Two sides of a triangle are 6 m and 7 m in length and the angle between them is increasing at a rate of 0.07 rad/s. How do you find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3?

1 Answer
Aug 27, 2015

The overall steps are:

  1. Draw a triangle consistent with the given information, labeling relevant information
  2. Determine which formulas make sense in the situation (Area of entire triangle based on two fixed-length sides, and trig relationships of right triangles for the variable height)
  3. Relate any unknown variables (height) back to the variable #(theta)# which corresponds to the only given rate #((d theta)/(dt))#
  4. Do some substitutions into a "main" formula (the area formula) so that you can anticipate using the given rate
  5. Differentiate and use the given rate to find the rate you are aiming for #((dA)/(dt))#

Let's write down the information given formally:

#(d theta)/(dt) = "0.07 rad/s"#

Then you have two fixed-length sides and an angle between them. The third length is a variable value, but it is technically an irrelevant length. What we want is #(dA)/(dt)#. There is no indication that this is a right triangle, however, so let's start by assuming that it's not at the moment.

A theoretically consistent triangle is:

Keep in mind that this is not proportionally representative of the true triangle. The area of this can be found most easily with:

#A = (B*h)/2#

where our base is of course #6#. What is #h#, though? If we draw a dividing line vertically from the apex down to the base, we automatically have a right triangle on the left side of the overall triangle, regardless of the length of side #x#:

Now we do have a right triangle. Notice, however, that our area formula has #h# but not #theta#, and we only know #(d theta)/(dt)#. So, we need to represent #h# in terms of an angle. Knowing that the only known side on the left-hand right triangle is the #7#-lengthed side:

#sintheta = h/7#

#7sintheta = h#

So far, we have:

#(d theta)/(dt) = "0.07 rad/s"# (1)

#A = (Bh)/2# (2)

#7sintheta = color(green)(h)# (3)

So, we can plug (3) into (2), differentiate (2) and implicitly acquire #(d theta)/(dt)#, and plug (1) into (2) to solve for #(dA)/(dt)#, our goal:

#A = (6*color(green)(7sintheta))/2 = 21sintheta#

#color(blue)((dA)/(dt)) = 21costheta((d theta)/(dt))#

#= 21costheta ("0.07 rad/s")#

Finally, at #theta = pi/3#, we have #cos(pi/3) = 1/2# and:

#= 10.5(0.07) = color(blue)("0.735 u"^2"/s")#

(note that #6*7# means the units become #"u"*"u" = "u"^2#, and #2# is not a side length so it had no units. Also, #"rad"# is usually considered to be left out, i.e. #"rad/s" => "1/s"#)