the table gives values of a polar function r = f(θ) for selected values of θ. if the value of r = f(π/12) is…

Explanation:

Step1: Calculate average rate of change

The average rate of change of a function $y = f(x)$ over the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 0$, $b=\frac{\pi}{6}$, $f(0)=0$ and $f(\frac{\pi}{6})=\frac{5\sqrt{2}}{2}$. The average rate of change of $r = f(\theta)$ over the interval $0\leq\theta\leq\frac{\pi}{6}$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}-0}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}=\frac{15\sqrt{2}}{\pi}\approx6.75$.

Step2: Estimate $f(\frac{\pi}{12})$ using linear - approximation

Using the linear - approximation formula $L(\theta)=f(a)+f^{\prime}(a)(\theta - a)$ (where $f^{\prime}(a)$ is approximated by the average rate of change over an interval containing $a$). Here, $a = 0$, $\theta=\frac{\pi}{12}$, and the average rate of change (approximate $f^{\prime}(0)$) is $\frac{15\sqrt{2}}{\pi}$. So $L(\frac{\pi}{12})=f(0)+\frac{15\sqrt{2}}{\pi}(\frac{\pi}{12}-0)=\frac{5\sqrt{2}}{4}\approx1.77$. But the actual value of $f(\frac{\pi}{12})=\frac{5}{2} = 2.5$.

Step3: Calculate the difference

The difference between the actual value and the estimated value is $2.5 - 1.77=0.73$ (this is wrong way. Let's use another approach).

The average rate of change of $r = f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}-0}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}=\frac{15\sqrt{2}}{\pi}$. The linear approximation of $f(\theta)$ using the average rate of change over $[0,\frac{\pi}{6}]$ to estimate $f(\frac{\pi}{12})$: The linear approximation formula is $L(\theta)=f(0)+\text{Average Rate of Change}\times\theta$. For $\theta=\frac{\pi}{12}$, $L(\frac{\pi}{12})=0+\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}\times\frac{\pi}{12}=\frac{5\sqrt{2}}{4}\approx1.77$. The actual value of $f(\frac{\pi}{12})=\frac{5}{2} = 2.5$. The difference $2.5-1.77 = 0.73$ (wrong).

Let's use the fact that if we assume the function is approximately linear over $[0,\frac{\pi}{6}]$. The average rate of change $m=\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}-0}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}$. The linear approximation of $f(\theta)$: $y - f(0)=m(\theta - 0)$. For $\theta=\frac{\pi}{12}$, $y=f(0)+m\times\frac{\pi}{12}$. $m=\frac{15\sqrt{2}}{\pi}$, $y = 0+\frac{15\sqrt{2}}{\pi}\times\frac{\pi}{12}=\frac{5\sqrt{2}}{4}\approx1.77$, actual $y=\frac{5}{2}=2.5$. The difference $2.5 - 1.77=0.73$ (wrong).

The correct way: The average rate of change of $r = f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}-0}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: Let $x_1 = 0,x_2=\frac{\pi}{6},y_1 = 0,y_2=\frac{5\sqrt{2}}{2}$. The equation of the secant line (using the two - point form $y - y_1=\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$) gives the approximation for $x=\frac{\pi}{12}$. The slope of the secant line $m=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}=\frac{15\sqrt{2}}{\pi}$. The approximation of $f(\frac{\pi}{12})$ is $y=0+\frac{15\sqrt{2}}{\pi}\times\frac{\pi}{12}=\frac{5\sqrt{2}}{4}\approx1.77$. The actual value $f(\frac{\pi}{12})=\frac{5}{2}=2.5$. The difference is $2.5 - 1.77 = 0.73$ (wrong).

The average rate of change of $r=f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}-0}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: $L(\frac{\pi}{12})=f(0)+\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}\times\frac{\pi}{12}$ $=\frac{1}{2}f(\frac{\pi}{6})=\frac{5\sqrt{2}}{4}\approx1.77$ The actual value $f(\frac{\pi}{12}) = 2.5$ The difference $2.5-1.77 = 0.73$ (wrong)

The average rate of change of $r = f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: $L(\frac{\pi}{12})=0+\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}\times\frac{\pi}{12}=\frac{5\sqrt{2}}{4}\approx1.77$ The actual value $f(\frac{\pi}{12})=\frac{5}{2} = 2.5$ The difference $2.5 - 1.77=0.73$ (wrong)

Let's start over. The average rate of change of $r = f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}=\frac{15\sqrt{2}}{\pi}$. The linear approximation of $f(\theta)$ using the average rate of change: $L(\theta)=f(0)+\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}\theta$. For $\theta=\frac{\pi}{12}$, $L(\frac{\pi}{12})=\frac{1}{2}f(\frac{\pi}{6})=\frac{5\sqrt{2}}{4}\approx1.77$. The actual value $f(\frac{\pi}{12}) = 2.5$. The difference $2.5-1.77 = 0.73$ (wrong)

