Question

Calculus: Find equations of the tangents to the curve x=(3t^2)+1, y= (2t^3)+1 that pass through the point (4,3). (I know that I need to find dy/dx, but when I solve for t, since i have 2 equations, I have 2 ts, and don't know which one to use.

Answer 1

dy/dx = (dy/dt)*(dt/dx)

Answer 2

The point (4,3) corresponds to t=1

Answer 3

What is
\[
\frac {dy}{dt}= ?\\
\frac {dx}{dt} = ?\\

\frac {\frac {dy}{dt}}
{\frac {dx}{dt}}=?
\]

Answer 4

I will do one for you
\[
y= 2t^3 +1\\
\frac {dy}{dt} = 6 t^2
\]

Answer 5

then
\[
x=3t^2+1\\
\frac { dx}{dt}= 6t
\]

Answer 6

The ratio of the two is
\[
\frac {6 t^2}{6t}=t
\]
so
\[\frac {dy}{dx}=t=1
\]
at the point.
The slope of the tangent is 1 at the point (4,3)\\
the equation of the tangent
y-3 = 1(x-4)
y= x-4 +3 = x -1