Application of Derivatives
85801
The angle of intersection of the curves \(y=x^{2}\), \(6 y=7-x^{3}\) at \((1,1)\) is :
1 \(\frac{\pi}{4}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{2}\)
4 none of these
Explanation:
(C) : Given,
Curve \(y=x^{2}\) and \(6 y=7-x^{3}\)
Intersection point \(=(1,1)\)
\(\begin{array}{ll}y=x^{2} \text { and } 6 y=7-x^{3} \\ m_{1}=\frac{d y}{d x}=2 x 6 \frac{d y}{d x}=-3 x^{2}\end{array}\)
\(\mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}_{(1,1)}}=2 \quad \mathrm{~m}_{2}=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{x}^{2}}{2}\)
\(\mathrm{~m}_{2}=\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2}\)
\(\tan \theta=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \cdot \mathrm{m}_{2}}\right|=\left\lvert\, \frac{\left.2-\left(\frac{-1}{2}\right) \right\rvert\,}{\left.1+2 \times \frac{-1}{2} \right\rvert\,}\right.\)
\(=\left|\frac{5 / 2}{0}\right|=\infty\)
\(\tan \theta=\infty\)
\(\theta=\tan \infty \infty\)
\(\theta=\frac{\pi}{2}\)