Answer :
To find the rate of depreciation of the car at [tex]\( t = 3 \)[/tex] years, we start with the given function for the resale value of the car:
[tex]\[ R(t) = 20000 \times (0.86)^t \][/tex]
The rate of depreciation is represented by the derivative of the resale value function [tex]\( R(t) \)[/tex] with respect to [tex]\( t \)[/tex], denoted as [tex]\( R'(t) \)[/tex].
Here's a step-by-step outline for calculating [tex]\( R'(t) \)[/tex]:
1. Differentiate the function [tex]\( R(t) = 20000 \times (0.86)^t \)[/tex]. The derivative of [tex]\( a^t \)[/tex] with respect to [tex]\( t \)[/tex] is [tex]\( a^t \ln(a) \)[/tex]. Applying this rule, we get:
[tex]\[ R'(t) = 20000 \times \ln(0.86) \times (0.86)^t \][/tex]
2. To determine [tex]\( R'(3) \)[/tex], substitute [tex]\( t = 3 \)[/tex] into the derivative function:
[tex]\[ R'(3) = 20000 \times \ln(0.86) \times (0.86)^3 \][/tex]
3. Calculate [tex]\( (0.86)^3 \)[/tex], which is approximately a number. Also, [tex]\( \ln(0.86) \)[/tex] is a known natural logarithm value.
4. Multiply these values together along with 20000 to find the rate of depreciation:
[tex]\[ R'(3) = -1918.64 \][/tex]
Thus, the rate of depreciation at [tex]\( t = 3 \)[/tex] years is approximately [tex]\( -1918.64 \)[/tex] dollars per year.
The correct multiple-choice answer is (C) -1918.64 dollars per year.
[tex]\[ R(t) = 20000 \times (0.86)^t \][/tex]
The rate of depreciation is represented by the derivative of the resale value function [tex]\( R(t) \)[/tex] with respect to [tex]\( t \)[/tex], denoted as [tex]\( R'(t) \)[/tex].
Here's a step-by-step outline for calculating [tex]\( R'(t) \)[/tex]:
1. Differentiate the function [tex]\( R(t) = 20000 \times (0.86)^t \)[/tex]. The derivative of [tex]\( a^t \)[/tex] with respect to [tex]\( t \)[/tex] is [tex]\( a^t \ln(a) \)[/tex]. Applying this rule, we get:
[tex]\[ R'(t) = 20000 \times \ln(0.86) \times (0.86)^t \][/tex]
2. To determine [tex]\( R'(3) \)[/tex], substitute [tex]\( t = 3 \)[/tex] into the derivative function:
[tex]\[ R'(3) = 20000 \times \ln(0.86) \times (0.86)^3 \][/tex]
3. Calculate [tex]\( (0.86)^3 \)[/tex], which is approximately a number. Also, [tex]\( \ln(0.86) \)[/tex] is a known natural logarithm value.
4. Multiply these values together along with 20000 to find the rate of depreciation:
[tex]\[ R'(3) = -1918.64 \][/tex]
Thus, the rate of depreciation at [tex]\( t = 3 \)[/tex] years is approximately [tex]\( -1918.64 \)[/tex] dollars per year.
The correct multiple-choice answer is (C) -1918.64 dollars per year.
