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# How do you find parametric equation for the tangent line to the curve with the given parametric equation at the specified point $x=1+10.\sqrt{t},y={{t}^{5}}-t,\text{ and }z={{t}^{5}}+t;\left( 11,0,2 \right)$ ?

Last updated date: 09th Aug 2024
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Hint: We are given 3 parametric equation that define our function as $x=1+10.\sqrt{t},y={{t}^{5}}-t,\text{ and }z={{t}^{5}}+t;$at point $\left( 11,0,2 \right)$ . We are asked to find parametric equation for the tangent line, we first find the parametric value of ‘t’ which will correspond to the point $\left( 11,0,2 \right)$ , then we will differentiate our function then find the value of that derivate at the parametric value of ‘t’, then we will get each required value and add the given point with the point obtained by derivative of our function along with ‘t’.

We are given 3 parametric equation of our function as –
$x=1+10.\sqrt{t},y={{t}^{5}}-t,\text{ and }z={{t}^{5}}+t.$
We have to find the parametric equation of the tangent line at the point $\left( 11,0,2 \right)$ .
Now solve this we will first find the value of the parametric variable ‘t’ corresponding to the point $\left( 11,0,2 \right)$ , to do so we can compare any of the three given parametric equations.
So, in $\left( 11,0,2 \right)$ we have $x=11,y=0\text{ and }z=2$
So, we will use $y={{t}^{5}}-t$ to find the value of ‘t’,
So, $0={{t}^{5}}-t$
By simplifying, we get –
$0=t\left( {{t}^{4}}-1 \right)$
So, we get –
Either $t=0$ or $t=1$
If $t=0$ then $z={{t}^{5}}+t$ never satisfy a given point as $z=2$ and $t=0$ will give $z=0$ .
So, the correct value of t’ is 1.
So we get –
$t=1$
Now we will differentiate our function as given our function is comprise if $x=1+10.\sqrt{t},y={{t}^{5}}-t,\text{ and }z={{t}^{5}}+t.$
So we differentiate these and find its value at $t=1$ .
So, $\dfrac{dx}{dt}=\dfrac{10}{2\sqrt{t}}$
$at\text{ }t=1$
$\dfrac{dx}{dt}=5$ ……………………. (1)
Or $x'\left( 4 \right)=5$
$\dfrac{dy}{dt}=5{{t}^{4}}-1$
$at\text{ }t=1$
$\dfrac{dy}{dt}=5-1=4$ or $4'\left( t \right)=4$ …………………………………… (2)
And lastly
$\dfrac{dz}{dt}=5{{t}^{4}}+1$
$at\text{ }t=1$
$\dfrac{dz}{dt}=5+1=6$ or $z'\left( 1 \right)=b$ ……………………………………. (3)
Now, we have got everything we need.
Now we will find our parametric equation of the tangent line.
We know given point $p\left( {{x}_{p}},{{y}_{p}},{{z}_{p}} \right)$ and direction $v\left( a,b,c \right)$ the line pass from that point with this director is
\begin{align} & x=xp+at \\ & y=yp+bt \\ & z=zp+ct \\ \end{align}
We have point $p\ne \left( 11,0,2 \right)$ and direction vector as $\left( 5,4,6 \right)$ (using 1, 2 and 3)
So, using these above, we get –
\begin{align} & x=11+5t \\ & y=0+4t=4t \\ & z=2+6t \\ \end{align}
This is our required parametric form of the tangent.

Note: Remember that if the directions $\left( x',y',z' \right)$ we found has term as common. So can simplify first and then use them in the tangent line to form the parametric equation.
So, it say $\left( x',y',z' \right)=\left( 2,4,6 \right)$ then –
\begin{align} & x=1+2t \\ & y=2+4t \\ & z=3+6t \\ \end{align} is same as \begin{align} & x=1+t \\ & y=2+2t \\ & z=3+3t \\ \end{align}
It was common, so it was cancelled and the equation still behaves the same.