Question

Find the equation of the tangents to the curve \(x = 3{t^3} + 1,y = 2{t^3} + 1\) that pass through the point \((4,3)\)?

Short answer

the equation of the tangent line of the curve can be given as: \(y = x - 1\)

Step-by-step solution

Find \(\frac{{dx}}{{dt}},\frac{{dy}}{{dt}},\frac{{dy}}{{dx}}\)

We are given that

\(\begin{array}{l}x = 3{t^2} + 1\\\frac{{dx}}{{dt}} = 6t\end{array}\)

\(\begin{array}{l}y = 2{t^3} + 1\\\frac{{dy}}{{dt}} = 6{t^2}\\\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\\\frac{{dy}}{{dx}} = \frac{{6{t^2}}}{{6t}}\\\frac{{dy}}{{dx}} = t\end{array}\)

Find the slope of the tangent line

Substitute \((x,y) = (4,3)\)

\(\begin{array}{l}x = 3{t^2} + 1\\4 = 3{t^2} + 1\\3{t^2} = 3\\t = 1\end{array}\)

\(\begin{array}{l}y = 2{t^3} + 1\\3 = 2{t^3} + 1\\{t^3} = \frac{2}{2}\\t = 1\end{array}\)

The slope of the tangent line is:

\(\frac{{dy}}{{dx}} = t\)

\(\frac{{dy}}{{dx}} = 1\)

Find the equation of the tangent line

Use the formula to find the equation of the tangent line: \(y - {y_1} = m(x - {x_1})\)

Substitute \(({x_1},{y_1}) = (4,3),m = 1\)

\(\begin{array}{l}y - {y_1} = m(x - {x_1})\\y - 3 = (x - 4)\\y = x - 1\end{array}\)

Hence, the equation of the tangent line of the curve can be given as: \(y = x - 1\)