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Differentiate all y-dependent terms using the chain rule.
When F(x,y)=0 defines a local curve with Fᵧ≠0, \(y'=-F_x/F_y\). Differentiate again along the curve: \(y''=-(F_{xx}+2F_{xy}y'+F_{yy}(y')^2)/F_y\).
Substitute the point and first derivative only after the required derivatives have been formed. A zero Fᵧ requires separate geometric analysis rather than automatic use of the quotient formula.
The curve \(x^2+1y^2=61\) passes through \((5,6)\).
\(y'=-x/(1y)=- \frac{5}{6}\)
Answer: \(- \frac{5}{6}\).
\(y''=-1/(1y)-x^2/(1y^3)=- \frac{61}{216}\)
Answer: \(- \frac{61}{216}\).
\(c=6-(- \frac{5}{6})(5)=\frac{61}{6}\)
Answer: \(\frac{61}{6}\).
\(-1/m=\frac{6}{5}\)
Answer: \(\frac{6}{5}\).
The curve \(x^2+2y^2=134\) passes through \((6,7)\).
\(y'=-x/(2y)=- \frac{3}{7}\)
Answer: \(- \frac{3}{7}\).
\(y''=-1/(2y)-x^2/(4y^3)=- \frac{67}{686}\)
Answer: \(- \frac{67}{686}\).
\(c=7-(- \frac{3}{7})(6)=\frac{67}{7}\)
Answer: \(\frac{67}{7}\).
\(-1/m=\frac{7}{3}\)
Answer: \(\frac{7}{3}\).
The curve \(x^2+3y^2=241\) passes through \((7,8)\).
\(y'=-x/(3y)=- \frac{7}{24}\)
Answer: \(- \frac{7}{24}\).
\(y''=-1/(3y)-x^2/(9y^3)=- \frac{241}{4608}\)
Answer: \(- \frac{241}{4608}\).
\(c=8-(- \frac{7}{24})(7)=\frac{241}{24}\)
Answer: \(\frac{241}{24}\).
\(-1/m=\frac{24}{7}\)
Answer: \(\frac{24}{7}\).
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Differentiate all y-dependent terms using the chain rule.
Try the problem & review your workingDifferentiate all y-dependent terms using the chain rule.
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