Rate of change calculus?

• Jul 21st 2011, 07:08 PM
homeylova223
Rate of change calculus?
I have two different questions

my first one is differentiate with respect to t

y=x^2-3x find dy/dt when x=3 and dx/dt=2

This is what I did
dy/dt=2x (dx/dt)-3 I am not sure if this is correct

my second one is

xy=4 find dy/dt when x=8 dx/dt=10

I am not sure how to begin this one
• Jul 21st 2011, 07:30 PM
pickslides
Re: Rate of change calculus?
Quote:

Originally Posted by homeylova223

my first one is differentiate with respect to t

y=x^2-3x find dy/dt when x=3 and dx/dt=2

This is what I did
dy/dt=2x (dx/dt)-3 I am not sure if this is correct

dy/dx = dy/dt * dt/dx

when x=3

2(3)-3 = dy/dt * 1/2

dy/dt = 6

For the second one make xy=4 so y=4/x find dy/dx
• Jul 21st 2011, 07:32 PM
skeeter
Re: Rate of change calculus?
Quote:

Originally Posted by homeylova223
I have two different questions

my first one is differentiate with respect to t

y=x^2-3x find dy/dt when x=3 and dx/dt=2

This is what I did
dy/dt=2x (dx/dt)-3(dx/dt)

my second one is

xy=4 find dy/dt when x=8 dx/dt=10

I am not sure how to begin this one

product rule ...

x(dy/dt) + y(dx/dt) = 0

...
• Jul 22nd 2011, 09:06 AM
homeylova223
Re: Rate of change calculus?
I used the product rule for xy=4 to get

x(dy/dt)+y(dx/dt)=0 then I plug x=8 dx/dt=10

8(dy/dt)+y(10)=0

How would I proceed now?
• Jul 22nd 2011, 09:10 AM
Siron
Re: Rate of change calculus?
You now y=4/x so y(10)=4/10, substituted in the equation:
8(dy/dt)+(4/10)=0
<-> ...
• Jul 22nd 2011, 09:13 AM
Re: Rate of change calculus?
Quote:

Originally Posted by homeylova223
I used the product rule for xy=4 to get

x(dy/dt)+y(dx/dt)=0 then I plug x=8 dx/dt=10

8(dy/dt)+y(10)=0

How would I proceed now?

$\displaystyle 8\frac{dy}{dt}=-10y\Rightarrow\frac{dy}{dt}=-\frac{10}{8}y$

and now use the fact that $\displaystyle xy=4$ to find "y" as you have "x".
• Jul 22nd 2011, 09:46 AM
homeylova223
Re: Rate of change calculus?
Hmm I understand up the part where you get

dy/dt=-10/8y

But what would I do knowing this
I am unsure where to plug it in.
• Jul 22nd 2011, 10:24 AM
Re: Rate of change calculus?
Quote:

Originally Posted by homeylova223
Hmm I understand up the part where you get

dy/dt=-10/8y

But what would I do knowing this
I am unsure where to plug it in.

$\displaystyle \frac{dy}{dt}=-\frac{10y}{8}$

Now you need the value of "y" to obtain a value for dy/dt.

Since xy=4, then 8y=4 since x=8.
This gives the value for "y" that allows you to find the value for dy/dt.