# increase in rate of area of triangle question

• Nov 12th 2006, 03:11 PM
dklee41
increase in rate of area of triangle question

Two sides of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06 radians/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3 (60 degrees)
• Nov 12th 2006, 03:24 PM
dklee41
and
The stiffness of a rectangular beam is jointly proportional to the breadth and the cube of the depth. Find the dimensions of the stiffest beam that can be cut from a log in the shape of a right-circular cylinder of radius a centimeters.
• Nov 12th 2006, 04:53 PM
Quick
Quote:

Originally Posted by dklee41

Two sides of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06 radians/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3 (60 degrees)

Area of a triangle is half the sine of an angle times the two connected sides. So you have: $A=\frac{1}{2}(4)(5)\sin\theta$

Change the variables to get: $y=\frac{1}{2}(20)\sin x=10\sin x$

Now your job is to find the slope when $x=\frac{\pi}{3}$ can you do that?

The graph is shown below:
• Nov 12th 2006, 06:00 PM
dklee41
what do you do with the rate of change of the angle .06 radians/sec?
• Nov 12th 2006, 06:12 PM
Soroban
[soze=3]Hello, dklee41![/size]

Quote:

The stiffness of a rectangular beam is jointly proportional
to the breadth and the cube of the depth.
Find the dimensions of the stiffest beam that can be cut from a log
in the shape of a right-circular cylinder of radius $a$ centimeters.

Code:

              * * *           *          *         *-------+-------*       *|      :      |*         |      y:      |       * |      :      | *       * |      *      | *       * |      : \    | *         |      y:  \a  |       *|      :    \ |*         *-------+-------*           *  x    x  *               * * *

The breadth is $2x$, the depth is $2y.$

Since $S \:=\:kbd^3$, we have: . $S \:= \:k(2x)(2y)^3\:=\:16kxy^3$ [1]

From the diagram we have: . $x^2 + y^2 \:=\:a^2\quad\Rightarrow\quad y \:=\:\sqrt{a^2 - x^2}$ [2]

Substitute [2] into [1]: . $S \;= \;16kx\left[\left(a^2-x^2\right)^{\frac{1}{2}}\right]^3 \;= \;16kx\left(a^2-x^2\right)^{\frac{3}{2}}$

And that is the function we must maximize . . .

• Nov 12th 2006, 06:29 PM
dklee41
what is does the variable k represent? and the maximum is found by the derivative correct? how would one go about finding the derivative of the last equation?