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?
Best 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 $(3,2)$: $$L(t) = (6,4) + t(3,2)$$