Post

Created by @nathanedwards
 at November 1st 2023, 6:05:50 pm.

Question:

Consider the equation of the circle given by the equation: [x^2 + y^2 = r^2] where rr is a positive constant.

(a) Find dydx\frac{{dy}}{{dx}} using implicit differentiation.

(b) Find the derivative d2ydx2\frac{{d^2y}}{{dx^2}} using implicit differentiation.

Answer:

(a) To find dydx\frac{{dy}}{{dx}} using implicit differentiation, we will differentiate both sides of the equation with respect to xx and apply the chain rule.

Differentiating both sides of the equation x2+y2=r2x^2 + y^2 = r^2 with respect to xx, we get:

2x+2ydydx=02x + 2y \cdot \frac{{dy}}{{dx}} = 0

To isolate dydx\frac{{dy}}{{dx}}, we rearrange the equation as:

2ydydx=2x2y \cdot \frac{{dy}}{{dx}} = -2x

Dividing both sides of the equation by 2y2y, we find:

dydx=xy\frac{{dy}}{{dx}} = -\frac{{x}}{{y}}

Thus, the derivative of yy with respect to xx is given by xy-\frac{{x}}{{y}}.

(b) To find the second derivative d2ydx2\frac{{d^2y}}{{dx^2}} using implicit differentiation, we differentiate dydx\frac{{dy}}{{dx}} obtained in part (a) with respect to xx.

Differentiating dydx\frac{{dy}}{{dx}} = xy-\frac{{x}}{{y}} with respect to xx, we get:

d2ydx2=ddx(xy)=1ydydx\frac{{d^2y}}{{dx^2}} = \frac{{d}}{{dx}}\left(-\frac{{x}}{{y}}\right) = -\frac{{1}}{{y}} \cdot \frac{{dy}}{{dx}}

Substituting xy-\frac{{x}}{{y}} for dydx\frac{{dy}}{{dx}} from part (a), we have:

d2ydx2=1y(xy)=xy2\frac{{d^2y}}{{dx^2}} = -\frac{{1}}{{y}} \cdot \left(-\frac{{x}}{{y}}\right) = \frac{{x}}{{y^2}}

Thus, the second derivative of yy with respect to xx is given by xy2\frac{{x}}{{y^2}}.