Question

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter.x = 6 + ln(t), y = t2 + 3, (6, 4)

Answer 1

Answer:

$y=2x-8$

Step-by-step explanation:

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

BY ELIMINATING THE PARAMETER

To eliminate the parameter we make $t$ the subject in one equation and put it inside the other.

We make $t$ the subject in $x=6+ln(t)$ because it is easier.

$\Rightarrow x-6=ln(t)$

$\Rightarrow {e}^{x-6}=e^{ln(t)}$

$\Rightarrow {e}^{x-6}=t$

Or

$t={e}^{x-6}$

We now substitute this into $y=t^2+3$.

This gives us;

$y=(e^{x-6})^2+3$.

$\Rightarrow y=e^{2(x-6)}+3$.

We have now eliminated the parameter.

The equation of the tangent at (6,4) is given by;

$y-y_1=m(x-x_1)$

where the gradient function is given by;

$\frac{dy}{dx}=2e^{2(x-6)}$

We substitute $x=6$ into the gradient function to obtain the gradient.

$\Rightarrow m=2e^{2(6-6)}$

$\Rightarrow m=2e^0$

$\Rightarrow m=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

WITHOUT ELIMINATING THE PARAMETER

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

For $x=6+ln(t)$

$\frac{dx}{dt}=\frac{1}{t}$

For $y=t^2+3$

$\frac{dy}{dt}=2t$

The slope is given by;

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{dy}{dx}=\frac{2t }{\frac{1}{t} }$

$\frac{dy}{dx}=2t^2$

At the point, (6,4), we plug in any of the values into the parametric equation and find the corresponding value for $t$.

Notice that

When $x=6$, $6=6+\ln(t)$

$6-6=\ln(t)$

$0=\ln(t)$

$e^0=e^\ln(t)$

$1=t$

when $y=4$, $4=t^2+3$

$4-3=t^2$

$1=t^2$

$t=\pm1$

But the slope is the same when we plug in any of these values for t.

$\frac{dy}{dx}=2(\pm1)^2=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$