Rates of Change [B&C Ex 22.4 Q8] (1 Viewer)

Pyrobooby · Mar 11, 2011

A point moves along the curve $y=\frac{1}{4} x^2$ . The abscissa of the point changes at the rate of 4 units per second. At what rate is the ordinate increasing when x=3?

My working:

$y= \frac{x^2}{4} \\ \therefore \frac{dy}{dx}= \frac{x}{2}\\ Given: \frac{dy}{dt}=4\\ \frac{dy}{dt}= \frac{dy}{dx} * \frac{dx}{dt} \\ \therefore \frac{dx}{dt}= \frac{( \frac{dy}{dt})}{( \frac{dy}{dx})} \\ \frac{dx}{dt}= \frac{8}{x}. \\ \text{When } x=3, \\ \frac{dx}{dt}= \frac{8}{3} \\ \therefore \text{the rate of the ordinate's increase is 8/3 units per second.}$

Yeah the answer says 6 units per second.

Drongoski · Mar 11, 2011

You need to find dy/dt and not dx/dt and abscissa refers to the x-coordinate and not the y-coordinate.

dy/dt = dy/dx * dx/dt = 3/2 * 4 = 6

Pyrobooby · Mar 11, 2011

Wouldn't dy/dt be the rate of the increasing abscissa?

Drongoski · Mar 11, 2011

No; it is dx/dt.

Long ago we called the x-part of (x,y) the abscissa and the y-part the ordinate.

Pyrobooby · Mar 11, 2011

I see, thank you!

Rates of Change [B&C Ex 22.4 Q8] (1 Viewer)

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