Question

A curve is such that . The point P (2,9) lies on the curve.

i.  A point moves on the curve in such a way that the x-coordinate is decreasing at a constant rate of 0.05 units per second. Find the rate of change of the y-coordinate when the point is at P.

ii.  Find the equation of the curve.

Solution

i.

We are given that point P moves along the curve in such a way that the x-coordinate is decreasing  at a constant rate of 0.05 units per second.

We are required to find the rate of change of the y-coordinate when P passes through (2, 9).

Rate of change of with respect to is derivative of with respect to  ;

Rate of change of with respect to is derivative of with respect to  ;

Since we are interested in rate of change of y-coordinate of P at point (2,9), we, therefore, first need  the derivative of the curve at point (2,9).

We are given that;

At point (2, 9);

We know that;

Therefore;

Since, x-coordinate is decreasing at a constant rate of 0.05 units per second;

Hence;

Hence, y-coordinate is decreasing at a constant rate of 0.35 units per second.

 

ii.

Next, we need to find equation of the curve.

We are given that;

We can find equation of the curve from its derivative through integration;

Therefore;

Rule for integration of  is:

Rule for integration of  is:

If a point  lies on the curve , we can find out value of . We substitute values of and in the equation obtained from integration of the derivative of the curve i.e. .

We are also given that curve passes through the point (2,9).

Substitution of x and y coordinates of point in above equation;

Therefore, equation of the curve is;