Rate of change in an Isosceles triangle

In summary, the area of an isosceles triangle increases at a rate of 10 degrees per minute. At what value of x will the triangle have a maximum area?
  • #1
Econguy
4
0
An Isosceles triangle has two equal sides of length 10cm. Let x be the angle between the two equal sides.

a. Express the area A of the triangle as a function of x in radians.

b. Suppose that x is increasing at the rate of 10 degrees per minute. How fast is A changing at the instant x = pi/3? At what value of x will the triangle have a maximum area?

I've set the triangle and drew a line to cut it in half. the angle is now x/2, and the base is b/2. The hypoteneuse of each triangle is 10.

cos x/2 = adj/hyp = h/10
sin x/2 = opp/hyp = b/2/10 = b/20

A = 100sinx/2 cosx/2
= 100 sinx/2

would that be the equation for part a?

any help for part b would be greatly appreciated.

thanks
 
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  • #2
Econguy said:
An Isosceles triangle has two equal sides of length 10cm. Let x be the angle between the two equal sides.

a. Express the area A of the triangle as a function of x in radians.

b. Suppose that x is increasing at the rate of 10 degrees per minute. How fast is A changing at the instant x = pi/3? At what value of x will the triangle have a maximum area?

I've set the triangle and drew a line to cut it in half. the angle is now x/2, and the base is b/2. The hypoteneuse of each triangle is 10.

cos x/2 = adj/hyp = h/10
sin x/2 = opp/hyp = b/2/10 = b/20

A = 100sinx/2 cosx/2
= 100 sinx/2
How did you go from "100 sin(x/2)cos(x/2)" to just "100 sin(x/2)"? What happened to the "cos(x/2)"? Perhaps you were trying to use the fact that sin(2a)= 2sin(a)cos(a).


would that be the equation for part a?

any help for part b would be greatly appreciated.

thanks
Once you have the correct formula for A as a function of x, then dA/dt= (dA/dx)(dx/dt) and you are told that dx/dt= 10 degrees per minute.
 
  • #3
HallsofIvy said:
How did you go from "100 sin(x/2)cos(x/2)" to just "100 sin(x/2)"? What happened to the "cos(x/2)"? Perhaps you were trying to use the fact that sin(2a)= 2sin(a)cos(a).

Yes, I used the double angle formula (sin(2a)= 2sin(a)cos(a)). Was that incorrect? I thought I could use that identity...?
 
  • #4
Econguy said:
HallsofIvy said:
How did you go from "100 sin(x/2)cos(x/2)" to just "100 sin(x/2)"? What happened to the "cos(x/2)"? Perhaps you were trying to use the fact that sin(2a)= 2sin(a)cos(a).

Yes, I used the double angle formula (sin(2a)= 2sin(a)cos(a)). Was that incorrect? I thought I could use that identity...?
You can use that identity, but you have to use it correctly.

100 sin(x/2) cos(x/2) = 50 * [2 sin(x/2)cos(x/2)] = 50 sin(2*x/2) = 50 sin(x)
 
  • #5
Hey guys, so I'd like to resurrect this thread since I'm also working on it.

So for part A) area=50sinx by double angle formula

B) My answer is:

dx/dt=10=0.1745rad/min

da/dt=50cosx(dx/dt)

=50cos(pi/3)(0.1745rad)

=4.36cm^2/min

For max area, using optimization techniques:

A'=50cosx >0 for 0<x<pi/2 and 3pi/2<x<2pi (since dimensions can't be negative)
A' <0 for pi/2<x<3pi/2
A'= 0 for x= pi/2 and 3pi/2
Using closed interval method:
A(pi/2)=50
A(3pi/2)=-50, therefore A will be max when x=pi/2

I have a feeling I did the first part of B) wrong, what do the pros think?
 
  • #6
Bump, any takers? Bribes? Love perhaps?
 

FAQ: Rate of change in an Isosceles triangle

1. What is the formula for finding the rate of change in an Isosceles triangle?

The formula for finding the rate of change in an Isosceles triangle is (change in y)/(change in x), where y represents the vertical side and x represents the horizontal side.

2. How do you calculate the slope of an Isosceles triangle?

The slope of an Isosceles triangle can be calculated by dividing the change in y by the change in x, or by using the formula (y2-y1)/(x2-x1) where (x1,y1) and (x2,y2) are two points on the triangle's base.

3. Can the rate of change in an Isosceles triangle be negative?

Yes, the rate of change in an Isosceles triangle can be negative. This means that the triangle is decreasing in size as the x-coordinate increases, or that the vertical side is decreasing in length as the horizontal side increases.

4. How does the rate of change affect the angles in an Isosceles triangle?

The rate of change does not directly affect the angles in an Isosceles triangle. However, if the triangle is changing in size at a constant rate, the angles will remain the same. If the rate of change is not constant, the angles may change slightly.

5. What real-life situations involve the concept of rate of change in an Isosceles triangle?

The concept of rate of change in an Isosceles triangle can be applied in various real-life situations, such as calculating the increase in height of a tree over time, determining the change in distance traveled by a moving object, or measuring the change in temperature over a period of time.

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