How Fast Does the X-Coordinate Move on a Complex Graph Curve?

[marshmallow1]

A point is moving on the graph of

3x^2 + 4y^3 = xyWhen the point is at P = (1/7, 1/7)

its y-coordinate is increasing at a speed of 3
units per second.
What is the speed of the x-coordinate at
that time and in which direction is the xcoordinate
moving?

 

[MarkFL]

Hello, and welcome to MHB! (Wave)

We are given the curve:

\(\displaystyle 3x^2+4y^3=xy\)

In order to answer the given question, it will be helpful to implicitly differentiate with respect to time $t$...in doing so, what do you get?

 

[marshmallow1]

hi thanks!

would you get

9x + 12y (dy/dx) = xy

I am not entirely sure how to differentiate in respect to timeAlso sorry the equation before was 3x^3+4y^3 = xy
the first exponent being wrong in my first post. either way it will help me to learn how its done

 

[MarkFL]

hi thanks!

would you get

9x + 12y (dy/dx) = xy

I am not entirely sure how to differentiate in respect to timeAlso sorry the equation before was 3x^3+4y^3 = xy
the first exponent being wrong in my first post. either way it will help me to learn how its done

We consider that both $x$ and $y$ are functions of time $t$, and so beginning with:

\(\displaystyle 3x^3+4y^3=xy\)

We apply on the left the power and chain rules, and on the right the product and chain rules to obtain:

\(\displaystyle 9x^2\d{x}{t}+12y^2\d{y}{t}=\d{x}{t}y+x\d{y}{t}\)

We are being asked to find \(\displaystyle \d{x}{t}\), and so solving for that, what do you obtain?

 

[MarkFL]

To follow up, we get:

\(\displaystyle \d{x}{t}=\frac{\d{y}{t}\left(x-12y^2\right)}{9x^2-y}\)

Plugging in the given values, we find:

\(\displaystyle \d{x}{t}=\frac{\dfrac{3}{7}\left(1-\dfrac{12}{7}\right)}{\dfrac{1}{7}\left(\dfrac{9}{7}-1\right)}=\frac{3(7-12)}{9-7}=-\frac{15}{2}\)

And so the x-coordinate is moving at 7.5 units/second to the left.