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Question
rates of change and tangent lines to question completed: 1 of 6 | my score: 1/6 pts (16.6 find the average rate of change of the function over the given interval. r(θ)=√(4θ + 1); 6,12 δr/δθ=□ (simplify your answer.)
Step1: Recall average - rate - of - change formula
The average rate of change of a function $y = R(\theta)$ over the interval $[a,b]$ is given by $\frac{\Delta R}{\Delta\theta}=\frac{R(b)-R(a)}{b - a}$. Here, $a = 6$, $b = 12$, and $R(\theta)=\sqrt{4\theta+1}$.
Step2: Calculate $R(12)$
Substitute $\theta = 12$ into $R(\theta)$:
$R(12)=\sqrt{4\times12 + 1}=\sqrt{48+1}=\sqrt{49}=7$.
Step3: Calculate $R(6)$
Substitute $\theta = 6$ into $R(\theta)$:
$R(6)=\sqrt{4\times6+1}=\sqrt{24 + 1}=\sqrt{25}=5$.
Step4: Calculate the average rate of change
$\frac{\Delta R}{\Delta\theta}=\frac{R(12)-R(6)}{12 - 6}=\frac{7 - 5}{6}=\frac{2}{6}=\frac{1}{3}$.
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$\frac{1}{3}$