Question

consider the parametric curve given by

$x = \cos(2t)$, $y = - 9\cos(t)$, $0 < t < \pi$

(a) find $dy/dx$ and $d^{2}y/dx^{2}$ in terms of $t$.

$dy/dx =$

$d^{2}y/dx^{2} =$

(b) using \less than\ and \greater than\ notation, list the $t$-interval where the curve is concave upward.

use upper - case \inf\ for positive infinity and upper - case
inf\ for negative infinity. if the curve is never concave upward, type an upper - case
\

in the answer field.

$t$-interval: $0 < t < \pi$

note: you can earn partial credit on this problem.

Explanation:

Part (a)

Step 1: Find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)

For \(y = -9\cos(t)\), the derivative with respect to \(t\) is:
\(\frac{dy}{dt}=9\sin(t)\) (using the derivative of \(\cos(t)\) is \(-\sin(t)\), so \(\frac{d}{dt}(-9\cos(t)) = 9\sin(t)\))

For \(x=\cos(2t)\), using the chain rule \(\frac{d}{dt}\cos(u)=-\sin(u)\cdot\frac{du}{dt}\) where \(u = 2t\), we get:
\(\frac{dx}{dt}=-2\sin(2t)\) (and \(\sin(2t)=2\sin(t)\cos(t)\), so \(\frac{dx}{dt}=-4\sin(t)\cos(t)\))

Step 2: Compute \(\frac{dy}{dx}\)

By the chain rule for parametric curves, \(\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

Substitute \(\frac{dy}{dt}=9\sin(t)\) and \(\frac{dx}{dt}=-4\sin(t)\cos(t)\) (after using double - angle formula \(\sin(2t) = 2\sin(t)\cos(t)\)):

\(\frac{dy}{dx}=\frac{9\sin(t)}{-4\sin(t)\cos(t)}\)

We can cancel out \(\sin(t)\) (note that \(\sin(t)
eq0\) for \(0 < t<\pi\) except at \(t = 0,\pi\) which are not in the open interval). So \(\frac{dy}{dx}=-\frac{9}{4\cos(t)}\) or \(-\frac{9}{4}\sec(t)\)

Step 3: Compute \(\frac{d^{2}y}{dx^{2}}\)

We know that \(\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\)

First, find \(\frac{d}{dt}(\frac{dy}{dx})\):

\(\frac{dy}{dx}=-\frac{9}{4}\sec(t)\), and the derivative of \(\sec(t)\) with respect to \(t\) is \(\sec(t)\tan(t)\), so \(\frac{d}{dt}(\frac{dy}{dx})=-\frac{9}{4}\sec(t)\tan(t)\)

We already have \(\frac{dx}{dt}=-4\sin(t)\cos(t)\)

Now, \(\frac{d^{2}y}{dx^{2}}=\frac{-\frac{9}{4}\sec(t)\tan(t)}{-4\sin(t)\cos(t)}\)

Simplify \(\sec(t)=\frac{1}{\cos(t)}\) and \(\tan(t)=\frac{\sin(t)}{\cos(t)}\):

$$
LATEXBLOCK0
$$

(We can cancel \(\sin(t)\) since \(\sin(t)
eq0\) for \(0 < t<\pi\))

Part (b)

A curve is concave upward when \(\frac{d^{2}y}{dx^{2}}>0\)

We have \(\frac{d^{2}y}{dx^{2}}=\frac{9}{16\cos^{3}(t)}\)

We want to find when \(\frac{9}{16\cos^{3}(t)}>0\)

Since \(\frac{9}{16}>0\), the sign of \(\frac{d^{2}y}{dx^{2}}\) is determined by the sign of \(\cos^{3}(t)\)

\(\cos^{3}(t)>0\) when \(\cos(t)>0\) (because if \(a>0\), \(a^{3}>0\); if \(a < 0\), \(a^{3}<0\))

\(\cos(t)>0\) for \(0 < t<\frac{\pi}{2}\) (since the cosine function is positive in the first quadrant)

Final Answers

(a)

\(\frac{dy}{dx}=\boldsymbol{-\frac{9}{4\cos(t)}}\) (or \(-\frac{9}{4}\sec(t)\))

