# Thread: related rates with triangle

1. ## related rates with triangle

basically I don't understand the suggestion on how this was set up

Let $\displaystyle \theta$ (in radians) be an angle in a right triangle, and let x and y, respectively, be the lengths of the sides adjacent to and opposite $\displaystyle \theta$ Suppose that x and y vary with time.

(a) How are $\displaystyle \frac{d\theta}{dt}$ , $\displaystyle \frac{dx}{dt}$ and $\displaystyle \frac{dy}{dt}$ Related

the book says that
$\displaystyle \frac{d\theta}{dt}=\frac{cos^2\theta}{x^2}\left( x \frac{dy}{dt}-y\frac{dr}{dt}\right)$

thanks for help

actually I don't know where the $\displaystyle \frac{cos^2\theta}{x^2}$ comes from I assume the hypotenuse is a constant

2. ## Re: related rates with triangle

$\displaystyle \frac{d\theta}{dt}=\frac{cos^2\theta}{x^2}\left( x \frac{dy}{dt}-y\frac{dr}{dt}\right)$

$\displaystyle \tan (\theta) = \frac{y}{x}$

Differentiate both sides.

Quotient rule. Etc.

(p.s. If x and y are changing, then the hypotenuse is NOT constant!)

3. ## Re: related rates with triangle

i did try that but still couldn't get this... will try again.. guess we don't need to worry about hyp if tan is used

don't know if you want to show how this was derived.

4. ## Re: related rates with triangle

Originally Posted by bigwave
basically I don't understand the suggestion on how this was set up

Let $\displaystyle \theta$ (in radians) be an angle in a right triangle, and let x and y, respectively, be the lengths of the sides adjacent to and opposite $\displaystyle \theta$ Suppose that x and y vary with time.

(a) How are $\displaystyle \frac{d\theta}{dt}$ , $\displaystyle \frac{dx}{dt}$ and $\displaystyle \frac{dy}{dt}$ Related

the book says that
$\displaystyle \frac{d\theta}{dt}=\frac{cos^2\theta}{x^2}\left( x \frac{dy}{dt}-y\frac{dr}{dt}\right)$

thanks for help

actually I don't know where the $\displaystyle \frac{cos^2\theta}{x^2}$ comes from I assume the hypotenuse is a constant
$\displaystyle \frac{d}{dt} \left[\tan{\theta} = \frac{y}{x}\right]$

$\displaystyle \sec^2{\theta} \cdot \frac{d\theta}{dt} = \frac{x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}}{x^2}$

$\displaystyle \frac{1}{\cos^2{\theta}} \cdot \frac{d\theta}{dt} = \frac{1}{x^2} \left(x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}\right)$

$\displaystyle \frac{d\theta}{dt} = \frac{\cos^2{\theta}}{x^2} \left(x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}\right)$

5. ## Re: related rates with triangle

Originally Posted by skeeter
$\displaystyle \frac{d}{dt} \left[\tan{\theta} = \frac{y}{x}\right]$

$\displaystyle \sec^2{\theta} \cdot \frac{d\theta}{dt} = \frac{x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}}{x^2}$

$\displaystyle \frac{1}{\cos^2{\theta}} \cdot \frac{d\theta}{dt} = \frac{1}{x^2} \left(x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}\right)$

$\displaystyle \frac{d\theta}{dt} = \frac{\cos^2{\theta}}{x^2} \left(x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}\right)$
Thanks, skeeter! I wasn't about to LaTex all that!

6. ## Re: related rates with triangle

wow, thx, I was close but not close enough

yep, latex is 80% of the process....

7. ## Re: related rates with triangle

Originally Posted by bigwave
wow, thx, I was close but not close enough

yep, latex is 80% of the process....
90% in this case ...