increase in rate of area of triangle question

**dklee41** · November 12th 2006, 02:11 PM

please help.

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)

**dklee41** · November 12th 2006, 02:24 PM

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.

**Quick** · November 12th 2006, 03:53 PM

Originally Posted by dklee41

please help.

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:

**dklee41** · November 12th 2006, 05:00 PM

what do you do with the rate of change of the angle .06 radians/sec?

**Soroban** · November 12th 2006, 05:12 PM

[soze=3]Hello, dklee41![/size]

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 . . .

**dklee41** · November 12th 2006, 05:29 PM

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?

Thread: increase in rate of area of triangle question

LinkBack

Thread Tools

Display

increase in rate of area of triangle question

and

Similar Math Help Forum Discussions

Rate of increase

[SOLVED] Percentage increase in area of rectangle given increase in perimeter

Wordy question, about rate of temperature increase.

Rate of change of area of triangle.

annual rate of increase in the price of this car

Search Tags