1. ## Related Rates

Equation; $y = 2(x^2 - 3x)$

Find;
a) dy/dt
b) dx/dt

Given;
a) x = 3, dx/dt = 2
b) x = 1, dy/dt = 5

How do I go about solving this? I'm confused as to the procedure I need to use. A step-by-step solution would be greatly appreciated.

2. Originally Posted by Archduke01
Equation; $y = 2(x^2 - 3x)$

Find;
a) dy/dt
b) dx/dt

Given;
a) x = 3, dx/dt = 2
b) x = 1, dy/dt = 5

How do I go about solving this? I'm confused as to the procedure I need to use. A step-by-step solution would be greatly appreciated.
$y = 2(x^2 - 3x)$
and we are assuming that the functions x = x(t) and y = y(t). So we do this implicitly:

$\frac{dy}{dt} = 4x \frac{dx}{dt} - 6 \frac{dx}{dt}$

Thus for part a you have x = 3 and dx/dt = 2 so
$\frac{dy}{dt} = 4(3)(2) - 6(2) = 12$

You try the second one.

-Dan

3. Originally Posted by topsquark
$y = 2(x^2 - 3x)$
and we are assuming that the functions x = x(t) and y = y(t). So we do this implicitly:

$\frac{dy}{dt} = 4x \frac{dx}{dt} - 6 \frac{dx}{dt}$

Thus for part a you have x = 3 and dx/dt = 2 so
$\frac{dy}{dt} = 4(3)(2) - 6(2) = 12$

You try the second one.

-Dan
Thank you. I just have one last question; how would you implicitly derive this function; $xy = 4$? It is of the same topic.

4. Originally Posted by Archduke01
Thank you. I just have one last question; how would you implicitly derive this function; $xy = 4$? It is of the same topic.
xy is a product. Thus:
$\frac{dx}{dt}y + x \frac{dy}{dt} = 0$

-Dan

5. I'm having trouble with something. The topic question was fine because it only had 3 variables, 2 of which were given, so you just had to isolate the 3rd one and solve for it. But for this one, there's 4 variables; $xy' + x'y = 0$. You're asked to find the value for y' while given x and x'. How could I solve that? I have 2 unknown variables. (y' and y)

BTW; x = 8, x' = 10.