# Math Insight

### Solutions to linear and quadratic Taylor polynomial problems

The following is a set of solutions to the linear and quadratic Taylor polynomial problems.

#### Problem 1

1. Since $f(3)=1$ and $f'(3)$, a linear approximation around $x=3$ is \begin{align*} p_1(x) &= f(3) + f'(3)(x-3)\\ &=1 + 5(x-3) \end{align*}
2. Given that, in addition, $f''(3)=-4$, the quadratic approximation is \begin{align*} p_1(x) &= f(3) + f'(3)(x-3)+ \frac{f''(3)}{2}(x-3)^2\\ &= 1 + 5(x-3)-2(x-3)^2 \end{align*}

#### Problem 2

1. A linear approximation around $q=-2$ is \begin{align*} L(q) &= m(-2)+m'(-2)(q+2)\\ &= -4 +0 (q+2)\\ &= -4 \end{align*}
2. A quadratic approximation around $q=-2$ is \begin{align*} Q(q) &= m(-2)+m'(-2)(q+2) + \frac{m''(-2)}{2}(q+2)^2\\ &= -4 +0 (q+2) + \frac{1}{2}(q+2)^2\\ &= -4 + \frac{1}{2}(q+2)^2 \end{align*}

#### Problem 3

Since $g(z)=e^{2-3z}$, its derivatives are \begin{align*} g'(z) &= -3e^{2-3z}\\ g''(z) &= 9e^{2-3z}. \end{align*} The function and its derivatives evaluated at $z=1$ are \begin{align*} g(1) &= e^{-1}\\ g'(1) &= -3e^{-1}\\ g''(1) &= 9e^{-1}. \end{align*} The second order Taylor polynomial around $z=1$ is \begin{align*} p_2(z) &= g(1) + g'(1)(z-1) + \frac{g''(1)}{2}(z-1)^2\\ &= e^{-1} -3e^{-1}(z-1) + \frac{9}{2}e^{-1}(z-1)^2. \end{align*}

#### Problem 4

The derivatives of $k(t)=t^4-t^2+t$ are \begin{align*} k'(t) &= 4t^3-2t+1\\ k''(t) &= 12t^2 -2. \end{align*} Evaluating the function and its derivatives at $t=-2$, \begin{align*} k(-2) &= (-2)^4 -(-2)^2 -2 = 18\\ k'(-2) &= 4(-2)^3-2(-2)+1 = -27\\ k''(-2) &= 12(-2)^2 -2 = 46, \end{align*} we calculate that the second order Taylor polynomial around $t=-2$ is \begin{align*} p_2(t) &= k(-2) + k'(-2)(t+2) + \frac{k''(-2)}{2}(t+2)^2\\ &= 18 -27(t+2) +23 (t+2)^2. \end{align*}

#### Problem 5

1. Let $C(t)$ be the temperature in degrees Celsius on planet Cook, where $t$ denotes years since 3012. The given information means that $C(0)=15$ and $C'(0) = 2$.
2. A linear approximation to $C(t)$ is \begin{align*} L(t) &= C(0) + C'(0)t\\ &=15 + 2t. \end{align*} The linear approximation predicts that the temperature of Cook will be $L(50) = 115$ degrees Celsius in year 3062 and $L(100) = 215$ degrees Celsius in year 3112.
3. The new measures would lead to $C''(0) = -0.03$. (Or maybe you could argue that it means $C''(t)=-0.03$ for all time $t$, but that wasn't the intent. You'll get the same answer either way.)
4. A quadratic approximation to $C(t)$ is \begin{align*} Q(t) &= C(0) + C'(0)t + \frac{C''(0)}{2}t^2\\ &=15 + 2t - 0.015 t^2. \end{align*} The quadratic approximation predicts that the temperature of Cook will be $$Q(50) = 115-0.015 (50)^2 = 77.5 \text{ degrees Celsius}$$ in year 3062 and $$Q(100)=215-0.015(100)^2 = 65 \text{ degrees Celsius}$$ in year 3112.

#### Problem 6

Let $r(k)$ be the width of the river in meters measured $k$ kilometers east of the city. The given information means that $r(10)=50$, $r'(10)=1$, and $r''(10) =0.1$. A quadratic Taylor polynomial approximation is \begin{align*} p_2(k) &= r(10) + r'(10)(k-10) + \frac{r''(10)}{2}(k-10)^2\\ &= 50 + (k-10) + 0.05(k-10)^2. \end{align*} According to this approximation, the width of the river 15 kilometers east of the city is \begin{align*} p_2(15) = 50 + 5 + 0.05(5)^2 = 56.25 \text{ meters}. \end{align*}