If the linearization of y= cube root of x at x=64 is used to approximate cube root of 66, what is the percentage error?
a) .01%
b).04%
c).4%
d)1%
e)4%
For the calculation of the error term ε, see response to:
http://www.jiskha.com/display.cgi?id=1291926261
In the given case,
h=2
a=64
f(a)=4
f(a)+hf'(a)=4.041667
f(x)=x^(1/3)=
f'(x)=1/(3*x^(2/3))
f"(x)=-2/(9*x^(5/3))
and
epsilon(x,h)=h²f"(x)/2!
=-2h²/(9*x^(5/3))
=-0.000434
actual error
=f(66)-4.041667
=-0.000427
Please double-check my calculations and make your choice for the correct answer.
Well, to find the percentage error, we first need to find the actual value of the cube root of 66. Let's do some math! The cube root of 66 is approximately equal to 4.0917.
Now, let's use the linearization of y = cube root of x to approximate the cube root of 66. The linearization of y = cube root of x at x = 64 is approximately y = 4 + 1/48 * (x - 64).
Substituting x = 66 into the linearization equation gives us y ≈ 4 + 1/48 * (66 - 64) = 4 + 1/48 * 2 = 4.0417.
To calculate the percentage error, we take the absolute value of the difference between the actual value and the approximate value, divide it by the actual value, and then multiply by 100%. Let's plug in the numbers: (|4.0917 - 4.0417| / 4.0917) * 100% ≈ 1.22%.
So, the correct answer is not one of the given options. The percentage error is approximately 1.22%. Keep up the good work!
To determine the percentage error when using the linearization of \(y = \sqrt[3]{x}\) at \(x = 64\) to approximate \(\sqrt[3]{66}\), we need to calculate the actual value and compare it to the approximation.
1. First, let's calculate the actual value of \(\sqrt[3]{66}\):
\(\sqrt[3]{66} \approx 4.0801\) (rounding to 4 decimal places).
2. Next, we need to find the approximation using the linearization at \(x = 64\). The equation of the linearization is:
\(L(x) = f'(a)(x - a) + f(a)\)
For \(f(x) = \sqrt[3]{x}\), we have \(f'(x) = \frac{1}{3\sqrt[3]{x^2}}\).
Plugging in \(a = 64\) and \(x = 66\), we get:
\(L(66) = \frac{1}{3\sqrt[3]{64^2}}(66 - 64) + \sqrt[3]{64}\)
Simplifying:
\(L(66) = \frac{1}{3\sqrt[3]{2^{12}}}(2) + 4\)
\(L(66) = \frac{1}{12} + 4\)
\(L(66) = 4.083\)
3. Now we can calculate the percentage error:
\(\text{Percentage error} = \left|\frac{\text{Actual value - Approximation}}{\text{Actual value}}\right| \times 100\)
\(\text{Percentage error} = \left|\frac{4.0801 - 4.083}{4.0801}\right| \times 100\)
\(\text{Percentage error} = \frac{0.0029}{4.0801} \times 100\)
\(\text{Percentage error} \approx 0.071\)
Therefore, the percentage error is approximately 0.071%, which is closest to option (a) .01%.
To find the answer, we need to follow these steps:
Step 1: Find the linearization of the function y = ∛x at x = 64.
The linearization of a function at a given point is the equation of the tangent line to the graph of the function at that point. It can be found using the formula:
L(x) = f(a) + f'(a)(x - a)
where f(a) is the value of the function at x = a and f'(a) is the derivative of the function evaluated at x = a.
Let's find the linearization of y = ∛x at x = 64:
f(x) = ∛x
f'(x) = 1/(3√x²)
f(64) = ∛64 = 4
f'(64) = 1/(3√64²) = 1/192
Therefore, the linearization is: L(x) = 4 + (1/192)(x - 64)
Step 2: Approximate ∛66 using the linearization.
We substitute x = 66 into the linearization equation:
L(66) = 4 + (1/192)(66 - 64)
= 4 + (1/192)(2)
= 4 + (1/96)
= 4.01041666667 (approx.)
Step 3: Calculate the percentage error.
The percentage error can be found using the formula:
Error = (Actual Value - Approximated Value) / Actual Value * 100
Error = (∛66 - 4.01041666667) / ∛66 * 100
Now, we need to find the value of ∛66 using a calculator or approximate it using a method like the Newton-Raphson method. Assuming ∛66 ≈ 4.0801 (approx.), we can calculate the percentage error:
Error = (4.0801 - 4.01041666667) / 4.0801 * 100
= 0.06968311144 / 4.0801 * 100
≈ 1.70689655172
Therefore, the percentage error is approximately 1%, which corresponds to option (d).