51af35…: “Behind a silver equation”

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Consider the curve $C = \left \{(x,y) \in \R ^2 : y=2x^2-7\right \}$ .

Find a Cartesian representation of the tangent line at $x=2$ . Express you final answer in the form $y=mx+b$ .

$y= \answer {8x-15}$

Now, require that $x(t) =t$ .

A parametric equation that traces out $C$ is given by

$\vec {r}(t) = \vector {\answer {t},\answer {2t^2-7}}.$

A parameterization of the tangent line to the curve where $x=2$ is

$\vecl (t)=\vector {\answer {t+2},\answer {8t+1}}.$

The curve passes through $x=2$ when $t=\answer {2}$ .

A vector $\vec {v}$ parallel to the tangent line at $x=2$ is found by evaluating when $t=\answer {2}$ .
A point $P_0$ on the tangent line is found by evaluating when $t=\answer {2}$ .

A parametric description of the tangent line can be found from

$\vecl (t) = \vec {v}t+\vec {P}_0$

Do the answers to Part I and Part II describe the same line? Intuitively, they should since the tangent line at the point on the curve where $x=2$ should not depend on how we choose to describe the curve.

To check this, we can determine if the set of parametric equations $x(t)=t+2$ and $y(t)=8t+1$ satisfy the equation $y=8x-15$ from the first part.

$8[x(t)]-15 = 8 [t+2] -15 = 8t+16-15 = 8t+1=y(t).$

Thus, the two equations describe the same line.