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

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

...

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?

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

<-> ...

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?

and now use the fact that to find "y" as you have "x".

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.

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.

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.