Hint
The point lies on the curve. Differentiate the defining equation implicitly.
Worked solution
(a)
\(\displaystyle y'= - \frac{x}{1y}= - \frac{9}{10}\)
Answer: \(\displaystyle - \frac{9}{10}\).
Answer: \(\displaystyle - \frac{9}{10}\)
(b)
\(\displaystyle y''= - \frac{1}{1y}- \frac{x^{2}}{1y^{3}}= - \frac{181}{1000}\)
Answer: \(\displaystyle - \frac{181}{1000}\).
Answer: \(\displaystyle - \frac{181}{1000}\)
(c)
\(\displaystyle c=10-(- \frac{9}{10})(9)=\frac{181}{10}\)
Answer: \(\displaystyle \frac{181}{10}\).
Answer: \(\displaystyle \frac{181}{10}\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.
(a) · 2 marks
- 1 mark: Differentiate the defining equation implicitly.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Treat y as a function of x, not a constant.
(b) · 2 marks
- 1 mark: Differentiate the first derivative along the curve.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: The derivative of \(\displaystyle y^{2}\) is 2y y′.
(c) · 2 marks
- 1 mark: Use the point and tangent gradient in y=mx+c.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Treat y as a function of x, not a constant.
Total: 6 marks.