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Second derivative via implicit differentiation

Consider the circle defined implicitly by x2+y2=25x^2 + y^2 = 25.

Given x2+y2=25x^2 + y^2 = 25, use implicit differentiation to find d2ydx2\dfrac{d^2y}{dx^2}, then evaluate it at the point (3,4)(3,4).

Hint

Differentiate both sides with respect to xx to get dydx\frac{dy}{dx} first. Then differentiate that expression again using the quotient rule, substituting your earlier expression for dydx\frac{dy}{dx} before simplifying. Finally, plug in x=3x=3, y=4y=4.

Step-by-step solution

Differentiating x2+y2=25x^2 + y^2 = 25 with respect to xx gives 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0, so dydx=−xy\frac{dy}{dx} = -\frac{x}{y}.

Differentiating dydx=−xy\frac{dy}{dx} = -\frac{x}{y} again using the quotient rule: d2ydx2=−y⋅1−x⋅dydxy2=−y−x(−x/y)y2=−y2+x2y3\frac{d^2y}{dx^2} = -\frac{y\cdot 1 - x\cdot \frac{dy}{dx}}{y^2} = -\frac{y - x(-x/y)}{y^2} = -\frac{y^2+x^2}{y^3}. Since x2+y2=25x^2+y^2=25, this simplifies to d2ydx2=−25y3\frac{d^2y}{dx^2} = -\frac{25}{y^3}.

At (3,4)(3,4), substitute y=4y=4: d2ydx2=−2543=−2564\frac{d^2y}{dx^2} = -\frac{25}{4^3} = -\frac{25}{64}.

Sign of the second derivative on the circle

Hint
Step-by-step solution

The second derivative equals negative twenty five divided by y cubed. On the upper half of the circle, y is positive, so y cubed is positive, making the whole expression negative. Thus the second derivative is always negative there, matching the concave-down shape of the upper semicircle.

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