Question

How do you find parametric equations for the tangent line to the curve with the given parametric equations $$x=7t^2-4$$ and $$y=7t^2+4$$ and $$z=6t+5$$ and $$(3,11,11)$$$$?$$

Answer 1

Hint: To get the equation for the tangent line to the curve. We have to firstly calculate the value for $$t$$ by substituting the value of $$x$$from the given point. Now differentiate all the three equations with respect to $$t$$, the coefficients of derivatives will give the generic vector and from that vector and the given point we can give the equation for the tangent line.

Complete step-by-step solution:
In the given question we have to find the parametric equation for the tangent line to the curve with given parameters but for that we have to first get the knowledge of parametric equations.
Parametric equations are the equations in which two or more variables are represented in the form of another same variable that equations are known as parametric equations.
For example, let us assume the equations
$$\begin{array}{rl}& x=4t\\ & y=10t\end{array}$$
These equations are the parametric equations because $$x$$ and $$y$$ variables are represented in the form of the same variable i.e. $$t$$. So here we can say that $$t$$ is working as the parameter.
Firstly we have to find the value of $$t$$ by simply putting the value $$x$$ or $$y$$ or we can say $$z$$ in their respective equations. So let us put the value of $$x$$ in the equation $$x=7t^2-4$$, we get
The given points are $$(3,11,11)$$ by this we can say that $$x=3$$. So we get
$$\begin{array}{rl}& \Rightarrow 3=7t^2-4\\ & \Rightarrow t^2=1\\ & \Rightarrow t=\pm 1\end{array}$$
The value for $$t$$ can be $$+1$$ or $$-1$$. But $$-1$$ is not possible because it is not satisfying the equation for $$z$$ i.e. $$z=6t+5$$. So $$t=-1$$ gets neglected.
The given equations are,
$$x=7t^2-4$$
$$y=7t^2+4$$
$$z=6t+5$$
Differentiate all the equations with respect to the variable $$t$$, we get

$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{dx}{dt}}={\displaystyle \frac{d}{dt}}(7t^2-4)\\ & \Rightarrow {\displaystyle \frac{dx}{dt}}=14t\end{array}$$
Similarly we can say that,
$$\begin{array}{rl}& {\displaystyle \frac{dy}{dt}}=14t\\ & {\displaystyle \frac{dz}{dt}}=6\end{array}$$
From the above equations we can say that for $$t=1$$ the generic vector to the curve is $$\overrightarrow{a}(14,14,6)$$
The line passing from the given point in the direction of the vector $$\overrightarrow{a}$$ gives the tangent to the curve and the equation for the line is
$$\begin{array}{rl}& x=3+14t\\ & y=11+14t\\ & z=11+6t\end{array}$$
But we can also express the direction $$(14,14,6)$$ as $$(7,7,3)$$. So the final equation of the line will become
$$\begin{array}{rl}& x=3+7t\\ & y=11+7t\\ & z=11+3t\end{array}$$

Note: Parametric equations are the most important and most useful concept because it is used to locate the complete position of any object in a graph. Parametric equations are used to draw all the types of curves but it is mostly used in the situation where the curves cannot be described by any function.