Related Rates and Area of a Triangle

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Homework Help Overview

The discussion revolves around a related rates problem involving a right triangle with a hypotenuse of 10 cm. One of the acute angles is decreasing at a rate of 5 degrees per second, and participants are exploring how this affects the area of the triangle when the angle is 30 degrees.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss finding the area of the triangle in terms of the changing angle and how to differentiate this to find the rate of change of the area. Some express confusion about the relationship between the angle and the area, while others suggest focusing on the general case rather than a specific angle.

Discussion Status

There is an ongoing exploration of the relationships between the triangle's dimensions and the area as the angle changes. Some participants have provided guidance on how to set up the problem, while others are questioning the implications of their calculations and the signs of their results.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also a focus on ensuring clarity in the definitions of variables and the relationships between them.

bondgirl007
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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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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:
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.
 
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.
 
Nice Job!
 
what is the answer to this problem, i want to check my work and answer! thanks.
 
The answer to this is 25/36 [tex]\pi[/tex]
 
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:
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.
 

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