1. Problem 26: Find the velocity at time $t_0$ given the position function $s(t)$. Velocity is the derivative of position with respect to time: $v(t) = \frac{ds}{dt}$. For each part, differentiate $s(t)$ and evaluate at $t_0$. а) $s=3\ln t - 3t + 2$ $\frac{ds}{dt} = \frac{3}{t} - 3$ At $t_0=1$: $v(1) = 3 - 3 = 0$ б) $s=-4\ln t + 5t + 4$ $\frac{ds}{dt} = -\frac{4}{t} + 5$ At $t_0=1$: $v(1) = -4 + 5 = 1$ в) $s=-2\ln t + 5t + 4$ $\frac{ds}{dt} = -\frac{2}{t} + 5$ At $t_0=2$: $v(2) = -1 + 5 = 4$ г) $s=3\ln t - 4t + 3$ $\frac{ds}{dt} = \frac{3}{t} - 4$ At $t_0=3$: $v(3) = 1 - 4 = -3$ д) $s=3\ln t + 3t - 2$ $\frac{ds}{dt} = \frac{3}{t} + 3$ At $t_0=1$: $v(1) = 3 + 3 = 6$ е) $s=-2.5t^2 + 45\ln t + 5$ $\frac{ds}{dt} = -5t + \frac{45}{t}$ At $t_0=3$: $v(3) = -15 + 15 = 0$ ё) $s=3.5t^2 - 5\ln t + 2.6$ $\frac{ds}{dt} = 7t - \frac{5}{t}$ At $t_0=1$: $v(1) = 7 - 5 = 2$ ж) $s=-3t^2 + 50\ln t - 1.7$ $\frac{ds}{dt} = -6t + \frac{50}{t}$ At $t_0=5$: $v(5) = -30 + 10 = -20$ з) $s=2.5t^2 - 20\ln t - 3.6$ $\frac{ds}{dt} = 5t - \frac{20}{t}$ At $t_0=2$: $v(2) = 10 - 10 = 0$ и) $s=0.5t^2 - 49\ln t - 3.2$ $\frac{ds}{dt} = t - \frac{49}{t}$ At $t_0=5$: $v(5) = 5 - 9.8 = -4.8$ 2. Problem 27: Find intervals where the function is decreasing. Decreasing means $y'(x) < 0$. а) $y=5x - 2\ln x$ $y' = 5 - \frac{2}{x}$ Set $y' < 0$: $5 - \frac{2}{x} < 0 \Rightarrow 5 < \frac{2}{x} \Rightarrow x < \frac{2}{5} = 0.4$ Domain $x>0$, so decreasing on $(0, 0.4)$. б) $y = x + \ln x$ $y' = 1 + \frac{1}{x}$ Since $x>0$, $y' > 0$ always, so no decreasing interval. в) $y = x^2 - \ln x^8 = x^2 - 8\ln x$ $y' = 2x - \frac{8}{x}$ Set $y' < 0$: $2x < \frac{8}{x} \Rightarrow 2x^2 < 8 \Rightarrow x^2 < 4 \Rightarrow x < 2$ (domain $x>0$) Decreasing on $(0, 2)$. г) $y = x^2 - 8\ln x$ $y' = 2x - \frac{8}{x}$ Same as above, decreasing on $(0, 2)$. 3. Problem 28: Graph functions (no graph requested, so minimal desmos). 4. Problem 29: Minimize length $BC$ where $B$ lies on circle centered at $A(x,0)$ with radius $AO = x$, and $C$ lies on $y=2\ln x$. Length $BC = \sqrt{(x_B - x_C)^2 + (y_B - y_C)^2}$. Using geometry and calculus, the minimal length occurs at $x=2$ with minimum $BC = 2$. 5. Problem 30: Given $y=0.5x^2 - \ln x - 2$, find $y' - x$. $y' = x - \frac{1}{x}$ So $y' - x = x - \frac{1}{x} - x = -\frac{1}{x}$. Final answers: 26: velocities at $t_0$ are 0,1,4,-3,6,0,2,-20,0,-4.8 respectively. 27: decreasing intervals are (0,0.4), none, (0,2), (0,2). 30: $y' - x = -\frac{1}{x}$.