Question

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

\(x = \ln t,y = 1 + {t^2};\quad t = 1\)

Short answer

The slope of the tangent line to the curve \(x = \ln t,y = 1 + {t^2}\) at the point \(t = 1\) is 2.

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,

\(x = \ln t,y = 1 + {t^2}\)

Calculate the difference between x and y in terms of t.

\(\begin{aligned}{l}\frac{{dx}}{{dt}} = \frac{1}{t}\\\frac{{dy}}{{dt}} = 2t\end{aligned}\)

The tangent's slope is given by,

\(\frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}}\)

Putting\(\frac{{dx}}{{dt}} = \frac{1}{t}\)and\(\frac{{dy}}{{dt}} = 2t\) in the above equation,

\(\begin{aligned}{c}\frac{{dy}}{{dx}} = \frac{{2t}}{{\left( {\frac{1}{t}} \right)}}\\ = 2{t^2}\end{aligned}\)

The \(t = 1\) is the given point. The x and y coordinates will then be as follows:

\(\begin{aligned}{c}x = \ln t\\ = \ln 1\\ = 0\\y = 1 + {t^2}\\ = 1 + {1^2}\\ = 2\end{aligned}\)

As a result, at\(t = 1\), the tangent's slope will be,

\(\begin{aligned}{c}\frac{{dy}}{{dx}} = 2{t^2}\\ = 2{(1)^2}\\ = 2\end{aligned}\)

That is, slope of the tangent to the curve\(x = \ln t,y = 1 + {t^2}\)at\({\rm{(0,2)}}\) is 2.

Therefore, the slope of the tangent line to the curve \(x = \ln t,y = 1 + {t^2}\) at the point \(t = 1\) is 2.