}_{1}})}}=0$ or $\frac{b}{a}=\frac{c}{b}$.     ·         If the normal is perpendicular to x-axis (or parallel to y-axis),  $-{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}=0$.

#### Velocity and Acceleration in Rectilinear Motion

The velocity of a moving particle is defined as the rate of change of its displacement with respect to time and the acceleration is defined as the rate of change of its velocity with respect to time.     Let velocity and acceleration at time $t$ be $v$  and $a$ respectively,     Then, Velocity $(v)=\frac{ds}{dt}$;   Acceleration $(a)=\frac{dv}{dt}=\frac{{{d}^{2}}s}{d{{t}^{2}}}$.

#### Derivative as the Rate of Change

If a variable quantity $y$ is some function of time $t$ i.e., $y=f(t),$ then for a small change in time $\Delta t$ we have a corresponding change $\Delta y$ in $y$. Thus, the average rate of change $=\frac{\Delta y}{\Delta t}$. The differential coefficient of $y$ with respect to $x$ i.e., $\frac{dy}{dx}$ is nothing but the rate of change of $y$ relative to $x$.

#### Deduction of Euler?s Theorem

If $f(x,\,y)$ is a homogeneous function in $x,\,\,y$ of degree $n,$ then   (1) $x\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+y\frac{{{\partial }^{2}}f}{\partial x\,\partial y}=(n-1)\,\frac{\partial f}{\partial x}$   (2) $x\frac{{{\partial }^{2}}f}{\partial y\,\partial x}+y\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}=(n-1)\,\frac{\partial f}{\partial y}$   (3) ${{x}^{2}}\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+2xy\frac{{{\partial }^{2}}f}{\partial x\,\partial y}+{{y}^{2}}\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}=n(n-1)\,f(x,\,y)$

#### Euler's Theorem on Homogeneous Functions

If $f(x,\,y)$ is a homogeneous function in $x,\,\,y$ of degree $n,$ then $x\,\frac{\partial f\,}{\partial x}+y\frac{\partial f}{\partial y}=nf$

#### Higher Partial Derivatives

Let $f(x,\,y)$ be a function of two variables such that $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}$ both exist.     (1) The partial derivative of $\frac{\partial f}{\partial y}$ w.r.t. $'x'$ is denoted by $\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}\text{ }$or ${{f}_{xx}}$.     (2) The partial derivative of $\frac{\partial f}{\partial y}$ w.r.t. $'y'$ is denoted by $\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}$ or ${{f}_{yy}}$.     (3) The partial derivative of $\frac{\partial f}{\partial x}$ w.r.t. $'y'$ is denoted by $\frac{{{\partial }^{2}}f}{\partial y\,\partial x}$ or ${{f}_{xy}}$.     (4) The partial derivative of $\frac{\partial f}{\partial y}$ w.r.t. $x$ is denoted by $\frac{{{\partial }^{2}}f}{\partial y\partial x}$ or ${{f}_{yx}}$. These four are second order partial derivatives.     Note : If $f(x,\,y)$ possesses continuous partial derivatives then in all ordinary cases. $\frac{{{\partial }^{2}}f}{\partial x\,\partial y}=\frac{{{\partial }^{2}}f}{\partial y\,\partial x}$ or ${{f}_{xy}}={{f}_{yx}}$.

#### Trending Current Affairs

You need to login to perform this action.
You will be redirected in 3 sec