6.0 Parametric Differentiation

Let $x(t)$ and $y(t)$ be the coordinates of the points of the curve expressed as function of a variable $t$. The first derivative of the parametric equation is given by,

$$ x = x\left( t \right)\;and\;y = y\left( t \right)\;then $$$$ \frac{{dx}}{{dt}}:derivative\;of\;x\;w.r.t\;t $$where,$$\frac{{dx}}{{dt}} \ne 0 $$$$ \frac{{dy}}{{dt}}:derivative\;of\;y\;w.r.t\;t $$Thus,$$\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} \bullet \frac{{dt}}{{dx}} $$$$ \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} $$

Question 1: $x = t{e^{ - t}},y = 2{t^2} + 1$ Find $\frac{{dy}}{{dx}}$

Solution: $$x = t{e^{ - t}}$$ Using Product differentiation,

Also, we know that, $$\frac{d}{dx}x^n = nx^{n-1}$$ $$\frac{d}{dx} e^x = e^x$$$$ \frac{{dx}}{{dt}} = t\frac{d}{{dt}}{e^{ - t}} + {e^{ - t}}\frac{d}{{dt}}t $$$$ \frac{{dx}}{{dt}} = - t{e^{ - t}} + {e^{ - t}}{t^{1 - 1}} $$

$$ \frac{{dx}}{{dt}} = - t{e^{ - t}} + {e^{ - t}} = \left( {t + 1} \right){e^{ - t}} \quad...\left( A \right) $$

$$y = 2{t^2} + 1$$ Using scalar and sum differentiation we know that, $$\frac{d}{dx}x^n = nx^{n-1}$$ $$\frac{d}{{dx}}k = 0\quad where\,k = constant$$$$ \frac{{dy}}{{dt}} = 2\frac{d}{{dt}}{t^2} + \frac{d}{{dt}}1 $$ $$ \frac{{dy}}{{dt}} = 2\left( {2{t^{2 - 1}}} \right) + 0 = 4t \quad...\left( B \right) $$

Thus,

$$ \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{4t}}{{\left( {t + 1} \right){e^{ - t}}}} $$

Question 2: $x = 3cost,y = 3sint$ Find $\frac{{dy}}{{dx}}$

Solution: $$x = 3cost$$ Using scalar differentiation

We know that, $$\frac{d}{dx} cos x = -sin x$$$$ \frac{{dx}}{{dt}} = 3\frac{d}{{dt}}\cos t $$$$ \frac{{dx}}{{dt}} = - 3\sin t \quad...\left( A \right) $$

$$y = 3sint$$ Using scalar differentiation we know that, $$\frac{d}{dx} sin x = cos x$$$$ \frac{{dx}}{{dt}} = 3\frac{d}{{dt}}\sin t $$$$ \frac{{dx}}{{dt}} = 3\cos t \quad...\left( B \right) $$

Thus, $$\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{3\cos t}}{{ - 3\sin t}} = - \frac{{\cos t}}{{\sin t}} = \cot t$$

Question 3: $x = \tanh t\;,y = 3t + 1$ Find $\frac{{dy}}{{dx}}$

$$ x = \tanh t\; $$$$ \frac{{dx}}{{dt}} = \frac{d}{{dt}}\tanh t $$

We know that, $$\frac{d}{dx} tanh x = sech^2 x$$$$ \frac{{dx}}{{dt}} = \sec {h^2}t\; \quad..(A) $$$$ y = 3t + 1 $$$$ \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {3t + 1} \right) $$

Using scalar and sum rule,

$$ \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {3t} \right) + \frac{d}{{dt}}\left( 1 \right) $$$$ \frac{{dy}}{{dt}} = 3\frac{d}{{dt}}\left( t \right) + \frac{d}{{dt}}\left( 1 \right) $$

We know that, $$\frac{d}{dx}x^n = nx^{n-1}$$ $$\frac{d}{{dx}}k = 0\quad where\,k = constant$$

$$ \frac{{dy}}{{dt}} = 3{t^{1 - 1}} + 0 $$$$ \frac{{dy}}{{dt}} = 3\;\quad..(B) $$

Thus, $$\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{3}{{\sec {h^2}t}} = 3\cos {h^2}t$$

Question 4: $x = {\cos ^{ - 1}}2\theta \;,y = {e^{2\theta + 3}}$ Find $\frac{{dy}}{{dx}}$

Solution:

$$ y = {e^{2\theta + 3}} $$$$ \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}{e^{2\theta + 3}} $$

We know that, $$\frac{d}{{dx}}{e^x} = {e^x}$$

Using Chain rule,

$$\frac{{dy}}{{d\theta }} = {e^{2\theta + 3}}\frac{d}{{d\theta }}\left( {2\theta + 3} \right)$$

Using sum and scalar rule,

$$\frac{{dy}}{{d\theta }} = {e^{2\theta + 3}}\left( {2\frac{d}{{d\theta }}\theta + \frac{d}{{d\theta }}3} \right)$$

We know that, $$\frac{d}{dx}x^n = nx^{n-1}$$

$$\frac{{dy}}{{d\theta }} = {e^{2\theta + 3}}\left( {2{\theta ^{1 - 1}} + 0} \right)$$$$\frac{{dy}}{{d\theta }} = 2{e^{2\theta + 3}}\; \quad...(A)$$$$x = {\cos ^{ - 1}}2\theta $$$$\frac{{dx}}{{d\theta }} = \frac{d}{{d\theta }}{\cos ^{ - 1}}2\theta $$

Using scalar rule,

$$\frac{{dx}}{{d\theta }} = \; - \frac{1}{{\sqrt {1 - 4{\theta ^2}} }}2\frac{d}{{d\theta }}\left( \theta \right)$$

We know that, $$\frac{d}{dx}x^n = nx^{n-1}$$

$$\frac{{dx}}{{d\theta }} = \; - \frac{{2{\theta ^{1 - 1}}}}{{\sqrt {1 - 4{\theta ^2}} }} = \; - \frac{2}{{\sqrt {1 - 4{\theta ^2}} }} \quad...(A)$$

Thus,

$$\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}} = \frac{{2{e^{2\theta + 3}}}}{{ - \frac{2}{{\sqrt {1 - 4{\theta ^2}} }}}} = - \frac{{2{e^{2\theta + 3}}\sqrt {1 - 4{\theta ^2}} }}{2}$$

$$\frac{{dy}}{{dx}} = - {e^{2\theta + 3}}\sqrt {1 - 4{\theta ^2}} $$