\(\frac{d^{2}y}{dx^{2}}=\boldsymbol{\frac{9}{16\cos^{3}(t)}}\)

(b)

The \(t\) - interval where the curve is concave upward is \(\boldsymbol{0 < t<\frac{\pi}{2}}\)

Answer:

Part (a)

Step 1: Find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)

For \(y = -9\cos(t)\), the derivative with respect to \(t\) is:
\(\frac{dy}{dt}=9\sin(t)\) (using the derivative of \(\cos(t)\) is \(-\sin(t)\), so \(\frac{d}{dt}(-9\cos(t)) = 9\sin(t)\))

For \(x=\cos(2t)\), using the chain rule \(\frac{d}{dt}\cos(u)=-\sin(u)\cdot\frac{du}{dt}\) where \(u = 2t\), we get:
\(\frac{dx}{dt}=-2\sin(2t)\) (and \(\sin(2t)=2\sin(t)\cos(t)\), so \(\frac{dx}{dt}=-4\sin(t)\cos(t)\))

Step 2: Compute \(\frac{dy}{dx}\)

By the chain rule for parametric curves, \(\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

Substitute \(\frac{dy}{dt}=9\sin(t)\) and \(\frac{dx}{dt}=-4\sin(t)\cos(t)\) (after using double - angle formula \(\sin(2t) = 2\sin(t)\cos(t)\)):

\(\frac{dy}{dx}=\frac{9\sin(t)}{-4\sin(t)\cos(t)}\)

We can cancel out \(\sin(t)\) (note that \(\sin(t)
eq0\) for \(0 < t<\pi\) except at \(t = 0,\pi\) which are not in the open interval). So \(\frac{dy}{dx}=-\frac{9}{4\cos(t)}\) or \(-\frac{9}{4}\sec(t)\)

Step 3: Compute \(\frac{d^{2}y}{dx^{2}}\)

We know that \(\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\)

First, find \(\frac{d}{dt}(\frac{dy}{dx})\):

\(\frac{dy}{dx}=-\frac{9}{4}\sec(t)\), and the derivative of \(\sec(t)\) with respect to \(t\) is \(\sec(t)\tan(t)\), so \(\frac{d}{dt}(\frac{dy}{dx})=-\frac{9}{4}\sec(t)\tan(t)\)

We already have \(\frac{dx}{dt}=-4\sin(t)\cos(t)\)

Now, \(\frac{d^{2}y}{dx^{2}}=\frac{-\frac{9}{4}\sec(t)\tan(t)}{-4\sin(t)\cos(t)}\)

Simplify \(\sec(t)=\frac{1}{\cos(t)}\) and \(\tan(t)=\frac{\sin(t)}{\cos(t)}\):

$$
LATEXBLOCK0
$$

(We can cancel \(\sin(t)\) since \(\sin(t)
eq0\) for \(0 < t<\pi\))

Part (b)

A curve is concave upward when \(\frac{d^{2}y}{dx^{2}}>0\)

We have \(\frac{d^{2}y}{dx^{2}}=\frac{9}{16\cos^{3}(t)}\)

We want to find when \(\frac{9}{16\cos^{3}(t)}>0\)

Since \(\frac{9}{16}>0\), the sign of \(\frac{d^{2}y}{dx^{2}}\) is determined by the sign of \(\cos^{3}(t)\)

\(\cos^{3}(t)>0\) when \(\cos(t)>0\) (because if \(a>0\), \(a^{3}>0\); if \(a < 0\), \(a^{3}<0\))

\(\cos(t)>0\) for \(0 < t<\frac{\pi}{2}\) (since the cosine function is positive in the first quadrant)

Final Answers

(a)

\(\frac{dy}{dx}=\boldsymbol{-\frac{9}{4\cos(t)}}\) (or \(-\frac{9}{4}\sec(t)\))

\(\frac{d^{2}y}{dx^{2}}=\boldsymbol{\frac{9}{16\cos^{3}(t)}}\)

(b)

The \(t\) - interval where the curve is concave upward is \(\boldsymbol{0 < t<\frac{\pi}{2}}\)