Find the points on the curve $x ( t ) = 4 t , y ( t ) = t ^ { 2 } + 3$ where the tangent line is horizontal.

A) $( 0,0 )$
B) $( - 3,3 )$
C) $( 3,0 )$
D) $( 0,3 )$
E) $( 0 , - 3 )$

Question

Find the points on the curve $x ( t ) = 4 t , y ( t ) = t ^ { 2 } + 3$ where the tangent line is horizontal.

A) $( 0,0 )$
B) $( - 3,3 )$
C) $( 3,0 )$
D) $( 0,3 )$
E) $( 0 , - 3 )$

Quizplus · Accepted Answer

The tangent line to a curve is horizontal where the derivative of $y$ with respect to $x$ is zero. To find this, we first find $\frac{dy}{dt}$ and $\frac{dx}{dt}$, and then use the chain rule to find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. Given $x(t) = 4t$ and $y(t) = t^2 + 3$, we have $\frac{dx}{dt} = 4$ and $\frac{dy}{dt} = 2t$. Thus, $\frac{dy}{dx} = \frac{2t}{4} = \frac{t}{2}$. Setting this equal to zero gives $t = 0$. Substituting $t = 0$ into the original equations for $x(t)$ and $y(t)$ gives $x(0) = 0$ and $y(0) = 3$, so the point is $(0, 3)$.

Find the Points on the Curve x(t)=4t,y(t)=t2+3x ( t ) = 4 t , y ( t ) = t ^ { 2 } + 3x(t)=4t,y(t)=t2+3

Find the Points on the Curve $x ( t ) = 4 t , y ( t ) = t ^ { 2 } + 3$