Rate of change in an Isosceles triangle

  • Thread starter Econguy
  • Start date
  • #1
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
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
964
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
4
0
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
35,120
6,861
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
96
0
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
96
0
Bump, any takers? Bribes? Love perhaps?
 

Related Threads on Rate of change in an Isosceles triangle

  • Last Post
Replies
1
Views
2K
Replies
5
Views
8K
Replies
2
Views
10K
  • Last Post
Replies
2
Views
4K
Replies
3
Views
6K
Replies
3
Views
3K
Replies
3
Views
1K
Replies
1
Views
4K
Top