Question #6e767

First, let's draw a diagram of this:

What we have are the change in the ladder's position in the #y# and the #x# direction over a period of time. We are focusing on the #y# derivative.

The actual rate at which the ladder falls over time is #(dy)/(dt)#, and this causes a change in the angle, #theta#, which also varies over a time period. so now we know that:

#(dy)/(dt) = (dy)/(d theta)*(d theta)/(dt)#

We already know #(dy)/(dt) = "-4 ft/sec"#. With that we are assuming down is negative. What we can do with this is relate #x#, #y#, and/or #z = 10# with #theta# using trig functions.

#costheta = y/10#

#10costheta = y#

Using implicit differentiation we get:

#-10sintheta(d theta)/(dt) = (dy)/(dt)#

#(d theta)/(dt) = 1/(-10sintheta)(dy)/(dt)#

The last thing we need is #theta#, but we can simply acquire the relationship #sintheta#. Let us write one more equation to find it using what we have for a second distance on the triangle:

#sintheta = x/10#

We are using the case #x = 8#, so we get:

#sintheta = 4/5#

Now we have everything we need. Our goal was #(d theta)/(dt)#, so:

#(d theta)/(dt) = 1/([-10*"4/5"] cancel"ft")(-4 cancel"ft""/s") = (-4)/(10*(-0.8))#

#"= 0.5 rad/s"#

#= color(blue)(28.6^o"/s")#