parametric equation tangent line?

Question

find the equation of the line tangent to
x(t)=2 cos (3t)
y(t)= 7 sin (4t)
0 less or equal t less or equal pi/2
where the graph croses the x-axis between x=-2 and x=-1

Answer 1

dy= -6sin(3t)dt
dx=28cos(4t)dt so dy/dx= slope = -3sin(3t)/14 cos(4t)
The cross of the x axis means y=0
sin4t=0 so 4t=kpi and t =k pi/4
we must find the value between -2 and -1
-2-8/pi and k<-4/pi
k>-2.54 and k<-1.27
So k = -2 as it is an integer so t=-pi/2
slope= 3/14
x(-pi/2)= 0
y(-pi/2)=0
so the tangent is y=3/14 *x

Answer 2

Q1. Yes -- this is the slope of the line tangent to the curve. Q2. Yes -- this is the equation of the tangent to the curve Q3. You want to make sure this tangent passes through (4,3). This is ensured by demanding that 3-[2t^3+1]) = t(4-[3t^2+1]) This now says that if t satisfies this equation, the line that is the tangent to the curve at (3t^2+1,2t^3+1) passes through (4,3). Q4. See comment above. This equation is a cubic in t so ought to have three roots (I'm not going to solve the problem as you've asked that we dont find the answers.) But we do know one solution -- t=1. Putting t=1 in the original definition of the curve shows that (4,3) is on the curve so the tangent to the curve at that point goes through it! Hope these comments help

Answer 3

If you can't work these sorts of things out, normally pi/2 is a good guess.