Question

If the parametric equation of a curve is given by x = e^t cos t, y = e^t sin t then the tangent to the curve at the point t = π/4 makes with the axis of x the angle:

Answer 1

Correct Answer - Option 4 : π/2 

Concept:

Let y = f(x) be the equation of a curve, then the slope of the tangent at any point say (x1, y1) is given by:

 \(m = {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {{x_1},\;\;{y_1}} \right)}}\)

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( x \right)\;{\rm{g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( x \right){\rm{\;g'}}\left( {\rm{x}} \right)\)
• Quotient Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {\frac{{{\rm{f}}\left( x \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}}} \right] = \frac{{{\rm{g}}\left( {\rm{x}} \right){\rm{\;f'}}\left( {\rm{x}} \right) - {\rm{f}}\left( x \right){\rm{\;g'}}\left( {\rm{x}} \right)}}{{{{\left[ {{\rm{g}}\left( {\rm{x}} \right)} \right]}^2}}}\)

​

Calculation:

Given that, 

x = et cos t, y = et sin t 

⇒ \(\frac{dx}{dt}\ =\ e^t(-sin\ t)\ +\ e^t(cos\ t)\)

⇒ \(\frac{dx}{dt}\ =\ e^t(cos\ t\ -\ sin\ t)\)     ---(1)

Similarly,

\(\frac{dy}{dt}\ =\ e^t(cos\ t\ +\ sin\ t)\)     ---(2)

On dividing equitation (2) with equation (1)

\(\frac{dy}{dx}\ =\ \frac{e^t (\sin t + \cos t)}{e^t (\cos t - \sin t)}\)

at t = π/4, 

\(\frac{dy}{dx}\ =\ \frac{\sqrt 2}{0}\ = \infty\)

Hence tangent is perpendicular to the x-axis.