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Max wrote an algorithm that searches for a specific term within a large set of terms. The following function gives the length of the search, in number of steps, over a set with n terms: S(n)=1.6*ln(0.9 n) What is the instantaneous rate of change of the search length for a set of 10 terms? Choose 1 answer: A 0.16 steps per term (B) 0.16 steps per second (C) 3.5 steps per term (D) 3.5 steps per second

Max wrote an algorithm that searches for a specific term within a large set of terms. The following function gives the length of the search, in number of steps, over a set with n n terms:\newlineS(n)=1.6⋅ln⁡(0.9n) S(n)=1.6 \cdot \ln (0.9 n) \newlineWhat is the instantaneous rate of change of the search length for a set of 1010 terms?\newlineChoose 11 answer:\newline(A) 00.1616 steps per term\newline(B) 00.1616 steps per second\newline(C) 33.55 steps per term\newline(D) 33.55 steps per second

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Q. Max wrote an algorithm that searches for a specific term within a large set of terms. The following function gives the length of the search, in number of steps, over a set with n n terms:\newlineS(n)=1.6⋅ln⁡(0.9n) S(n)=1.6 \cdot \ln (0.9 n) \newlineWhat is the instantaneous rate of change of the search length for a set of 1010 terms?\newlineChoose 11 answer:\newline(A) 00.1616 steps per term\newline(B) 00.1616 steps per second\newline(C) 33.55 steps per term\newline(D) 33.55 steps per second
  1. Find Derivative: To find the instantaneous rate of change, we need to differentiate the function S(n)S(n) with respect to nn.
  2. Apply Chain Rule: Differentiate S(n)=1.6⋅ln⁡(0.9n)S(n) = 1.6\cdot\ln(0.9n) using the chain rule.\newlinedSdn=1.6⋅(10.9n)⋅(0.9)\frac{dS}{dn} = 1.6 \cdot \left(\frac{1}{0.9n}\right) \cdot (0.9)
  3. Simplify Derivative: Simplify the derivative. dSdn=1.6×(0.90.9n)\frac{dS}{dn} = 1.6 \times \left(\frac{0.9}{0.9n}\right)
  4. Cancel Out Terms: Cancel out the 0.90.9 in the numerator and denominator.dSdn=1.6n\frac{dS}{dn} = \frac{1.6}{n}
  5. Substitute n=10n=10: Substitute n=10n = 10 into the derivative to find the instantaneous rate of change at n=10n = 10.dSdn=1.610\frac{dS}{dn} = \frac{1.6}{10}
  6. Calculate Value: Calculate the value. dSdn=0.16\frac{dS}{dn} = 0.16 steps per term

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