Question

If $x = \sin t$ and $y = \tan t$ then $\dfrac{{dy}}{{dx}}$ is equal to
$\left( 1 \right){\text{ }}{\cos ^3}t$
$\left( 2 \right){\text{ }}\dfrac{1}{{{{\cos }^3}t}}$
$\left( 3 \right){\text{ }}\dfrac{1}{{{{\cos }^2}t}}$
$\left( 4 \right){\text{ }}\dfrac{1}{{{{\sin }^2}t}}$

Answer 1

So in this question what we have to do is first differentiate the given equation of $x$ and $y$ with respect to $t$ because it’s mandatory as the terms contains $t$ in them. Then after differentiating both the equations with respect to $t$ we put those values in formula, $\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}$ .Then on further solving we can get the final answer of the question.

Complete step-by-step solution:
Here we will use the chain rule because x and y are the parametric equations. The parametric equations are in the form of $x = h\left( t \right)$ and $y = g\left( t \right)$ .The chain rule states that the derivative $\dfrac{{dy}}{{dx}}$ is the ratio of $\dfrac{{dy}}{{dt}}$ to $\dfrac{{dx}}{{dt}}$ that is
$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}$ ----------- (i)
So it is given that $x = \sin t$ .On differentiating it with respect to $t$ we get
$\dfrac{{dx}}{{dt}} = \dfrac{{d\left( {\sin t} \right)}}{{dt}}$
Because we know that derivate of $\sin x$ is $\cos x$ .Therefore, the required value of $\dfrac{{dx}}{{dt}}$ will be
$\dfrac{{dx}}{{dt}} = \cos t$ ---------- (ii)
Next we have $y = \tan t$ .Therefore on differentiating it with respect to $t$ we get
$\dfrac{{dy}}{{dt}} = \dfrac{{d\left( {\tan t} \right)}}{{dt}}$
Because the derivative of $\tan x$ is ${\sec ^2}x$ .Therefore the above expression becomes
$\dfrac{{dy}}{{dt}} = {\sec ^2}t$ -------- (iii)
Now substitute the values of equation (ii) and (iii) in equation (i) we get
$\dfrac{{dy}}{{dx}} = \dfrac{{{{\sec }^2}t}}{{\cos t}}$
We know that the reciprocal of $\sec x$ is equal to $\dfrac{1}{{\cos x}}$ that means ${\sec ^2}t$ is equal to $\dfrac{1}{{{{\cos }^2}t}}$ .So the above equation can be written as
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\cos }^2}t.\cos t}}$
As in the denominator there are the same terms so we can add their powers. By doing this we will get
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\cos }^3}t}}$
Hence, the correct option is $\left( 2 \right){\text{ }}\dfrac{1}{{{{\cos }^3}t}}$

Note: Remember the chain rule because chain rule is applied every time in these types of questions. Don’t get confused whenever you get questions like this. Just remember the formula of the rule, you will be able to answer these types of questions. Always differentiate the parametric equations first with respect to the suitable variable and then step forward to the other steps.