How Fast Does the Area of a Triangle Change with Increasing Angle?

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Discussion Overview

The discussion revolves around determining the rate at which the area of a triangle changes as the angle between two fixed-length sides increases. Participants explore the application of calculus, specifically derivatives, to solve the problem, while also debating the appropriate formula for area and the implications of constant side lengths.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant presents the problem of finding the rate of change of the area of a triangle given two sides and an increasing angle, expressing uncertainty about which area formula to use.
  • Another participant suggests using the formula for the area of a triangle, \(A=\frac{4\times 5\times \sin\theta}{2}\), and indicates the need to find \(\frac{dA}{dt}\) when \(\theta=\frac{\pi}{3}\).
  • A further reply reiterates the area formula and proposes differentiating it with respect to time, leading to a calculation of \(\frac{dA}{dt}\) but yielding a result that conflicts with a provided answer.
  • Some participants challenge the differentiation process, noting that the side lengths should be considered as constants in the derivative, while others argue that they are factors that should remain in the differentiation.

Areas of Agreement / Disagreement

Participants express differing views on the differentiation of the area formula, particularly regarding the treatment of constant side lengths. There is no consensus on the correct approach to the problem or the resulting calculations.

Contextual Notes

There are unresolved assumptions regarding the application of differentiation rules and the treatment of constants in the context of the area formula. The discussion reflects varying interpretations of mathematical principles without reaching a definitive resolution.

tmt1
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Two sides of a triangle are 4 m and 5 m in length and the angle between them is increasing at a rate of 0.06 rad/s. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is $$\pi/3$$

When the angle is $$pi/3$$, the third side is equal to : $$\sqrt{21}$$ based on the law of cosines.I'm not sure which formula for Area I should use to figure this problem out.
 
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tmt said:
Two sides of a triangle are 4 m and 5 m in length and the angle between them is increasing at a rate of 0.06 rad/s. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is $$\pi/3$$

When the angle is $$pi/3$$, the third side is equal to : $$\sqrt{21}$$ based on the law of cosines.I'm not sure which formula for Area I should use to figure this problem out.

Hi tmt, :)

You could use derivatives to solve this question. Suppose the angle between the two given sides is $\theta$. Then the area of the triangle is given by,

\[A=\frac{4\times 5\times \sin\theta}{2}\]

You have to find $\frac{dA}{dt}$ when $\theta=\frac{\pi}{3}$. Try to do it from here. :)
 
Sudharaka said:
Hi tmt, :)

You could use derivatives to solve this question. Suppose the angle between the two given sides is $\theta$. Then the area of the triangle is given by,

\[A=\frac{4\times 5\times \sin\theta}{2}\]

You have to find $\frac{dA}{d\theta}$ when $\theta=\frac{\pi}{3}$. Try to do it from here. :)

\[A=\frac{4\times 5\times \sin\theta}{2}\]So I differentiate this with respect to time in seconds.

$\d{A}{t} = \frac{1}{2} \cos\theta \d{\theta}{t}$

I plug this in:

$\cos \pi / 2 = 1/2$

and I know that

\d{\theta}{t} = 0.06 radians/second

so

$\d{A}{t} = 0.06/ 4 m^2 / second$

or

0.015 squared/ second, but the answer in the text is 0.3 metres squared a second?
 
tmt said:
$\d{A}{t} = \frac{1}{2} \cos\theta \d{\theta}{t}$

Double check this one. You've left out the side lengths from your derivative
 
SamJohannes said:
Double check this one. You've left out the side lengths from your derivative

The side length are constant so when they are differentiated they go to 0, don't they?
 
tmt said:
The side length are constant so when they are differentiated they go to 0, don't they?

Not when they are factors...recall:

$$\d{}{x}\left(k\cdot f(x)\right)=k\cdot\d{}{x}\left(f(x)\right)$$
 

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