Find the points on the curve $x ( t ) = \frac { t ^ { 3 } } { 1 + t } , y ( t ) = \frac { 1 } { 1 + t }$ where the tangent line is vertical.

A) $\left( \frac { 27 } { 4 } , - 2 \right)$
B) $( 0,1 ) , \left( \frac { 27 } { 4 } , - 2 \right)$
C) $( 0,1 )$
D) $( 0,1 ) , \left( \frac { 27 } { 4 } , 2 \right)$
E) $( 0,1 ) , \left( - \frac { 27 } { 4 } , 2 \right)$

Question

Find the points on the curve $x ( t ) = \frac { t ^ { 3 } } { 1 + t } , y ( t ) = \frac { 1 } { 1 + t }$ where the tangent line is vertical.

A) $\left( \frac { 27 } { 4 } , - 2 \right)$
B) $( 0,1 ) , \left( \frac { 27 } { 4 } , - 2 \right)$
C) $( 0,1 )$
D) $( 0,1 ) , \left( \frac { 27 } { 4 } , 2 \right)$
E) $( 0,1 ) , \left( - \frac { 27 } { 4 } , 2 \right)$

Quizplus · Accepted Answer

The tangent line is vertical when the derivative of x(t) with respect to t is 0. So, we have:$$\frac{d}{dt}\left(\frac{t^3}{1+t}\right)=\frac{(1+t)\frac{d}{dt}(t^3)-t^3\frac{d}{dt}(1+t)}{(1+t)^2}=\frac{(1+t)(3t^2)-t^3(1)}{(1+t)^2}=\frac{3t^3+3t^2-t^3}{(1+t)^2}=\frac{2t^3+3t^2}{(1+t)^2}$$Setting this equal to 0 and solving for t, we get:$$2t^3+3t^2=0$$$$t^2(2t+3)=0$$$$t=0,-\frac{3}{2}$$So, the points on the curve where the tangent line is vertical are:$$\left(\frac{0^3}{1+0}, \frac{1}{1+0}\right)=(0,1)$$$$\left(\frac{(-\frac{3}{2})^3}{1+(-\frac{3}{2})}, \frac{1}{1+(-\frac{3}{2})}\right)=\left(\frac{27}{4},-2\right)$$

Find the Points on the Curve x(t)=t31+t,y(t)=11+tx ( t ) = \frac { t ^ { 3 } } { 1 + t } , y ( t ) = \frac { 1 } { 1 + t }x(t)=1+tt3​,y(t)=1+t1​

Find the Points on the Curve $x ( t ) = \frac { t ^ { 3 } } { 1 + t } , y ( t ) = \frac { 1 } { 1 + t }$