(A) : Given, Curves are \(y=3^{x}\) And, \(y=5^{x}\) Intersect at the point \((0,1)\). Now, differentiating equation (i) and (ii) w.r.t.x, We get - \(\frac{d y}{d x}=3^{x} \log 3 \text { and } \frac{d y}{d x}=5^{x} \log 5\) \(\left(\frac{d y}{d x}\right)_{(0,1)}=\log 3 \text { and }\left(\frac{d y}{d x}\right)_{(0,1)}=\log 5\) \(\mathrm{m}_{1}=\log 3\) and \(\mathrm{m}_{2}=\log 5\) Angle between these curves is given by \(\tan \theta=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) \(\tan \theta=\frac{\log 3-\log 5}{1+\log 3 \log 5}\) \(\theta=\tan ^{-1}\left(\frac{\log 3-\log 5}{1+\log 3 \log 5}\right)\)
VITEEE-2010
Application of Derivatives
85784
The angle at which the curve \(y=x^{2}\) and the curve \(x=\frac{5}{3} \cos t, y=\frac{5}{4} \sin t\) intersect is
1 \(\tan ^{-1} \frac{2}{41}\)
2 \(\tan ^{-1} \frac{41}{2}\)
3 \(-\tan ^{-1} \frac{2}{41}\)
4 \(2 \tan ^{-1} \frac{41}{2}\)
Explanation:
(B) : Given, \(y=x^{2} \tag{i}\) \(x=\frac{5}{3} \cos t, \quad y=\frac{5}{4} \sin t\) \(\frac{3}{5} x=\cos t, \quad \frac{4}{5} y=\sin t\) \(\sin ^{2} \mathrm{t}+\cos ^{2} \mathrm{t}=\frac{9}{25} \mathrm{x}^{2}+\frac{16}{25} \mathrm{y}^{2} \quad\left(\because \sin ^{2} \mathrm{t}+\cos ^{2} \mathrm{t}=1\right)\) \(9 x^{2}+16 y^{2}=25 \tag{ii}\) The intersection point at equation (i) and (ii) are (1,1) and \((-1,1)\) Now, \(\mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) \(\mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=2\) And, \(m_{2}=\frac{d y}{d x_{(1,1)}}=\frac{-9}{16}\) \(\theta =\tan ^{-1}\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\tan ^{-1}\left|\frac{2-\left(-\frac{9}{16}\right)}{1+2 \times \frac{-9}{16}}\right|\) \(=\left|\frac{41}{16} \times \frac{-16}{2}\right|=\left[\frac{-41}{2}\right]\) \(\theta =\tan ^{-1} \frac{41}{2}\)
UPSEE-2017
Application of Derivatives
85785
Angle of intersection of the curves \(r=\) \(\sin \theta+\cos \theta\) and \(r=2 \sin \theta\) is equal to
(A) : Given, Curves are \(y=3^{x}\) And, \(y=5^{x}\) Intersect at the point \((0,1)\). Now, differentiating equation (i) and (ii) w.r.t.x, We get - \(\frac{d y}{d x}=3^{x} \log 3 \text { and } \frac{d y}{d x}=5^{x} \log 5\) \(\left(\frac{d y}{d x}\right)_{(0,1)}=\log 3 \text { and }\left(\frac{d y}{d x}\right)_{(0,1)}=\log 5\) \(\mathrm{m}_{1}=\log 3\) and \(\mathrm{m}_{2}=\log 5\) Angle between these curves is given by \(\tan \theta=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) \(\tan \theta=\frac{\log 3-\log 5}{1+\log 3 \log 5}\) \(\theta=\tan ^{-1}\left(\frac{\log 3-\log 5}{1+\log 3 \log 5}\right)\)
VITEEE-2010
Application of Derivatives
85784
The angle at which the curve \(y=x^{2}\) and the curve \(x=\frac{5}{3} \cos t, y=\frac{5}{4} \sin t\) intersect is
1 \(\tan ^{-1} \frac{2}{41}\)
2 \(\tan ^{-1} \frac{41}{2}\)
3 \(-\tan ^{-1} \frac{2}{41}\)
4 \(2 \tan ^{-1} \frac{41}{2}\)
Explanation:
(B) : Given, \(y=x^{2} \tag{i}\) \(x=\frac{5}{3} \cos t, \quad y=\frac{5}{4} \sin t\) \(\frac{3}{5} x=\cos t, \quad \frac{4}{5} y=\sin t\) \(\sin ^{2} \mathrm{t}+\cos ^{2} \mathrm{t}=\frac{9}{25} \mathrm{x}^{2}+\frac{16}{25} \mathrm{y}^{2} \quad\left(\because \sin ^{2} \mathrm{t}+\cos ^{2} \mathrm{t}=1\right)\) \(9 x^{2}+16 y^{2}=25 \tag{ii}\) The intersection point at equation (i) and (ii) are (1,1) and \((-1,1)\) Now, \(\mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) \(\mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=2\) And, \(m_{2}=\frac{d y}{d x_{(1,1)}}=\frac{-9}{16}\) \(\theta =\tan ^{-1}\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\tan ^{-1}\left|\frac{2-\left(-\frac{9}{16}\right)}{1+2 \times \frac{-9}{16}}\right|\) \(=\left|\frac{41}{16} \times \frac{-16}{2}\right|=\left[\frac{-41}{2}\right]\) \(\theta =\tan ^{-1} \frac{41}{2}\)
UPSEE-2017
Application of Derivatives
85785
Angle of intersection of the curves \(r=\) \(\sin \theta+\cos \theta\) and \(r=2 \sin \theta\) is equal to
