Find the following for the given parametric equations.
x = t^2 and y= sin 2t. The point A is an intersection of C with the x-axis
Find, in terms of pi, the x coordinate of A
Find dy/dx in terms of t, t>0
Show that an equation of the tangent to C at A is 4x - 2pi*y = pi^2
Hi! Im really stuck on this question and cant to part a or c! Would really appreciate it if you could help me!
Thankyou sooooo much!
1 Answer | Add Yours
It is given that x = t^2 and y = sin 2t for the points of a curve. The y-coordinate of the point A where the curve intersects the x-axis is 0.
sin 2t = 0
=> 2t = `pi`
=> t = `pi/2`
x = t^2 = `(pi/2)^2` = `pi^2/4`
`dy/dt = 2*cos 2t`
`dx/dt = 2t`
`dy/dx = (2*cos 2t)/(2t) = (cos 2t)/t`
The slope of a tangent drawn at any point to the curve is the value of `dy/dx` at that point. At A, `dy/dx` = `1/(pi/2)`
The equation of the tangent is `(y - 0)/(x - pi^2/4) = 1/(pi/2)`
=> `(pi*y)/2 = x - pi^2/4`
=> `4x - 2*pi*y = pi^2`
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