Gradient of the parametric curve at one of it's X-intercepts

[P-isabel]

So the question is asking to use the derivative to find the gradient of the curve at one of its x-intercepts.
My teacher has said that I first need to find a suitable value for t and then evaluate to get a gradient.
The derivative I have found is: 3cos3t/-15sin(3t)
I am unsure of how to find a suitable value for t

 

[Deleted member 4993]

So the question is asking to use the derivative to find the gradient of the curve at one of its x-intercepts.
My teacher has said that I first need to find a suitable value for t and then evaluate to get a gradient.
The derivative I have found is: 3cos3t/-15sin(3t)
I am unsure of how to find a suitable value for t

Which curve ?

 

[Otis]

... 3cos3t/-15sin(3t) ...

Hello. Did you see Dr. Peterson's reply yesterday (in your other thread), where he said you can simplify by cancelling common factors? He showed we need grouping symbols around the denominator, too. The same applies above.

Also, it's good that you expressed sine using function notation; you ought to do the same with cosine. And, a negative sign in a ratio can move out front. Putting it all together:

-cos(3t)/[5 sin(3t)] \(\;\) or \(\;\) -1/5 cot(3t)

?