Question

Use a graph to estimate the coordinates of the lowest point and the leftmost point on the curve \(x = {t^4} - 2t,y = {t^4} + t\).Then use calculus to find the exact coordinates.

Short answer

The exact coordinates of the leftmost point is: \(\left( { - \frac{3}{{2\sqrt(3){2}}},\frac{3}{{2\sqrt(3){2}}}} \right)\) and the lowest point is: \(\left( {\frac{9}{{4\sqrt(3){4}}}, - \frac{3}{{4\sqrt(3){4}}}} \right)\).

Step-by-step solution

Sketch the graph of the given curve

The graph of the given curve \(x = {t^4} - 2t,y = t + {t^4}\) can be given by as follows:

From the graph, we can see the leftmost point and the lower point can be given as:: \((1.4, - 0.5),( - 1.2,1.2)\) respectively

Find the point for vertical tangent line

Substitute \(\frac{{dx}}{{dt}} = 0\) to find the vertical line

\(\begin{array}{l}x = {t^4} - 2t\\\frac{{dx}}{{dt}} = 4{t^3} - 2\\\frac{{dx}}{{dt}} = 0\\4{t^3} - 2 = 0\\4{t^3} = 2\\{t^3} = \frac{1}{2}\,\\t = \frac{1}{{\sqrt(3){2}}}\end{array}\)

Substitute \(t = \frac{1}{{\sqrt(3){2}}}\) and obtained point

\(\begin{array}{l}(x,y) = ({t^4} - 2t,t + {t^4})\\(x,y) = \left( {{{\left( {\frac{1}{{\sqrt(3){2}}}} \right)}^4} - 2\left( {\frac{1}{{\sqrt(3){2}}}} \right),\frac{1}{{\sqrt(3){2}}} + {{\left( {\frac{1}{{\sqrt(3){2}}}} \right)}^4}} \right)\\(x,y) = \left( { - \frac{3}{{2\sqrt(3){2}}},\frac{3}{{2\sqrt(3){2}}}} \right)\\(x,y) = \left( { - 1.1906,1.1906} \right)\end{array}\)

Find the point for horizontal tangent line

Substitute \(\frac{{dy}}{{dt}} = 0\) to find the vertical line

\(\begin{array}{l}y = t + {t^4}\\\frac{{dy}}{{dt}} = 1 - 4{t^3}\\\frac{{dy}}{{dt}} = 0\\1 - 4{t^3} = 0\\4{t^3} = - 1\\{t^3} = - \frac{1}{4}\,\\t = - \frac{1}{{\sqrt(3){4}}}\end{array}\)

Substitute \(t = - \frac{1}{{\sqrt(3){4}}}\) and obtained point

\(\begin{array}{l}(x,y) = ({t^4} - 2t,t + {t^4})\\(x,y) = \left( {{{\left( { - \frac{1}{{\sqrt(3){4}}}} \right)}^4} - 2\left( { - \frac{1}{{\sqrt(3){4}}}} \right),\frac{{ - 1}}{{\sqrt(3){4}}} + {{\left( { - \frac{1}{{\sqrt(3){4}}}} \right)}^4}} \right)\\(x,y) = \left( {\frac{9}{{4\sqrt(3){4}}}, - \frac{3}{{4\sqrt(3){4}}}} \right)\\(x,y) = \left( {1.4174, - 0.4725} \right)\end{array}\)

Hence, the exact coordinates of the leftmost point of the given curve is: \(\left( { - \frac{3}{{2\sqrt(3){2}}},\frac{3}{{2\sqrt(3){2}}}} \right)\) and the lowest point is: \(\left( {\frac{9}{{4\sqrt(3){4}}}, - \frac{3}{{4\sqrt(3){4}}}} \right)\)