Question

Find the slope of the tangent line to the given curve at the point corresponding to the specified value of the parameter.

\(r = {e^{ - \theta }};\theta = \pi \)

Short answer

The slope of the tangent line to the curve \(r = {e^{ - \theta }}\) at the point \(\theta = \pi \) is \( - 1\).

Step-by-step solution

Definition of Concept

The slope or gradient of a line in mathematics is a number that describes the line's direction as well as its steepness.

Find the slope of the tangent line to the curve

Considering the given information:

The parametric equation of the curve is given as,

\(r = {e^{ - \theta }}\)

The Cartesian equation of the curve is\(x = r\cos \theta ,y = r\sin \theta \).

Put\(r = {e^{ - \theta }}\)in\(x = r\cos \theta ,y = r\sin \theta \)to obtain the Cartesian coordinates as follows.

\(\begin{aligned}{c}x = r\cos \theta \\ = {e^{ - \theta }}\cos \theta \\y = r\sin \theta \\ = {e^{ - \theta }}\sin \theta \end{aligned}\)

Differentiate\(x = {e^{ - \theta }}\cos \theta \)with respect to\({\rm{\theta }}\).

\(\begin{aligned}{c}\frac{{dx}}{{d\theta }} = {e^{ - \theta }}\frac{d}{{d\theta }}(\cos \theta ) + \cos \theta \frac{d}{{d\theta }}\left( {{e^{ - \theta }}} \right)\\ = {e^{ - \theta }}( - \sin \theta ) + \cos \theta \left( { - {e^{ - \theta }}} \right)\\ = - {e^{ - \theta }}\cos \theta - {e^{ - \theta }}\sin \theta \\ = - {e^{ - \theta }}(\cos \theta + \sin \theta )\end{aligned}\)

Differentiate\(y = {e^{ - \theta }}\sin \theta \)with respect to\({\rm{\theta }}\).

\(\begin{aligned}{c}\frac{{dy}}{{d\theta }} = {e^{ - \theta }}\frac{d}{{d\theta }}(\sin \theta ) + \sin \theta \frac{d}{{d\theta }}\left( {{e^{ - \theta }}} \right)\\ = {e^{ - \theta }}(\cos \theta ) + \sin \theta \left( { - {e^{ - \theta }}} \right)\\ = {e^{ - \theta }}\cos \theta - {e^{ - \theta }}\sin \theta \\ = - {e^{ - \theta }}(\sin \theta - \cos \theta )\end{aligned}\)

The tangent's slope is given by,

\(\frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}}\)

Putting\(\frac{{dx}}{{d\theta }} = - {e^{ - \theta }}(\cos \theta + \sin \theta )\)and\(\frac{{dy}}{{d\theta }} = - {e^{ - \theta }}(\sin \theta - \cos \theta )\) in the above equation,

\(\begin{aligned}{c}\frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}}\\ = \frac{{ - {e^{ - \theta }}(\sin \theta - \cos \theta )}}{{ - {e^{ - \theta }}(\cos \theta + \sin \theta )}}\\ = \frac{{\sin \theta - \cos \theta }}{{\cos \theta + \sin \theta }}\end{aligned}\)

The given point is, \(\theta = \pi \)

As a result, the tangent's slope at \(\theta = \pi \) will be,

\(\begin{aligned}{c}\frac{{dy}}{{dx}} = \frac{{\sin \theta - \cos \theta }}{{\cos \theta + \sin \theta }}\\ = \frac{{\sin \pi - \cos \pi }}{{\cos \pi + \sin \pi }}\\ = \frac{{0 - ( - 1)}}{{( - 1) + 0}}\\ = \frac{1}{{ - 1}}\\ = - 1\end{aligned}\)

Therefore, the slope of the tangent line to the curve \(r = {e^{ - \theta }}\) at the point \(\theta = \pi \) is \( - 1\).