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.
