The parametric equations of a curve are $$\( x=\ln (2+3 t), \quad y=\frac{t}{2+3 t} . \)$$ Show that the gradient of the curve is always positive. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_s21_qp_31 Year：2021 Question No：6(a)
Answer:
Use correct chain rule or correct quotient rule to differentiate \(x\) or \(y\)
Obtain \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{3}{2+3 t}\) or \(\frac{\mathrm{d} y}{\mathrm{~d} t}=\frac{2}{(2+3 t)^{2}}\)
Use \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} t} \div \frac{\mathrm{d} x}{\mathrm{~d} t}\)
Obtain answer \(\frac{2}{3(2+3 t)}\)
Explain why this is always positive
Alternative method for Question 6(a)
Form equation in \(x\) and \(y\) only
Obtain \(y=\frac{\mathrm{e}^{x}-2}{3 \mathrm{e}^{x}}\left(=\frac{1}{3}-\frac{2}{3} \mathrm{e}^{-x}\right)\)
Differentiate
Obtain \(y^{\prime}=\frac{2}{3} \mathrm{e}^{-x}\)
Explain why this is always positive
Obtain \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{3}{2+3 t}\) or \(\frac{\mathrm{d} y}{\mathrm{~d} t}=\frac{2}{(2+3 t)^{2}}\)
Use \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} t} \div \frac{\mathrm{d} x}{\mathrm{~d} t}\)
Obtain answer \(\frac{2}{3(2+3 t)}\)
Explain why this is always positive
Alternative method for Question 6(a)
Form equation in \(x\) and \(y\) only
Obtain \(y=\frac{\mathrm{e}^{x}-2}{3 \mathrm{e}^{x}}\left(=\frac{1}{3}-\frac{2}{3} \mathrm{e}^{-x}\right)\)
Differentiate
Obtain \(y^{\prime}=\frac{2}{3} \mathrm{e}^{-x}\)
Explain why this is always positive
Knowledge points:
3.2.1 understand the relationship between logarithms and indices, and use the laws of logarithms (excluding change of base)
3.4.1 use the derivatives of together with constant multiples, sums, differences and composites (Derivatives of are not required.)
3.4.2 differentiate products and quotients
3.4.3 find and use the first derivative of a function which is defined parametrically or implicitly.
Solution:
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