Question

If \[x = \sin^{ - 1}\left( {3t - 4{t^3}} \right)\], and \[y = \cos^{ - 1}\left( {\sqrt {1 - {t^2}} } \right)\]. Then find the value of \[\dfrac{{dy}}{{dx}}\].
A. \[\dfrac{1}{2}\]
B. \[\dfrac{2}{3}\]
C. \[\dfrac{1}{3}\]
D. \[\dfrac{2}{5}\]

Answer 1

Hint: Substitute \[t = \sin\theta \] in the given trigonometric equations and simplify the equations in terms of \[t\]. After that, differentiate both equations with respect to \[t\]. In the end, take the ratio of \[\dfrac{{dy}}{{dt}}\] to the \[\dfrac{{dx}}{{dt}}\] and get the required answer.

Formula used:
1. \[\sin3\theta = 3\sin\theta - 4\sin^{3}\theta \]
2. \[sin^{2}x + \cos^{2}x = 1\]
3. \[\dfrac{d}{{dx}}\left( {sin^{ - 1}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}\]

Complete step by step solution:
The given trigonometric equations are \[x = \sin^{ - 1}\left( {3t - 4{t^3}} \right)\] and \[y = \cos^{ - 1}\left( {\sqrt {1 - {t^2}} } \right)\].
Now substitute \[t = \sin\theta \] in the equation \[x = \sin^{ - 1}\left( {3t - 4{t^3}} \right)\].
\[x = \sin^{ - 1}\left( {3\sin\theta - 4\sin^{3}\theta } \right)\] [ Since \[\sin3\theta = 3\sin\theta - 4\sin^{3}\theta \]]
\[ \Rightarrow \]\[x = sin^{ - 1}\left( {\sin 3\theta } \right)\]
\[ \Rightarrow \]\[x = 3\theta \] [ Since \[\sin^{ - 1}\left( {sin x} \right) = x\]]
Resubstitute the value of \[\theta \].
\[x = 3\sin^{ - 1}t\]
Now differentiate the above equation with respect to \[t\].
Apply the formula \[\dfrac{d}{{dx}}\left( {sin^{ - 1}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}\].
\[\dfrac{{dx}}{{dt}} = 3\left( {\dfrac{1}{{\sqrt {1 - {t^2}} }}} \right)\]

Now substitute \[t = \sin\theta \] in the equation \[y = \cos^{ - 1}\left( {\sqrt {1 - {t^2}} } \right)\].
\[y = \cos^{ - 1}\left( {\sqrt {1 - \sin^{2}\theta } } \right)\]
\[ \Rightarrow \]\[y = \cos^{ - 1}\left( {\sqrt {\cos^{2}\theta } } \right)\] [ Since \[\sin^{2}x + \cos^{2}x = 1\]]
\[ \Rightarrow \]\[y = \cos^{ - 1}\left( {\cos\theta } \right)\] [ Since \[\cos^{ - 1}\left( {\cos x} \right) = x\]]
\[ \Rightarrow \]\[y = \theta \]
Resubstitute the value of \[\theta \].
\[y = sin^{ - 1}t\]
Now differentiate the above equation with respect to \[t\].
Apply the formula \[\dfrac{d}{{dx}}\left( {sin^{ - 1}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}\].
\[\dfrac{{dy}}{{dt}} = \dfrac{1}{{\sqrt {1 - {t^2}} }}\]

Now take the ratio of \[\dfrac{{dy}}{{dt}}\] to the \[\dfrac{{dx}}{{dt}}\].
\[\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}\]
Substitute the values in the above equation.
\[\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{1}{{\sqrt {1 - {t^2}} }}}}{{3\left( {\dfrac{1}{{\sqrt {1 - {t^2}} }}} \right)}}\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{3}\]
Hence the correct option is C.

Note: Students are often confused with the formula \[\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}\] and \[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \cdot \dfrac{{dx}}{{dt}}\]. But the correct formulas are \[\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}\] and \[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \cdot \dfrac{{dt}}{{dx}}\].