Question

72–75. equations of tangent lines
a. find an equation of the line tangent to the following curves at the given value of x.
b. use a graphing utility to plot the curve and the tangent line.

1. y = 4 sin x cos x; x = $\frac{pi}{3}$
2. y = 1 + 2 sin x; x = $\frac{pi}{6}$

Explanation:

1. For problem 72:

• Step - by - step solution for part (a):
• Step1: Simplify the function using double - angle formula
• Recall that \(2\sin x\cos x=\sin(2x)\), so \(y = 4\sin x\cos x = 2\sin(2x)\).
• Step2: Find the derivative of the function
• Using the chain - rule, if \(y = 2\sin(2x)\), let \(u = 2x\), then \(y = 2\sin u\). The derivative of \(y\) with respect to \(u\) is \(\frac{dy}{du}=2\cos u\), and the derivative of \(u\) with respect to \(x\) is \(\frac{du}{dx}=2\). By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\), so \(\frac{dy}{dx}=4\cos(2x)\).
• Step3: Find the slope of the tangent line at \(x = \frac{\pi}{3}\)
• Substitute \(x=\frac{\pi}{3}\) into \(\frac{dy}{dx}\): \(\frac{dy}{dx}\big|_{x = \frac{\pi}{3}}=4\cos(2\cdot\frac{\pi}{3})=4\cos(\frac{2\pi}{3})\). Since \(\cos(\frac{2\pi}{3})=-\frac{1}{2}\), the slope \(m = 4\times(-\frac{1}{2})=-2\).
• Step4: Find the \(y\) - coordinate at \(x=\frac{\pi}{3}\)
• Substitute \(x = \frac{\pi}{3}\) into \(y = 2\sin(2x)\): \(y=2\sin(2\cdot\frac{\pi}{3})=2\sin(\frac{2\pi}{3})=\sqrt{3}\).
• Step5: Use the point - slope form \(y - y_1=m(x - x_1)\) to find the equation of the tangent line
• Here \(x_1=\frac{\pi}{3}\), \(y_1 = \sqrt{3}\), and \(m=-2\). So \(y-\sqrt{3}=-2(x - \frac{\pi}{3})\), which simplifies to \(y=-2x+\frac{2\pi}{3}+\sqrt{3}\).
• For part (b):
• Using a graphing utility (such as a graphing calculator like TI - 84 Plus or software like Desmos), enter the function \(y = 2\sin(2x)\) and the tangent - line function \(y=-2x+\frac{2\pi}{3}+\sqrt{3}\). The curve \(y = 2\sin(2x)\) is a sinusoidal curve, and the tangent line \(y=-2x+\frac{2\pi}{3}+\sqrt{3}\) will touch the curve at the point \((\frac{\pi}{3},\sqrt{3})\).

1. For problem 73:

• Step - by - step solution for part (a):
• Step1: Find the derivative of the function
• Given \(y = 1+2\sin x\), the derivative \(\frac{dy}{dx}=2\cos x\).
• Step2: Find the slope of the tangent line at \(x=\frac{\pi}{6}\)
• Substitute \(x = \frac{\pi}{6}\) into \(\frac{dy}{dx}\): \(\frac{dy}{dx}\big|_{x=\frac{\pi}{6}}=2\cos(\frac{\pi}{6})\). Since \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\), the slope \(m = 2\times\frac{\sqrt{3}}{2}=\sqrt{3}\).
• Step3: Find the \(y\) - coordinate at \(x=\frac{\pi}{6}\)
• Substitute \(x=\frac{\pi}{6}\) into \(y = 1 + 2\sin x\): \(y=1+2\sin(\frac{\pi}{6})=1 + 2\times\frac{1}{2}=2\).
• Step4: Use the point - slope form \(y - y_1=m(x - x_1)\) to find the equation of the tangent line
• Here \(x_1=\frac{\pi}{6}\), \(y_1 = 2\), and \(m=\sqrt{3}\). So \(y - 2=\sqrt{3}(x-\frac{\pi}{6})\), which simplifies to \(y=\sqrt{3}x-\frac{\sqrt{3}\pi}{6}+2\).
• For part (b):
• Using a graphing utility (like Desmos or a graphing calculator), enter the function \(y = 1+2\sin x\) and the tangent - line function \(y=\sqrt{3}x-\frac{\sqrt{3}\pi}{6}+2\). The curve \(y = 1+2\sin x\) is a sinusoidal curve shifted up by 1 unit, and the tangent line \(y=\sqrt{3}x-\frac{\sqrt{3}\pi}{6}+2\) will touch the curve at the point \((\frac{\pi}{6},2)\).

Answer:

For problem 72:
a. \(y=-2x+\frac{2\pi}{3}+\sqrt{3}\)
b. Use a graphing utility to plot \(y = 2\sin(2x)\) and \(y=-2x+\frac{2\pi}{3}+\sqrt{3}\)

For problem 73:
a. \(y=\sqrt{3}x-\frac{\sqrt{3}\pi}{6}+2\)
b. Use a graphing utility to plot \(y = 1+2\sin x\) and \(y=\sqrt{3}x-\frac{\sqrt{3}\pi}{6}+2\)