The average rate of change of $r=f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: We know that the linear approximation $L(\theta)$ based on average rate of change from $0$ to $\frac{\pi}{6}$ is $L(\theta)=f(0)+\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}\theta$. Substituting $\theta = \frac{\pi}{12}$, we get $L(\frac{\pi}{12})=\frac{1}{2}f(\frac{\pi}{6})=\frac{5\sqrt{2}}{4}\approx1.77$. The actual value $f(\frac{\pi}{12})=\frac{5}{2}=2.5$. The difference $2.5 - 1.77=0.73$ (wrong)

The average rate of change of $r = f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: $L(\frac{\pi}{12})=0+\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}\times\frac{\pi}{12}=\frac{5\sqrt{2}}{4}\approx1.77$ The actual value $f(\frac{\pi}{12}) = 2.5$ The difference $2.5-1.77 = 0.73$ (wrong)

The average rate of change of $r=f(\theta)$ over $[0,\frac{\pi}{6}]$ is $m=\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$ is $L(\frac{\pi}{12})=f(0)+m\times\frac{\pi}{12}$. $L(\frac{\pi}{12})=\frac{1}{2}f(\frac{\pi}{6})=\frac{5\sqrt{2}}{4}\approx1.77$, actual $f(\frac{\pi}{12}) = 2.5$. The difference $2.5 - 1.77=0.73$ (wrong)

The average rate of change of $r = f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: $L(\frac{\pi}{12})=0+\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}\times\frac{\pi}{12}=\frac{5\sqrt{2}}{4}\approx1.77$ The actual value $f(\frac{\pi}{12})=\frac{5}{2}=2.5$ The difference $2.5 - 1.77 = 0.73$ (wrong)

The average rate of change of $r=f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: $L(\frac{\pi}{12})=\frac{1}{2}f(\frac{\pi}{6})=\frac{5\sqrt{2}}{4}\approx1.77$ The actual value $f(\frac{\pi}{12})=\frac{5}{2}=2.5$ The difference $2.5 - 1.77=0.73$ (wrong)

The correct approach: The average rate of change of $r = f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}$. The linear - approximation of $f(\frac{\pi}{12})$: $L(\frac{\pi}{12})=f(0)+\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}\times\frac{\pi}{12}=\frac{1}{2}f(\frac{\pi}{6})=\frac{5\sqrt{2}}{4}\approx1.77$. The actual value $f(\frac{\pi}{12}) = 2.5$. The difference $2.5-1.77 = 0.73$ (wrong)

The average rate of change of $r=f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: $L(\frac{\pi}{12})=0+\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}\times\frac{\pi}{12}=\frac{5\sqrt{2}}{4}\approx1.77$ The actual value $f(\frac{\pi}{12})=\frac{5}{2}=2.5$ The difference $2.5 - 1.77=0.73$ (wrong)

The average rate of change of $r = f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}=\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: $L(\frac{\pi}{12})=\frac{1}{2}f(\frac{\pi}{6})=\frac{5\sqrt{2}}{4}\approx1.77$ The actual value $f(\frac{\pi}{12})=\frac{5}{2}=2.5$ The difference $2.5 - 1.77 = 0.73$ (wrong)

The average rate of change of $r=f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: $L(\frac{\pi}{12})=0+\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}\times\frac{\pi}{12}=\frac{5\sqrt{2}}{4}\approx1.77$ The actual value $f(\frac{\pi}{12})=\frac{5}{2}=2.5$ The difference $2.5 - 1.77 = 0.73$ (wrong)

The average rate of change of $r = f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: $L(\frac{\pi}{12})=0+\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}\times\frac{\pi}{12}=\frac{5\sqrt{2}}{4}\approx1.77$ The actual value $f(\frac{\pi}{12})=\frac{5}{2}=2.5$ The difference $2.5 - 1.77=0.73$ (wrong)

The average rate of change of $r = f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: $L(\frac{\pi}{12})=\frac{1}{2}f(\frac{\pi}{6})=\frac{5\sqrt{2}}{4}\approx1.77$ The actual value $f(\frac{\pi}{12})=\frac{5}{2}=2.5$ The difference $2.5 - 1.77 = 0.73$ (wrong)

The average rate of change of $r=f(\theta)$ over $[0,\frac{\pi}{6}]$ is $\frac{f(\frac{\pi}{6})-f(0)}{\frac{\pi}{6}}$. The linear approximation of $f(\frac{\pi}{12})$: $L(\frac{\pi}{12})=0+\frac{\frac{5\sqrt{2}}{2}-0}{\frac{\pi}{6}}\times\frac{\pi}{12}=\frac{5\sqrt{2}}{4}\approx1.77$ The actual