Related Rates and Area of a Triangle

In summary: But after this, I don't know how to get to the answer 25/36 pi cm/s.Can you please show me the differentiation part?Also, the units for the answer should be cm^2/s.In summary, to find the rate of decrease of the area of a right triangle when one of the acute angles is decreasing at a rate of 5 degrees/s, we can use the formula A = 1/2 b*h and differentiate both sides with respect to time. This results in dA/dt = 50cos(2theta)*dtheta/dt, where theta is the changing angle. Substituting theta = pi/6 and dtheta/dt = -5 degrees/s, we get dA/d
  • #1
bondgirl007
73
0

Homework Statement



The length of the hypotenuse of a right triangle is 10 cm. One of the acute angles is decreasing at a rate of 5 degrees/s. how fast is the area decreasing when this angle is 30 degrees?

Homework Equations



The Attempt at a Solution



I got the a and b using the cos and sin of the 30degrees. For a, I got 5[tex]\sqrt{3}[/tex] and for b, I got 5.

I know that a will stay constant when the angle is decreasing so da/dt is 0. For the db/dt I got (-50[tex]\sqrt{3}[/tex])/2

How am I supposed to find the area from this?

Any help would be appreciated!
 
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  • #2
Try to start this problem by finding the Area of the triangle in terms of the angle that changing. After you do this, can you find the rate of change of the area in terms of the rate of change of the angle using some calculus?

To sum up:

[tex]\frac{d\phi}{dt}= 5 [/tex] where [tex]\phi[/tex] is the angle that is changing.

Now, can you you find A in terms of phi, the angle?

If so, you should be able to find [tex]\frac{dA}{dt}[/tex] in terms of [tex]\frac{d\phi}{dt}[/tex] using some calculus.

Does this help?

EDIT: I just realized you used "a" as a variable. Here, by "A," I mean the Area of the triangle. Just so you aren't confused.
 
Last edited:
  • #3
bondgirl007 said:

Homework Statement



The length of the hypotenuse of a right triangle is 10 cm. One of the acute angles is decreasing at a rate of 5 degrees/s. how fast is the area decreasing when this angle is 30 degrees?


Homework Equations






The Attempt at a Solution



I got the a and b using the cos and sin of the 30degrees. For a, I got 5[tex]\sqrt{3}[/tex] and for b, I got 5.

I know that a will stay constant when the angle is decreasing so da/dt is 0. For the db/dt I got (-50[tex]\sqrt{3}[/tex])/2

How am I supposed to find the area from this?

Any help would be appreciated!
Don't start with what happens when [itex]\theta= 30 degrees[/itex]. This is a "dynamic" problem. What happens for any right triangle with hypotenuse 10 cm? Since you are asked, in particular, about the area, and area of a right triangle is "1/2 base times height, what is the height of a right triangl with hypotenuse 10 and angle [itex]\theta[/itex]? What is the base of a right triangle with hypotenuse 10 and angle [itex]\theta[/itex]? And so what is the area? Now differentiate both sides of that equation with respect to time.
 
  • #4
Thanks for the replies. I got the right answer by doing

A = 1/2 b*h
A = 1/2 (10cos [itex]\theta[/itex]* 10sin [itex]\theta[/itex])

After that I differentiated to find dA/dt.
 
  • #5
Nice Job!
 
  • #6
what is the answer to this problem, i want to check my work and answer! thanks.
 
  • #7
The answer to this is 25/36 [tex]\pi[/tex]
 
  • #8
bondgirl007 said:
The answer to this is 25/36 [tex]\pi[/tex]
mmph ... i did it the same method as you did and i am not getting that answer

[tex]A=\frac{1}{2}xy[/tex]

[tex]x=10\cos{30}[/tex]
[tex]y=10\sin{30}[/tex]

[tex]A=\frac{1}{2}10\sin{\theta}10\cos{\theta}[/tex]

[tex]A=50\sin{\theta}\cos{\theta}[/tex]

[tex]\frac{1}{2}\sin{2\theta}=\sin{\theta}\cos{\theta}[/tex]

[tex]A=25\sin{2\theta}[/tex]

[tex]\frac{dA}{dt}=50\cos{2\theta}\frac{d\theta}{dt}[/tex]

[tex]\frac{dA}{dt}=-250\cos{60}\frac{cm}{s}[/tex]

[tex]\frac{dA}{dt}=-125\frac{cm}{s}[/tex]

argh!
 
Last edited:
  • #9
I got the same thing you got robo but my question is would it be negative? Because the rate is decreasing so shouldn't it be -5 degrees/sec?.
My question to Bondgirl007 is what units is your answer in?
 
  • #10
ace123 said:
I got the same thing you got robo but my question is would it be negative? Because the rate is decreasing so shouldn't it be -5 degrees/sec?.
My question to Bondgirl007 is what units is your answer in?
yes it would be negative, my bad. but hmm ...
 
  • #11
rocophysics said:
mmph ... i did it the same method as you did and i am not getting that answer

[tex]A=\frac{1}{2}xy[/tex]

[tex]x=10\cos{30}[/tex]
[tex]y=10\sin{30}[/tex]

[tex]A=\frac{1}{2}10\sin{\theta}10\cos{\theta}[/tex]

Till here, I have the same answer as you but after this, I differentiated dA/dt and applied the Product Law to the sin[tex]\theta[/tex]cos[tex]\theta[/tex] and then I substituted pi/6 for [tex]\theta[/tex] and multiplied by that will give me a 25. And d[tex]\theta[/tex]/dt is [tex]\pi[/tex]/36 cm.
 

1. What is the concept of "related rates" in mathematics?

The concept of related rates in mathematics involves finding the rate of change of one quantity with respect to another related quantity. This is typically done by using the chain rule from calculus.

2. How are related rates applied to the area of a triangle?

Related rates can be applied to the area of a triangle by considering the triangle as a changing shape, with one side increasing or decreasing in length. By using the formula for the area of a triangle and the chain rule, the rate of change of the area can be determined.

3. What is the formula for finding the area of a triangle?

The formula for finding the area of a triangle is A = 1/2 * base * height, where A is the area, base is the length of the triangle's base, and height is the perpendicular height of the triangle.

4. How can the Pythagorean Theorem be used to find related rates in a right triangle?

The Pythagorean Theorem can be used to find related rates in a right triangle by considering the triangle as a changing shape, with one of the sides increasing or decreasing in length. By using the Pythagorean Theorem and the chain rule, the rates of change of the other two sides can be determined.

5. Can related rates be applied to any shape, or only triangles?

Related rates can be applied to any shape, as long as the relationship between the changing quantities can be expressed mathematically. However, triangles are a common and useful shape to apply related rates to, as they often involve changing angles and sides.

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