# Find an equation of the tangent line to this parametric curve

I am trying to solve this problem:

Find an equation of the tangent line to the curve $x=t^3 +5, y=t^2 +3$ at the point corresponding to $t=1$.

Here is my work (which is wrong according the the online submittion software) First I need a point $(x,y)$. So I plug $t=1$ in the parametric equations to get $x=6$ and $y=4$.

Next, I need $m$, the slope. So I follow the formula $$(dy/dt)/(dx/dt)$$ and I end up with $2t/3t^2$ which simplifies to $2/3t$. I plug in $t=1$ and get $2/3$ for my slope. Then I simply plug my point $(6,4)$ and my $m$ into the tangent line equation $$y_1 - y=m(x_1 -x)$$My answer is therefore $$y=2/3(x-6)+4$$But apparently that answer is wrong. Any ideas why?

• Where did that $1/2$ in the last step come from? – dxiv Nov 6 '16 at 23:02
• Was a typo. Thanks. – Dragneel Nov 6 '16 at 23:07
• The online software probably wanted the answer in the form $y=mx+b$ in which case the answer it wanted was $y=\frac{2}{3}x$. – John Wayland Bales Nov 6 '16 at 23:10
• Your work is correct for the curve as given. One possibility is, a single line has infinitely many correct equations. Another is that the original question was copied incorrectly. Do you know "the answer is wrong" because an automated software system doesn't accept it as correct, or for some other reason...? – Andrew D. Hwang Nov 6 '16 at 23:10
• @JohnWaylandBales That might be the case. I can not check at the moment, but thanks for the input. – Dragneel Nov 6 '16 at 23:16

## 1 Answer

I think it would be better to consider the parametric curve $r(t) = (t^3 + 5,t^2 + 3)$. Then, the tangent vector to the curve at $t = 1$ will be $r'(1)$. With that in mind, $r'(t) = (3 t^2, 2t)$ and $r'(1) = (3,2)$. As you said, we need the point at $t=1$, which is $(6,4)$. Now, we just need the equation of a line that passes through $(6,4)$ with direction $(3,2)$: $$L(t) = (6,4) + t(3,2)$$