The question: A car leaves an intersection traveling east. Its position t sec later is given by $x = t^2 + t$ ft. At the same time, another car leaves the same intersection heading north, traveling $y = t^2 + 5t$ ft in t sec. Find the rate at which the distance between the two cars will be changing 5 sec later. (Round your answer to one decimal place.) So what I have from this is $$x=30,y=50,\frac{dx}{dt}=10,\frac{dy}{dt}=15,z=\sqrt{3,400}$$ I found $\frac{dx}{dt}$and $\frac{dy}{dt}$ by differentiating $x = t^2 + t$ and $y = t^2 + 5t$ and then plugging in 5 for $t$. (Not sure if right)
So I have $x^2+y^2=z^2$ so $$2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}$$ and after plugging in the values I THINK I know I end up with $$\frac{60(10)+100(15)}{2\sqrt{3,400}}$$ which is around 18.007 but the answer is 18.5 rounded. So where am I going wrong? All help is appreciated!
Good approach, but after doing some related rate problems myself, I found that the best way to solve these problems is just to simplify everything.
So, you obviously noticed that we need to use the Pythagorean Theorem,
$$a^2+b^2=c^2$$
In this case,
$$a = x = t^2 + t$$ $$b = y = t^2 + 5t$$
and we can let d represent c to become,
$$c = d$$
So, we can now simplify your equation to:
$$x^2 + y^2 = d^2$$
which is,
$$(t^2+t)^2+(t^2+5t)^2=d^2$$
after simplying this, we get:
$$2t^4+12t^3+26t^2=d^2$$
NOW we can differentiate to get: $$8t^3+36t^2+52t=2d(dd/dt)$$
now, solve for dd/dt after plugging in t = 5. d can be found by plugging in t = 5 from this equation in the first part of the problem:
$$(t^2+t)^2+(t^2+5t)^2=d^2$$
Hope this helped!