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The curve given by the parametric equations;
x=36−t^2, y=t^3−25t
is symmetric about the x-axis. (If t gives us the point (x,y), then −t will give (x,−y) ).
a. At which x value is the tangent to this curve horizontal?
b. At which t value is the tangent to this curve vertical?

Respuesta :

Answer:

x=36, t = ±5/sqrt 3

Step-by-step explanation:

we are given a curve reprsented by parametric equations.

[tex]x=36-t^2,\\ y=t^3-25t[/tex]

Also given that this is symmetric about x axis.

For finding a tangent we first find slope i.e. dy/dx

If dy/dx has dy/dt in numerator and dx/dt in the denominator

Hence for a tangent to be horizontal slope =0

i.e.dy/dt =0

Similarly for a tangent to be vertical slope=undefined

Or dx/dt =0

a) [tex]\frac{dx}{dt} =-2t =0\\t=0\\x=36[/tex]

when x= 36, the tangent is horizontal

b) [tex]\frac{dy}{dt} =0\\3t^2-25 =-\\t=\frac{5}{\sqrt{3} } ,-frac{5}{\sqrt{3} }[/tex]

For these t values tangent is vertical.

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