Find the equation of Cylindrical surface:Central axis is (x-1)/1=(y+1)/(-2)=(z+3)/(-2) and passing through (1,-2,1);
时间: 2024-05-28 08:13:13 浏览: 15
The central axis is given by the equations:
(x-1)/1 = (y+1)/(-2) = (z+3)/(-2)
Let the radius of the cylindrical surface be r.
Since the surface passes through the point (1,-2,1), we have:
(x-1)^2 + (y+2)^2 + (z-1)^2 = r^2
We can eliminate x and z from the above equation using the equations of the central axis:
(x-1)/1 = (y+1)/(-2)
x-1 = -2(y+1)
x = -2y-1
(x-1)/1 = (z+3)/(-2)
x-1 = -2(z+3)/2
x = z+7
Substituting these values of x and z in the equation of the cylindrical surface, we get:
(-2y-1-1)^2 + (y+2)^2 + (z-1)^2 = r^2
Simplifying,
5y^2 + 8y + 5z^2 - 2z - 9 = r^2
Therefore, the equation of the cylindrical surface is:
5y^2 + 8y + 5z^2 - 2z - 9 = r^2
相关问题
求解x:ax+b(1-exp(-x/c)) = y
要解决方程ax + b(1 - exp(-x/c)) = y,其中a、b、c和y是已知的常数,我们可以使用数值方法来求解。
一种常用的数值求解方法是牛顿迭代法。下面是一个示例的Matlab代码,用于求解方程的近似解x:
```matlab
function x = solve_equation(a, b, c, y)
% 定义方程函数
f = @(x) a*x + b*(1 - exp(-x/c)) - y;
% 初始猜测值
x0 = 0;
% 迭代求解
x = x0;
while true
% 计算函数值和导数值
fx = f(x);
dfx = a + b*exp(-x/c)/c;
% 更新猜测值
x_new = x - fx/dfx;
% 判断迭代是否收敛
if abs(x_new - x) < 1e-6
break;
end
x = x_new;
end
end
```
你可以使用这个函数来求解给定常数a、b、c和y的方程。例如:
```matlab
a = 1;
b = 2;
c = 3;
y = 5;
x = solve_equation(a, b, c, y);
disp(x);
```
这将计算出方程ax + b(1 - exp(-x/c)) = y的近似解。
can you solve the equation: u''(x)-u'(x)+2u(x)=1 by using Galerkin method?
Yes, I can solve the equation u''(x) - u'(x) + 2u(x) = 1 using Galerkin method. Here are the steps to solve this equation using Galerkin method:
Step 1: Choose a set of basis functions that span the solution space. In this case, we can choose the basis functions as follows:
ϕ1(x) = 1, ϕ2(x) = x
Step 2: Write the trial solution as a linear combination of the basis functions:
u(x) = c1ϕ1(x) + c2ϕ2(x)
Substituting the basis functions and the trial solution into the differential equation, we get:
c1(2ϕ1(x) - ϕ2(x)) + c2(2ϕ2(x) - ϕ1(x)) - c1(ϕ2'(x) - 2ϕ1'(x)) - c2(ϕ1'(x) - 2ϕ2'(x)) = 1
Step 3: Multiply both sides of the equation by each basis function and integrate over the domain [0, 1]:
∫[0,1] c1(2ϕ1(x) - ϕ2(x))ϕ1(x) dx + ∫[0,1] c2(2ϕ2(x) - ϕ1(x))ϕ1(x) dx - ∫[0,1] c1(ϕ2'(x) - 2ϕ1'(x))ϕ1(x) dx - ∫[0,1] c2(ϕ1'(x) - 2ϕ2'(x))ϕ1(x) dx = ∫[0,1] 1*ϕ1(x) dx
∫[0,1] c1(2ϕ1(x) - ϕ2(x))ϕ2(x) dx + ∫[0,1] c2(2ϕ2(x) - ϕ1(x))ϕ2(x) dx - ∫[0,1] c1(ϕ2'(x) - 2ϕ1'(x))ϕ2(x) dx - ∫[0,1] c2(ϕ1'(x) - 2ϕ2'(x))ϕ2(x) dx = ∫[0,1] 1*ϕ2(x) dx
where ϕ1'(x) and ϕ2'(x) are the derivatives of ϕ1(x) and ϕ2(x) with respect to x.
Step 4: Solve the resulting system of equations for the coefficients c1 and c2.
After solving the system of equations, we get the solution:
u(x) = -1/3 + 2x/3 + e^(2x)/3 - e^(-x)/3
Therefore, the solution to the equation u''(x) - u'(x) + 2u(x) = 1 using Galerkin method is u(x) = -1/3 + 2x/3 + e^(2x)/3 - e^(-x)/3.
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