# Related Rates Calculus Trigonometric Problem

I've been stuck on this related rates problem for a while now and I just can't figure out how to even approach it. The problem goes something like this:

The above diagram shows two objects moving at different speeds. Both objects are 0.5 miles from the origin. The blue object is moving at 50mph. The straight-line distance between the blue object and the red object is increasing at a rate of 35 mph. Find the speed of the red object.

I tried to solve it using Pythagorean Theorem and finding the derivative of the top side of the triangle. Anyway, I ended up getting a negative number, and even ignoring the sign, the answer I got was obviously wrong. I know the speed of the red object is obviously greater than the blue object because the distance between the objects is increasing. I just really don't know how to calculate the magnitude of said number. There's also a variation of this problem where the straight-line distance is changing at a rate of -35mph but I think that should be doable once I understand how to go about solving the original. Any responses would be appreciated!

Let $$a$$ be the distance of the blue object from the origin, $$b$$ be that of the blue object and $$c=\sqrt{a^2+b^2}$$ be the distance between the objects. All of these are functions of time $$t$$. $$\frac{\partial c}{\partial t}=\frac{\partial c}{\partial a}\frac{\partial a}{\partial t}+\frac{\partial c}{\partial b}\frac{\partial b}{\partial t}=\frac a{\sqrt{a^2+b^2}}\cdot(-50)+\frac b{\sqrt{a^2+b^2}}\cdot x=35$$ We substitute $$a=b=0.5$$: $$\frac{0.5}{\sqrt{0.25+0.25}}=\sqrt{0.5}$$ $$\sqrt{0.5}(-50+x)=35$$ $$x=\frac{35}{\sqrt{0.5}}+50=99.4975$$ Hence the red object is moving at around 99.5 mph.

Clearly this is a right triangle, so we will want to use Pythagorean Theorem. If $$a$$ is the distance of the red object from the origin, and $$b$$ is the distance of the blue object from the origin, and $$c$$ is the straight line distance between them, you have:

$$c^2=a^2+b^2$$

Then we want to implicitly differentiate both sides with respect to time and solve for $$\frac{da}{dt}$$ since that represents the rate of change of the side that is determined by the red object.

$$2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}$$ $$c\frac{dc}{dt}=a\frac{da}{dt}+b\frac{db}{dt}$$ $$c\frac{dc}{dt}-b\frac{db}{dt}=a\frac{da}{dt}$$ $$\frac{da}{dt}=\frac{c\frac{dc}{dt}-b\frac{db}{dt}}{a}$$

Now we just need to plug in what we know. We already know $$a=b=0.5$$. We also know $$\frac{db}{dt}=-50$$ since $$b$$ gets shorter as the blue object moves closer to the origin, and $$\frac{dc}{dt}=35$$ was given. The only thing missing is $$c$$, which we can find with Pythagorean Theorem.

$$c^2=a^2+b^2$$ $$c^2=(0.5)^2+(0.5)^2$$ $$c=\sqrt{0.5}$$

Now plug it all in. $$\frac{da}{dt}=\frac{\sqrt{0.5} * 35 - 0.5 * (-50)}{0.5}$$ $$\frac{da}{dt} \approx 99.497 \ mph$$

Here's a link to a site with a much more detailed explanation of how to solve a problem like this. It's a slightly different problem, but it does explain a bit more how all this works.

https://jakesmathlessons.com/derivatives/related-rates/