solve the equation dl1/dx + dl2/dx = 0 for x in the interval 0,d. x = md/(m + n) use the expression for x to…

solve the equation dl1/dx + dl2/dx = 0 for x in the interval 0,d. x = md/(m + n) use the expression for x to find θ1 in terms of m, n, and d. θ1 = tan^(-1)((m + n)/d) use the expression for x to find θ2 in terms of m, n, and d. θ2 = tan^(-1)( )

solve the equation dl1/dx + dl2/dx = 0 for x in the interval 0,d. x = md/(m + n) use the expression for x to find θ1 in terms of m, n, and d. θ1 = tan^(-1)((m + n)/d) use the expression for x to find θ2 in terms of m, n, and d. θ2 = tan^(-1)( )

Answer

Explanation:

Step1: Recall the relationship between variables

Assume there are right - triangle relationships for $\theta_1$ and $\theta_2$ related to $x$. Given $x = \frac{md}{m + n}$, and assume some geometric setup where we use tangent function for angles in right - triangles.

Step2: Analyze the relationship for $\theta_2$

If we consider the complementary geometric situation to that of $\theta_1$, and assume the total length along a certain axis is $d$. The value related to $\theta_2$ in terms of tangent inverse can be found from the remaining part of the length. Since $x=\frac{md}{m + n}$, the remaining length is $d - x=d-\frac{md}{m + n}=\frac{nd}{m + n}$. So $\theta_2=\tan^{-1}\left(\frac{m + n}{n}\right)$

Answer:

$\theta_2=\tan^{-1}\left(\frac{m + n}{n}\right)$