Related Rates of a triangle angle

[courtrigrad]

Hello all

Just came across a few questions on related rates and would like some verification on whether I am doing these correctly:

1. Let $$\theta$$ be an acute angle in a right triangle, and let x and y, respectively be the sides adjacent and opposite of $$\theta$$. Suppose that x and y vary with time? How are $$\frac{d\theta}{dt} \frac{dx}{dt} \frac{dy}{dt}$$ related? Well I set up a relationship where $$tan \theta = \frac{y}{x}$$ So $$\theta = \arctan(\frac{y}{x})$$ Hence $$\frac{d\theta}{dt} = d(\arctan(\frac{y}{x})$$ Is this right?

 

[BobG]

So far. Do you know the derivatives for inverse trig functions?

If not, it might be better to set your equation up like this:

$$tan \theta=\frac{y}{x}$$

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

Find the derivative for the left side (remembering to apply the chain rule). Find the derivative for the right side using the quotient rule. Isolate $$\frac{d \theta}{dt}$$ on one side of the equation and simplify.

You get the same answer either way (at least, you do if you know your trig identities). I'm not sure the arctangent saves you any steps and the inverse functions are hard to memorize on top of everything else you have to remember (unless you're doing the inverse trig functions, in which case you're probably expected to remember them for at least a couple of weeks).

 

[courtrigrad]

ok so $$\frac{d}{dt}(tan \theta)=\frac{d}{dt} (\frac{y}{x})$$ and this equals $$sec^{2}x \frac{d\theta}{dt} = \frac{x(\frac{dy}{dt}) + y(\frac{dx}{dt})}{x^2}$$ So do I just solve for $$\frac{d \theta}{dt}$$ from this equation?

Thanks

 

[BobG]

Yes.

But, to isolate $$\frac{d \theta}{dt}$$, you need to multiply both sides by $$\frac{1}{sec^2 \theta}$$, which happens to be $$cos^2 \theta$$

Your typical related rates problem for this will give you your intial x and y positions, or one of them and your angle (from the two provided, you can find the third). They'll also give you your two velocities, which are the derivatives of your position components. Once you have all 5, you plug into the equation to find out how fast your angle is changing.

 

[learningphysics]

ok so $$\frac{d}{dt}(tan \theta)=\frac{d}{dt} (\frac{y}{x})$$ and this equals $$sec^{2}x \frac{d\theta}{dt} = \frac{x(\frac{dy}{dt}) + y(\frac{dx}{dt})}{x^2}$$ So do I just solve for $$\frac{d \theta}{dt}$$ from this equation?

Thanks

I believe the above should be

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

 

[courtrigrad]

ok so the final solution would be: $$\frac{d\theta}{dt} = cos^2\theta \frac{x(\frac{dy}{dt}) - y(\frac{dx}{dt})}{x^2}$$

is this right?