Expert: Scotto - 10/16/2009

Question
QUESTION: Find the equations of the tangents to the given curve that pass through the point (15, 9).

x = 9t^2 + 6
y = 6t^3 + 3

I correctly got the answer that the tangent at larger t is         y = x - 6



but I dont know how to find the tangent at the smaller t. I think that it does not exist.

t can = 1 or -1 which is found from 15 = 9t^2 +6 but the problem is that t= -1 does not work when plugged into the y equation!! y will not equal 9!!

ANSWER: At x = 15, take 15 = 9t² + 6, so 9 = 9t², so t = ±1.
At t=1, y = 6+3 = 9, so t = 1 was taken.

The slope is given by dy/dx.  Here, take (dy/dt)/(dx/dt).

Since dy/dt = 18t² and dx/dt = 18t, and we have dy/dx = (dy/dt)/(dx/dt) = 18t²/(18t) = t.



---------- FOLLOW-UP ----------

QUESTION: i think that we are both wrong. The point (15,9) is not even in the curve its only on the tangent lines. so can you please further explain the procedure

Answer
If t=1, then x = 9(1²) + 6 = 9 + 6 = 15 and y = 6(1³) + 3 = 6 + 3 = 9.
The point at t=1 is (15,9).

The slope is given by dy/dx.  Since both x and y are functions of t, it is
(dy/dt)/(dx/dt), since the 'dt's both cancel.

Note that dy/dt = 18t² and dx/dt = 18t.

Since t=1, dy/dy = 18 and dx/dt = 18.  This means dy/dx = 1.

When (dy/dt) is divided by (dx/dt), it is just like variables.  The dt's cancel, leaving dy/dx.
This is another form of the chain rule.  If we have f(t) = f(x)x(t), then
df/dt = (df/dx)(dx/dt), and the dx's cancel.