Find parametric equations for the tangent line to the curve with parametric equations $x = 3 t$, $y = 5 t ^ { 2 } , z = 4 t ^ { 3 }$ at the point with $t = 1$.
A) $x = 1 + 3 t , y = 1 + 10 t , z = 1 + 12 t$
B) $x = 3 + 3 t , y = 5 + 10 t , z = 4 + 12 t$
C) $x = 1 + 3 t , y = 1 + 5 t , z = 1 + 4 t$
D) $x = 3 + 3 t , y = 5 + 5 t , z = 4 + 4 t$

Question

Quizplus · Accepted Answer

To find the tangent line at the point with $t=1$, we need to first find the coordinates of the point. Plugging in $t=1$ into the parametric equations, we get:

$x=3(1)=3$
$y=5(1)^2=5$
$z=4(1)^3=4$

So the point is $(3,5,4)$.

To find the direction vector of the tangent line, we need to take the derivative of each component and evaluate it at $t=1$:

$$\vec{v}=\left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right)|_{t=1}=(3(1)^0,10(1)^1,12(1)^2)=(3,10,12)$$

So the direction vector is $\vec{v}=(3,10,12)$.

Now we can use the point-slope form of the equation of a line to find the equation of the tangent line:

$$\vec{r}=\vec{r_0}+t\vec{v}$$
$$\begin{pmatrix}x\y\z\end{pmatrix}=\begin{pmatrix}3\5\4\end{pmatrix}+t\begin{pmatrix}3\10\12\end{pmatrix}$$
$$x=3+3t,\quad y=5+10t,\quad z=4+12t$$

So the correct choice is B, which gives the desired parametric equations.

Find Parametric Equations for the Tangent Line to the Curve x=3tx = 3 tx=3t

Find Parametric Equations for the Tangent Line to the Curve $x = 3 t$