Answer:
[tex]y(t) = \frac{1}{t}(x - 18}) = 12[/tex]
Step-by-step explanation:
The tangent of a curve is given by the following first degree function:
[tex]y - y_{0} = m(x - x_{0})[/tex]
In which [tex](x_{0}, y_{0})[/tex] is a function in which the curve pass through and m is the slope, given by the following formula:
[tex]m = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
In this problem, we have that:
[tex]x = 9t^{2} + 9[/tex].
So:
[tex]\frac{dx}{dt} = 18t[/tex]
And
[tex]y = 6t^{3} + 6[/tex]
So:
[tex]\frac{dy}{dt} = 18t^{2}[/tex].
Which means that
[tex]m = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{18t}{18t^{2}} = \frac{1}{t}[/tex]
The curve passes through (18,12). This means that [tex]x_{0} = 18, y_{0} = 12[/tex].
So the equation is:
[tex]y - y_{0} = m(x - x_{0})[/tex]
[tex]y - 12 = \frac{1}{t}(x - 18})[/tex]
[tex]y(t) = \frac{1}{t}(x - 18}) = 12[/tex]
