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 June 6th, 2012, 06:15 AM #1 Member   Joined: Jan 2012 Posts: 72 Thanks: 0 Rate of change An isosceles triangle has fixed base of length b cm. The other 2 equal sides of the triangle are each decreasing at the constant rate of 3cm per second. How fast is the area changing when the triangle is equilateral? Leave your answer in terms of b.
 June 6th, 2012, 07:35 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Rate of change Let s be the equal sides of the isosceles triangle and h be the height. We then have the area A is: $A=\frac{1}{2}bh$ By Pythagoras, we have: $h=\sqrt{s^2-$$\frac{b}{2}$$^2}=\frac{1}{2}\sqrt{4s^2-b^2}$ Hence: $A=\frac{1}{4}b\sqrt{4s^2-b^2}$ Implicitly differentiate with respect to time t: $\frac{dA}{dt}=\frac{1}{4}b$$\frac{8s}{2\sqrt{4s^2-b^2}}\cdot\frac{ds}{dt}$$=\frac{bs}{\sqrt{4s^2-b^2}}\cdot\frac{ds}{dt}$ We are given $\frac{ds}{dt}=-3\text{ \frac{cm}{s}}$ thus: $\frac{dA}{dt}=-\frac{3bs}{\sqrt{4s^2-b^2}}\:\text{\frac{cm^2}{s}}$ Now, when  we have: $\frac{dA}{dt}\|_{s=b}=-\frac{3b^2}{\sqrt{4b^2-b^2}}\:\text{\frac{cm^2}{s}}=-\sqrt{3}b\text{ \frac{cm^2}{s}}$
 June 6th, 2012, 08:23 PM #3 Member   Joined: Jan 2012 Posts: 72 Thanks: 0 Re: Rate of change thanks

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# the two equal.sides of an . isosceles triangle with.fixed base.b are decreasing

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