The graph shows the solution to which system of equations?
1 answer:
You might be interested in
Let the curve C be the intersection of the cylinder
and the plane
The projection of C on to the x-y plane is the ellipse
To see clearly that this is an ellipse, le us divide through by 16, to get
or
,
We can write the following parametric equations,
for
Since C lies on the plane,
it must satisfy its equation.
Let us make z the subject first,
This implies that,
We can now write the vector equation of C, to obtain,
The length of the curve of the intersection of the cylinder and the plane is now given by,
But
Therefore the length of the curve of the intersection intersection of the cylinder and the plane is 24.0878 units correct to four decimal places.
Answer:
The values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.
Step-by-step explanation:
Note: This question is not complete. The complete question is therefore provided before answering the question as follows:
Determine the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001. f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0
The explanation of the answer is now provided as follows:
Given:
f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0 …………….. (1)
= (x) = (e^z /4!)x^4
Since the aim is (x) < 0.001, this implies that:
(e^z /4!)x^4 < 0.0001 ………………………………….. (2)
Multiply both sided of equation (2) by (1), we have:
e^4x^4 < 0.024 ……………………….......……………. (4)
Taking 4th root of both sided of equation (4), we have:
|xe^(z/4) < 0.3936 ……………………..........…………(5)
Dividing both sides of equation (5) by e^(z/4) gives us:
|x| < 0.3936 / e^(z/4) ……………….................…… (6)
In equation (6), when z > 0, e^(z/4) > 1. Therefore, we have:
|x| < 0.3936 -----> 0 < x < 0.3936
Therefore, the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.