Question 1203157
<font color=black size=3>
Plug in theta = pi/6
{{{T(theta) = ((2V)/(g))*sin(theta)}}}


{{{T(pi/6) = ((2V)/(g))*sin(pi/6)}}}


{{{T(pi/6) = ((2V)/(g))*(1/2)}}} Use the unit circle


{{{T(pi/6) = V/g}}} 



Repeat for theta = pi/4
{{{T(theta) = ((2V)/(g))*sin(theta)}}}


{{{T(pi/4) = ((2V)/(g))*sin(pi/4)}}}


{{{T(pi/4) = ((2V)/(g))*(sqrt(2)/2)}}} Use the unit circle


{{{T(pi/4) = (V*sqrt(2))/(g)}}}


The difference between those results is
{{{T(pi/6) - T(pi/4) = V/g - (V*sqrt(2))/(g)}}}


{{{T(pi/6) - T(pi/4) = (V - V*sqrt(2))/(g)}}}


{{{T(pi/6) - T(pi/4) = V(1 - sqrt(2))/(g)}}}


Then divide this over the distance from theta = pi/6 to theta = pi/4
{{{(T(pi/6) - T(pi/4))/(pi/6 - pi/4) = matrix(1,3,(T(pi/6) - T(pi/4))," divide ", (pi/6 - pi/4))}}}


{{{(T(pi/6) - T(pi/4))/(pi/6 - pi/4) = matrix(1,3,V(1 - sqrt(2))/(g)," divide ", (pi/6 - pi/4))}}}


{{{(T(pi/6) - T(pi/4))/(pi/6 - pi/4) = matrix(1,3,V(1 - sqrt(2))/(g)," divide ", (2pi/12 - 3pi/12))}}}


{{{(T(pi/6) - T(pi/4))/(pi/6 - pi/4) = matrix(1,3,V(1 - sqrt(2))/(g)," divide ", (-pi/12))}}}


{{{(T(pi/6) - T(pi/4))/(pi/6 - pi/4) = matrix(1,3,V(1 - sqrt(2))/(g)," * ", (-12/pi))}}}


{{{(T(pi/6) - T(pi/4))/(pi/6 - pi/4) = -12V(1 - sqrt(2))/(g*pi)}}}


The average rate of change over the interval [pi/6, pi/4] is {{{-12V(1 - sqrt(2))/(g*pi)}}}

</font>