Respuesta :
By the chain rule, the first derivative is
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}[/tex]
We have
[tex]x=\cos2t\implies\dfrac{\mathrm dx}{\mathrm dt}=-2\sin2t[/tex]
[tex]y=\cos t\implies\dfrac{\mathrm dy}{\mathrm dt}=-2\sin t[/tex]
so that
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{-2\sin t}{-2\sin2t}=\dfrac{\sin t}{2\sin t\cos t}=\boxed{\dfrac{\sec t}2}[/tex]
The second derivative is
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{\mathrm dy}{\mathrm dx}\right][/tex]
where [tex]\dfrac{\mathrm dy}{\mathrm dx}[/tex] is a function of [tex]t[/tex]; denote it by [tex]f(t)[/tex]. Then by the chain rule,
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}[/tex]
Since
[tex]f=\dfrac{\sec t}2\implies\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\sec t\tan t}2[/tex]
we end up with
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac{\sec t\tan t}2}{-2\sin2t}=\boxed{-\dfrac{\sec^3t}8}[/tex]
The curve is concave upward whenever the second derivative is positive. We have this for
[tex]-\dfrac{\sec^3t}8>0\implies\sec^3t=\dfrac1{\cos^3t}<0\implies\cos^3t<0[/tex]
We have [tex]\cos t=0[/tex] for [tex]t=\dfrac{n\pi}2[/tex] where [tex]n[/tex] is any integer; for the given interval, this happens at [tex]t=\dfrac\pi2[/tex]. For [tex]0<t<\dfrac\pi2[/tex], we have [tex]\cos t>0[/tex], while for [tex]\boxed{\dfrac\pi2<t<\pi}[/tex], we have [tex]\cos t<0[/tex]; it's the latter that we care about, since over this interval we have [tex]\sec^3t<0[/tex].
The value of 't' at which the curve is concave upward is [tex]\rm \pi/2<t<\pi[/tex] and this can be determined by using the properties of the trigonometric function.
Given :
- x = cos(2t),
- y = cos(t), 0 < t < π
First, differentiate 'x' with respect to 't'.
[tex]\rm \dfrac{dx}{dt} = -2\;sin\;2t[/tex]
Now, differentiate 'y' with respect to 't'.
[tex]\rm \dfrac{dy}{dt} = -2\;sin\;t[/tex]
Now, the value of dy/dx is given by:
[tex]\rm \dfrac{dy}{dx} =\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{sin\;t}{sin\;2t}=\dfrac{sec\;t}{2}[/tex]
Now, the expression of [tex]\rm d^2y/dx^2[/tex] is given by:
[tex]\rm \dfrac{d^2y}{dx^2}=\dfrac{\dfrac{df}{dt}}{\dfrac{dx}{dt}}[/tex] --- (1)
where f is the function of 't'.
[tex]\rm f =\dfrac{sec\;t}{2}[/tex]
[tex]\rm \dfrac{df}{dt}=\dfrac{sec\;t\times tan\;t}{2}[/tex]
Substitute the value of df/dt in the expression (1).
[tex]\rm \dfrac{d^2y}{dx^2}=\dfrac{\dfrac{sec\;t\times tan\;t}{2}}{-2sin\;2t}=-\dfrac{sec^3t}{8}[/tex]
For the curve to be concave upward the expression of the second derivative should be greater than zero.
[tex]\rm -\dfrac{sec^3t}{8}>0[/tex]
[tex]\rm cos^3t<0[/tex]
Now, the value of 't' at which (cos t < 0) is given by:
[tex]\rm \dfrac{\pi}{2}<t<\pi[/tex]
For more information, refer to the link given below:
https://brainly.com/question/13077606
