The approximation $(1+x)^k \approx 1+k x$ is commonly used by engineers for quick calculations.
(a) Derive this result and use it to make a rough estimate of $(1.001)^{36}$.
$(1.001)^{36} \approx$ $\square$ Enter value to 6 decimal places.
(b) The error between the computed and estimated values of $(1+x)^k$ is given by;
$E(x)=\left|(1+x)^k-(1+k x)\right|$
Using your answer to part (a) compute $E(1.001)$ to 6 decimal places.
$E(1.001)=\square$
Answer (2)
Approximate ( 1.001 ) 36 using the formula ( 1 + x ) k ≈ 1 + k x , where x = 0.001 and k = 36 , resulting in 1 + 36 ( 0.001 ) = 1.036 .
Calculate the exact value of ( 1.001 ) 36 which is approximately 1.036637 .
Compute the error E ( 0.001 ) by finding the absolute difference between the exact value and the approximation: ∣1.036637 − 1.036∣ = 0.000637 .
State the final results: ( 1.001 ) 36 ≈ 1.036 and E ( 1.001 ) = 0.000637 .
Explanation
Problem Setup We are given the approximation ( 1 + x ) k ≈ 1 + k x and asked to use it to estimate ( 1.001 ) 36 and then calculate the error in the approximation.
Estimating (1.001)^36 (a) Using the approximation ( 1 + x ) k ≈ 1 + k x , we have x = 0.001 and k = 36 . Therefore, ( 1.001 ) 36 ≈ 1 + 36 ( 0.001 ) = 1 + 0.036 = 1.036 .
Calculating the Error (b) The error is given by E ( x ) = ∣ ( 1 + x ) k − ( 1 + k x ) ∣ . We need to compute E ( 0.001 ) = ∣ ( 1.001 ) 36 − ( 1 + 36 ( 0.001 )) ∣ .
We have already calculated the approximation as 1.036 . Now we need to find the actual value of ( 1.001 ) 36 .
Using a calculator, we find that ( 1.001 ) 36 ≈ 1.036637199 .
Therefore, E ( 0.001 ) = ∣1.036637199 − 1.036∣ = 0.000637199 ≈ 0.000637 .
Final Answer Therefore, the estimate of ( 1.001 ) 36 is approximately 1.036 , and the error E ( 1.001 ) is approximately 0.000637 .
Examples
Engineers often use approximations like ( 1 + x ) k ≈ 1 + k x for quick estimations in scenarios such as calculating the effect of small changes in dimensions or material properties on the overall performance of a structure. For instance, if you're designing a bridge and need to quickly estimate how a slight increase in the thickness of a support beam (say, 0.1%) affects its load-bearing capacity, this approximation can provide a fast and reasonably accurate answer without needing complex calculations. This allows for rapid decision-making during the design phase.
The approximation of ( 1.001 ) 36 is 1.036 and the error in that approximation is 0.000637 .
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