Derivatives of Parametric Functions
The relationship between the variables x and y can be defined in parametric form using two equations:
where the variable t is called a parameter. For example, two functions
describe in parametric form the equation of a circle centered at the origin with the radius \(R.\) In this case, the parameter \(t\) varies from \(0\) to \(2 \pi.\)
Find an expression for the derivative of a parametrically defined function. Suppose that the functions \(x = x\left( t \right)\) and \(y = y\left( t \right)\) are differentiable in the interval \(\alpha \lt t \lt \beta \) and \(x'\left( t \right) \ne 0.\) Moreover, we assume that the function \(x = x\left( t \right)\) has an inverse function \(t = \varphi \left( x \right).\)
By the inverse function theorem we can write:
The original function \(y\left( x \right)\) can be considered as a composite function:
Then its derivative is given by
This formula allows to find the derivative of a parametrically defined function without expressing the function \(y\left( x \right)\) in explicit form.
In the examples below, find the derivative of the parametric function.
Solved Problems
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Example 1
\[x = {t^2},\;y = {t^3}.\]
Example 2
\[x = 2t + 1,\;y = 4t - 3.\]
Example 3
\[x = {e^{2t}},\;y = {e^{3t}}.\]
Example 4
\[x = at,\;y = b{t^2}.\]
Example 5
\[x = {\sin ^2}t,\;y = {\cos ^2}t.\]
Example 6
\[x = \sinh t,\;y = \cosh t.\]
Example 7
\[x = a\cos t,\;y = b\sin t.\]
Example 8
Find the derivative \(\frac{{dy}}{{dx}}\) for the function \(x = \sin 2t,\) \(y = -\cos t\) at the point \(t =\frac{\pi }{6}.\)
Example 9
\(x = 2{t^2} + t + 1,\) \(y = 8{t^3} + 3{t^2} + 2.\)
Example 10
\[x = \sqrt {1 - {t^2}} ,\;y = \arcsin t.\]
Example 1.
\[x = {t^2},\;y = {t^3}.\]
Solution.
We find the derivatives of \(x\) and \(y\) with respect to \(t:\)
Hence,
Example 2.
\[x = 2t + 1,\;y = 4t - 3.\]
Solution.
Consequently,
Example 3.
\[x = {e^{2t}},\;y = {e^{3t}}.\]
Solution.
Hence, the derivative \(\frac{{dy}}{{dx}}\) is given by
Example 4.
\[x = at,\;y = b{t^2}.\]
Solution.
In this example, the derivatives with respect to \(t\) are given by
Hence,
Example 5.
\[x = {\sin ^2}t,\;y = {\cos ^2}t.\]
Solution.
Differentiate with respect to the parameter \(t:\)
Then
Example 6.
\[x = \sinh t,\;y = \cosh t.\]
Solution.
Calculate the derivatives:
Then the derivative \(\frac{{dy}}{{dx}}\) is given by
Example 7.
\[{x^2} + {y^2} - 2x - 4y = 4\]
Solution.
These equations describe an ellipse centered at the origin with semi-axes \(a\) and \(b\). Differentiate the variables \(x\) and \(y\) with respect to \(t:\)
The derivative \(\frac{{dy}}{{dx}}\) depends on \(t\) as follows:
Here the parameter \(t\) varies from \(-\pi\) to \(\pi\). However the derivative \(\frac{{dy}}{{dx}}\) becomes infinite at the points \(t = 0, \pm \pi .\) Therefore, the domain can be represented as \(0 \lt \left| t \right| \lt \pi .\)
Example 8.
Find the derivative \(\frac{{dy}}{{dx}}\) for the function \(x = \sin 2t,\) \(y = -\cos t\) at the point \(t =\frac{\pi }{6}.\)
Solution.
Compute the derivatives with respect to \(t:\)
So, the derivative \(\frac{{dy}}{{dx}}\) is given by
Compute the derivative at \(t = \frac{\pi }{6}:\)
Example 9.
\(x = 2{t^2} + t + 1,\) \(y = 8{t^3} + 3{t^2} + 2.\)
Solution.
Differentiate both equations with respect to the parameter \(t:\)
Consequently, the derivative \(\frac{{dy}}{{dx}}\) is given by
Example 10.
\[x = \sqrt {1 - {t^2}} ,\;y = \arcsin t.\]
Solution.
The derivative \(\frac{{dy}}{{dx}}\) is expressed by the formula
where the parameter \(t\) can take values satisfying the conditions \(\left| t \right| \lt 1,\;t \ne 0.\)