Question

suppose a curve is traced by the parametric equations
$x = 5(\sin(t)+\cos(t))$ and $y = 44 - 15\cos^{2}(t)-30\sin(t)$ with $0\leq t\leq\pi$.
at what point, $(x,y)$, on this curve is the tangent line horizontal?
$(x,y)=$

Explanation:

Step1: Find derivatives of \(x\) and \(y\) with respect to \(t\)

For \(x = 5(\sin(t)+\cos(t))\), using the sum - rule \((u + v)^\prime=u^\prime + v^\prime\) and \((\sin t)^\prime=\cos t\), \((\cos t)^\prime=-\sin t\), we have \(\frac{dx}{dt}=5(\cos(t)-\sin(t))\).
For \(y = 44-15\cos^{2}(t)-30\sin(t)\), using the chain - rule \((u^{n})^\prime = nu^{n - 1}u^\prime\) (\(u = \cos t\), \(n = 2\)) and \((\sin t)^\prime=\cos t\), we get \(\frac{dy}{dt}=-15\times2\cos(t)(-\sin(t))-30\cos(t)=30\sin(t)\cos(t)-30\cos(t)=30\cos(t)(\sin(t)-1)\).

Step2: Use the formula \(\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) and set \(\frac{dy}{dx} = 0\)

Since \(\frac{dy}{dx}=\frac{30\cos(t)(\sin(t)-1)}{5(\cos(t)-\sin(t))}\), for \(\frac{dy}{dx}=0\), we need \(\frac{dy}{dt} = 0\) (and \(\frac{dx}{dt}
eq0\)).
Set \(\frac{dy}{dt}=30\cos(t)(\sin(t)-1)=0\).

• Case 1: \(\cos(t)=0\). Then \(t=\frac{\pi}{2}\) (since \(0\leq t\leq\pi\)).
• When \(t = \frac{\pi}{2}\), \(x = 5(\sin(\frac{\pi}{2})+\cos(\frac{\pi}{2}))=5(1 + 0)=5\).
• \(y=44-15\cos^{2}(\frac{\pi}{2})-30\sin(\frac{\pi}{2})=44-0 - 30=14\).
• Case 2: \(\sin(t)-1 = 0\), i.e., \(\sin(t)=1\), then \(t=\frac{\pi}{2}\) (same as above).

Answer:

\((x,y)=(5,14)\)