Question

Find $ \dfrac{{dy}}{{dx}} $ given that
 $
x = a{\cos ^2}t\\
 y = a{\sin ^2}t
 $

Answer 1

Hint: Here, function depends on both variables x and y so, we will partially differentiate the function x and y with respect to t. Then we will use the value of derivative of x and y to find the value of $ \dfrac{{dy}}{{dx}} $ .

Complete step-by-step answer:
Given, function is:
 $
x = a{\cos ^2}t\\
y = a{\sin ^2}t
 $
Here, function in the form of values of x and y separately but they depend on the common variable t and we have to find the value of $ \dfrac{{dy}}{{dx}} $ .
In this case, First we will differentiate x with respect to t,
$
\dfrac{d}{{dt}}x = \dfrac{d}{{dt}}a{\cos ^2}t\\
\dfrac{{dx}}{{dt}} = a\dfrac{d}{{dt}}{\cos ^2}t
$
Here, a is constant so differentiating the remaining value we get,
Where, the derivative of cost is –sin t
$
\Rightarrow \dfrac{{dx}}{{dt}} = a \times 2\cos t \times \left( { - \sin t} \right)\\
\Rightarrow \dfrac{{dx}}{{dt}} = - 2a\cos t\sin t
$
Now, we will differentiate y with respect to t,
$
\Rightarrow \dfrac{d}{{dt}}y = \dfrac{d}{{dt}}a{\sin ^2}t\\
\Rightarrow \dfrac{{dy}}{{dt}} = a\dfrac{d}{{dt}}{\sin ^2}t
 $
Here, a is constant again so differentiating the remaining value we get,
Where, the derivative of sint is cost
 $
\dfrac{{dy}}{{dt}} = a \times 2\sin t \times \cos t\\
\dfrac{{dy}}{{dt}} = 2a\sin t\cos t
 $
Now, we will use these derivatives in the further solution.
To get the value of $ \dfrac{{dy}}{{dx}} $ from given function, first multiply and divide it by $ dt $ then we get,
$
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dx}} \times \dfrac{{dt}}{{dt}}\\
 = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}}\\
 = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}
$
Now we have to substitute the value of $ \dfrac{{dy}}{{dt}} $ and $ \dfrac{{dx}}{{dt}} $ then we get,
$
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2a\sin t\cos t}}{{ - 2a\sin t\cos t}}\\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{ - 1}}\\
\Rightarrow \dfrac{{dy}}{{dx}} = - 1
 $
Hence the derivative of y with respect to x is -1.

Note: In this type of problem, you can also use a chain method for solving the problem. First make the chain form of the given derivative and then calculate the derivative of the given different function. It is compulsory that one part of each derivative must be common with the second function’s derivative so that they can cancel out easily and we will get the value of the derivative.