(A) : Given, Curves are \(y=3^{x}\) And, \(y=5^{x}\) Intersect at the point \((0,1)\). Now, differentiating equation (i) and (ii) w.r.t.x, We get - \(\frac{d y}{d x}=3^{x} \log 3 \text { and } \frac{d y}{d x}=5^{x} \log 5\) \(\left(\frac{d y}{d x}\right)_{(0,1)}=\log 3 \text { and }\left(\frac{d y}{d x}\right)_{(0,1)}=\log 5\) \(\mathrm{m}_{1}=\log 3\) and \(\mathrm{m}_{2}=\log 5\) Angle between these curves is given by \(\tan \theta=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) \(\tan \theta=\frac{\log 3-\log 5}{1+\log 3 \log 5}\) \(\theta=\tan ^{-1}\left(\frac{\log 3-\log 5}{1+\log 3 \log 5}\right)\)
VITEEE-2010
Application of Derivatives
85784
The angle at which the curve \(y=x^{2}\) and the curve \(x=\frac{5}{3} \cos t, y=\frac{5}{4} \sin t\) intersect is
1 \(\tan ^{-1} \frac{2}{41}\)
2 \(\tan ^{-1} \frac{41}{2}\)
3 \(-\tan ^{-1} \frac{2}{41}\)
4 \(2 \tan ^{-1} \frac{41}{2}\)
Explanation:
(B) : Given, \(y=x^{2} \tag{i}\) \(x=\frac{5}{3} \cos t, \quad y=\frac{5}{4} \sin t\) \(\frac{3}{5} x=\cos t, \quad \frac{4}{5} y=\sin t\) \(\sin ^{2} \mathrm{t}+\cos ^{2} \mathrm{t}=\frac{9}{25} \mathrm{x}^{2}+\frac{16}{25} \mathrm{y}^{2} \quad\left(\because \sin ^{2} \mathrm{t}+\cos ^{2} \mathrm{t}=1\right)\) \(9 x^{2}+16 y^{2}=25 \tag{ii}\) The intersection point at equation (i) and (ii) are (1,1) and \((-1,1)\) Now, \(\mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) \(\mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=2\) And, \(m_{2}=\frac{d y}{d x_{(1,1)}}=\frac{-9}{16}\) \(\theta =\tan ^{-1}\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\tan ^{-1}\left|\frac{2-\left(-\frac{9}{16}\right)}{1+2 \times \frac{-9}{16}}\right|\) \(=\left|\frac{41}{16} \times \frac{-16}{2}\right|=\left[\frac{-41}{2}\right]\) \(\theta =\tan ^{-1} \frac{41}{2}\)
UPSEE-2017
Application of Derivatives
85785
Angle of intersection of the curves \(r=\) \(\sin \theta+\cos \theta\) and \(r=2 \sin \theta\) is equal to
(A) : Given, Curves are \(y=3^{x}\) And, \(y=5^{x}\) Intersect at the point \((0,1)\). Now, differentiating equation (i) and (ii) w.r.t.x, We get - \(\frac{d y}{d x}=3^{x} \log 3 \text { and } \frac{d y}{d x}=5^{x} \log 5\) \(\left(\frac{d y}{d x}\right)_{(0,1)}=\log 3 \text { and }\left(\frac{d y}{d x}\right)_{(0,1)}=\log 5\) \(\mathrm{m}_{1}=\log 3\) and \(\mathrm{m}_{2}=\log 5\) Angle between these curves is given by \(\tan \theta=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) \(\tan \theta=\frac{\log 3-\log 5}{1+\log 3 \log 5}\) \(\theta=\tan ^{-1}\left(\frac{\log 3-\log 5}{1+\log 3 \log 5}\right)\)
VITEEE-2010
Application of Derivatives
85784
The angle at which the curve \(y=x^{2}\) and the curve \(x=\frac{5}{3} \cos t, y=\frac{5}{4} \sin t\) intersect is
1 \(\tan ^{-1} \frac{2}{41}\)
2 \(\tan ^{-1} \frac{41}{2}\)
3 \(-\tan ^{-1} \frac{2}{41}\)
4 \(2 \tan ^{-1} \frac{41}{2}\)
Explanation:
(B) : Given, \(y=x^{2} \tag{i}\) \(x=\frac{5}{3} \cos t, \quad y=\frac{5}{4} \sin t\) \(\frac{3}{5} x=\cos t, \quad \frac{4}{5} y=\sin t\) \(\sin ^{2} \mathrm{t}+\cos ^{2} \mathrm{t}=\frac{9}{25} \mathrm{x}^{2}+\frac{16}{25} \mathrm{y}^{2} \quad\left(\because \sin ^{2} \mathrm{t}+\cos ^{2} \mathrm{t}=1\right)\) \(9 x^{2}+16 y^{2}=25 \tag{ii}\) The intersection point at equation (i) and (ii) are (1,1) and \((-1,1)\) Now, \(\mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) \(\mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=2\) And, \(m_{2}=\frac{d y}{d x_{(1,1)}}=\frac{-9}{16}\) \(\theta =\tan ^{-1}\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\tan ^{-1}\left|\frac{2-\left(-\frac{9}{16}\right)}{1+2 \times \frac{-9}{16}}\right|\) \(=\left|\frac{41}{16} \times \frac{-16}{2}\right|=\left[\frac{-41}{2}\right]\) \(\theta =\tan ^{-1} \frac{41}{2}\)
UPSEE-2017
Application of Derivatives
85785
Angle of intersection of the curves \(r=\) \(\sin \theta+\cos \theta\) and \(r=2 \sin \theta\) is equal to
