# Thread: Simple real world calculus problem

1. ## Simple real world calculus problem

How do you do part a) ?

2. ## Re: Simple real world calculus problem

Originally Posted by princessmath

How do you do part a) ?
Hi there,

You have x^2 + y^2 = 200

differentiating them implicitly gives you

(x^2)' + (y^2)' = (200)' which gives you

2x(dx/dt) + 2y = 0 ...

now, plug in the values and solve for dx/dt,

hope this help,

dokrbb

3. ## Re: Simple real world calculus problem

Originally Posted by dokrbb
Hi there,

You have x^2 + y^2 = 200

differentiating them implicitly gives you

(x^2)' + (y^2)' = (200)' which gives you

2x(dx/dt) + 2y = 0 ...

now, plug in the values and solve for dx/dt,

hope this help,

dokrbb
This is incorrect. x and y are both functions of t, so when you differentiate both sides, you should get

\displaystyle \begin{align*} x^2 + y^2 &= 100 \\ \frac{d}{dt} \left( x^2 + y^2 \right) &= \frac{d}{dt} \left( 100 \right) \\ \frac{d}{dx}\left( x^2 \right) \frac{dx}{dt} + \frac{d}{dy}\left( y^2 \right) \frac{dy}{dt} &= 0 \\ 2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} &= 0 \\ 2x\,\frac{dx}{dt} + 2 \, \sqrt{ 100 - x^2 } \,\frac{dy}{dt} &= 0 \end{align*}

NOW you can substitute in the known values and solve for $\displaystyle \frac{dx}{dt}$.