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 \(\vec {r}(t)\)\(\vec {r}'(t)\) when \(t=\answer {2}\).
- A point \(P_0\) on the tangent line is found by evaluating \(\vec {r}(t)\)\(\vec {r}'(t)\) 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.