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Application of Derivatives

85800 The angle between the curves, $y = x^{2}$ and $y^{2} - x$ $= 0$ at the point $(1, 1)$ is

\frac{π}{2}

\tan^{- 1} \frac{4}{3}

\frac{π}{3}

\frac{π}{4}

\tan^{- 1} \frac{3}{4}

Explanation:

(E) : Differentiate the equation $y = x^{2}$ with respect to $x$
$y = x^{2}$
$\frac{d y}{d x} = 2 x$
$m_{2} = 2$
Again, differentiate the equation
$y^{2} - x = 0 with respect to x$
$y^{2} - x = 0$
$y^{2} = x$
$2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y}$
$m_{1} = \frac{1}{2}$
We know that
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
$\tan θ = | \frac{2 - \frac{1}{2}}{1 + (\frac{1}{2} \times 2)} | \tan θ = | \frac{\frac{3}{2}}{2} |$
$\tan θ = \frac{3}{2} \times \frac{1}{2} = 3 / 4$
$θ = \tan^{- 1} (\frac{3}{4})$

Kerala CEE-2010

Application of Derivatives

85801 The angle of intersection of the curves $y = x^{2}$ , $6 y = 7 - x^{3}$ at $(1, 1)$ is :

\frac{π}{4}

\frac{π}{3}

\frac{π}{2}

4 none of these

Explanation:

(C) : Given,
Curve $y = x^{2}$ and $6 y = 7 - x^{3}$
Intersection point $= (1, 1)$
$\begin{array}{ll} y = x^{2} and 6 y = 7 - x^{3} \\ m_{1} = \frac{d y}{d x} = 2 x 6 \frac{d y}{d x} = - 3 x^{2} \end{array}$
$m_{1} = \frac{dy}{{dx}_{(1, 1)}} = 2 m_{2} = \frac{dy}{dx} = \frac{- x^{2}}{2}$
$m_{2} = \frac{dy}{dx} = - \frac{1}{2}$
$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} \cdot m_{2}} | = | \frac{2 - (\frac{- 1}{2}) |}{1 + 2 \times \frac{- 1}{2} |}$
$= | \frac{5 / 2}{0} | = \infty$
$\tan θ = \infty$
$θ = \tan \infty \infty$
$θ = \frac{π}{2}$

BCECE-2006

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Application of Derivatives

85800 The angle between the curves, $y = x^{2}$ and $y^{2} - x$ $= 0$ at the point $(1, 1)$ is

\frac{π}{2}

\tan^{- 1} \frac{4}{3}

\frac{π}{3}

\frac{π}{4}

\tan^{- 1} \frac{3}{4}

Explanation:

Kerala CEE-2010

Application of Derivatives

85801 The angle of intersection of the curves $y = x^{2}$ , $6 y = 7 - x^{3}$ at $(1, 1)$ is :

\frac{π}{4}

\frac{π}{3}

\frac{π}{2}

4 none of these

Explanation:

BCECE-2006

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