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Alex learned 80 new vocabulary words for his English exam. The following function gives the number of words he remembers after 
t days:

W(t)=80(1-0.1 t)^(3)
What is the instantaneous rate of change of the number of words Alex remembers after 5 days?
Choose 1 answer:
(A) 
10 days
(B) 
10 words per day
(C) -6 days
(D) -6 words per day

Alex learned 8080 new vocabulary words for his English exam. The following function gives the number of words he remembers after t t days:\newlineW(t)=80(10.1t)3 W(t)=80(1-0.1 t)^{3} \newlineWhat is the instantaneous rate of change of the number of words Alex remembers after 55 days?\newlineChoose 11 answer:\newline(A) 10 \mathbf{1 0} days\newline(B) 10 \mathbf{1 0} words per day\newline(C) 6-6 days\newline(D) 6-6 words per day

Full solution

Q. Alex learned 8080 new vocabulary words for his English exam. The following function gives the number of words he remembers after t t days:\newlineW(t)=80(10.1t)3 W(t)=80(1-0.1 t)^{3} \newlineWhat is the instantaneous rate of change of the number of words Alex remembers after 55 days?\newlineChoose 11 answer:\newline(A) 10 \mathbf{1 0} days\newline(B) 10 \mathbf{1 0} words per day\newline(C) 6-6 days\newline(D) 6-6 words per day
  1. Take Derivative of W(t)W(t): To find the instantaneous rate of change, we need to take the derivative of W(t)W(t) with respect to tt.
  2. Apply Chain Rule: Differentiate W(t)=80(10.1t)3W(t) = 80(1 - 0.1t)^{3}. Using the chain rule, ddt[80(10.1t)3]=80×3×(10.1t)2×(0.1)\frac{d}{dt}[80(1 - 0.1t)^{3}] = 80 \times 3 \times (1 - 0.1t)^{2} \times (-0.1).
  3. Simplify Derivative: Simplify the derivative to get W(t)=24×(10.1t)2.W'(t) = -24 \times (1 - 0.1t)^{2}.
  4. Substitute t=5t = 5: Substitute t=5t = 5 into W(t)W'(t) to find the instantaneous rate of change after 55 days.\newlineW(5)=24×(10.1×5)2W'(5) = -24 \times (1 - 0.1\times5)^{2}.
  5. Calculate W(5)W'(5): Calculate W(5)=24×(10.5)2W'(5) = -24 \times (1 - 0.5)^{2}.\newlineW(5)=24×(0.5)2W'(5) = -24 \times (0.5)^{2}.
  6. Final Result: W(5)=24×0.25W'(5) = -24 \times 0.25.\newlineW(5)=6W'(5) = -6.

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