[Linear Approximations and Differentials]
I tripled checked my work but I'm still not getting the correct answer.
y=3x^2 x=3 & delta x =0.2
I get 3(3.2)^2 - 3(3) = 21.72
Let Δx = 0.2,
Δy
=3(3.2)² - 3(3)²
=30.72-27
=3.72
Δy/Δx
=3.72/0.2
=18.6
Theoretical value
dy/dx = 6x = 18.
Oh I'm missing the last ^2 at the end. Thanks!
You're welcome!
To find the linear approximation, we need to use the tangent line to the curve at the given point. The equation of the tangent line can be found using the point-slope form.
First, let's find the slope of the tangent line, which is equal to the derivative of the function y with respect to x evaluated at the given point.
y = 3x^2
Taking the derivative of y with respect to x:
dy/dx = 6x
Now, evaluate the derivative at x = 3:
dy/dx = 6(3) = 18
Now, we have the slope of the tangent line. To find the equation of the tangent line, we need to find the y-intercept. We can use the point-slope form:
y - y₁ = m(x - x₁)
where (x₁, y₁) is the given point (3, y) and m is the slope we calculated (18).
Plugging in the values:
y - 3^2 = 18(x - 3)
Simplifying further:
y - 9 = 18x - 54
y = 18x - 45
Now, let's use this equation to approximate the value when x = 3.2:
y ≈ 18(3.2) - 45
y ≈ 57.6 - 45
y ≈ 12.6
So, the linear approximation of y when x = 3.2 is approximately 12.6.