1. Problem 43: Find the equation of the tangent line to the graph of $y = x^2 \sin 4x$ at $x = \frac{\pi}{4}$. 2. To find the tangent line, we need the point and the slope at $x = \frac{\pi}{4}$. The slope is the derivative $y'$. 3. Use the product rule: If $y = u v$, then $y' = u' v + u v'$. Here, $u = x^2$, $v = \sin 4x$. 4. Compute derivatives: $u' = 2x$, $v' = 4 \cos 4x$ (chain rule). 5. So, $$y' = 2x \sin 4x + x^2 (4 \cos 4x) = 2x \sin 4x + 4x^2 \cos 4x.$$ 6. Evaluate $y$ and $y'$ at $x = \frac{\pi}{4}$: $$y\left(\frac{\pi}{4}\right) = \left(\frac{\pi}{4}\right)^2 \sin \left(4 \cdot \frac{\pi}{4}\right) = \frac{\pi^2}{16} \sin \pi = 0.$$ $$y'\left(\frac{\pi}{4}\right) = 2 \cdot \frac{\pi}{4} \sin \pi + 4 \left(\frac{\pi}{4}\right)^2 \cos \pi = \frac{\pi}{2} \cdot 0 + 4 \cdot \frac{\pi^2}{16} \cdot (-1) = -\frac{\pi^2}{4}.$$ 7. The slope of the tangent line at $x=\frac{\pi}{4}$ is $m = -\frac{\pi^2}{4}$ and the point is $\left(\frac{\pi}{4}, 0\right)$. 8. Equation of tangent line: $$y - y_1 = m (x - x_1)$$ $$y - 0 = -\frac{\pi^2}{4} \left(x - \frac{\pi}{4}\right)$$ $$\boxed{y = -\frac{\pi^2}{4} x + \frac{\pi^3}{16}}.$$ --- 1. Problem 44: Find $x$ where the tangent line to $y = \sin 3x + \cos 3x$ is horizontal. 2. Horizontal tangent means slope $y' = 0$. 3. Compute derivative: $$y' = 3 \cos 3x - 3 \sin 3x = 3(\cos 3x - \sin 3x).$$ 4. Set $y' = 0$: $$3(\cos 3x - \sin 3x) = 0 \implies \cos 3x = \sin 3x.$$ 5. Divide both sides by $\cos 3x$ (assuming $\cos 3x \neq 0$): $$\cancel{\cos 3x} = \tan 3x \cancel{\cos 3x} \implies 1 = \tan 3x.$$ 6. So, $$\tan 3x = 1.$$ 7. General solution for $\tan \theta = 1$ is $$\theta = \frac{\pi}{4} + k\pi, \quad k \in \mathbb{Z}.$$ 8. Substitute back: $$3x = \frac{\pi}{4} + k\pi \implies x = \frac{\pi}{12} + \frac{k\pi}{3}, \quad k \in \mathbb{Z}.$$ 9. Therefore, all $x$ where tangent is horizontal are $$\boxed{x = \frac{\pi}{12} + \frac{k\pi}{3}, \quad k \in \mathbb{Z}}.$$ --- 1. Problem 45(a): Given $s(t) = 12 + \cos(\pi t)$, find velocity $v(t) = s'(t)$ at $t=0, \frac{1}{2}, 1, \frac{3}{2}, 2$. 2. Derivative: $$v(t) = s'(t) = -\pi \sin(\pi t)$$ by chain rule. 3. Evaluate: - $v(0) = -\pi \sin 0 = 0$ - $v\left(\frac{1}{2}\right) = -\pi \sin \frac{\pi}{2} = -\pi$ - $v(1) = -\pi \sin \pi = 0$ - $v\left(\frac{3}{2}\right) = -\pi \sin \frac{3\pi}{2} = -\pi (-1) = \pi$ - $v(2) = -\pi \sin 2\pi = 0$ --- 1. Problem 45(b): Find intervals where cork is rising (velocity $>0$) and falling (velocity $<0$). 2. Since $$v(t) = -\pi \sin(\pi t),$$ sign depends on $\sin(\pi t)$. 3. $\sin(\pi t) > 0$ on intervals $(2k, 2k+1)$ for integers $k$, so $v(t) = -\pi \sin(\pi t) < 0$ there (falling). 4. $\sin(\pi t) < 0$ on intervals $(2k+1, 2k+2)$, so $v(t) > 0$ there (rising). 5. For $t \geq 0$, cork is rising on intervals $$\boxed{(1,2), (3,4), (5,6), \ldots}$$ and falling on intervals $$\boxed{(0,1), (2,3), (4,5), \ldots}.$$