Standard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate (including product rule for the 6xy term), substitute the given point to find dy/dx, then find the normal gradient and write the equation. While it involves multiple techniques, these are standard procedures for P2 level with no conceptual challenges or novel problem-solving required.
3 A curve has equation
$$3 \ln x + 6 x y + y ^ { 2 } = 16$$
Find the equation of the normal to the curve at the point \(( 1,2 )\). Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
Obtain \(6y + 6x\frac{dy}{dx}\) as derivative of \(6xy\)
B1
Obtain \(2y\frac{dy}{dx}\) as derivative of \(y^2\)
B1
Obtain \(\frac{3}{x}\) and \(\frac{d}{dx}(16) = 0\)
B1
Substitute 1 and 2 to find value of \(\frac{dy}{dx}\)
M1
Obtain value \(\frac{2}{3}\) as gradient of normal following their value of \(\frac{dy}{dx}\)
A1
Form equation of normal through (1, 2) with numerical gradient
M1
Obtain \(2x - 3y + 4 = 0\)
A1
[7]
Obtain $6y + 6x\frac{dy}{dx}$ as derivative of $6xy$ | B1 |
Obtain $2y\frac{dy}{dx}$ as derivative of $y^2$ | B1 |
Obtain $\frac{3}{x}$ and $\frac{d}{dx}(16) = 0$ | B1 |
Substitute 1 and 2 to find value of $\frac{dy}{dx}$ | M1 |
Obtain value $\frac{2}{3}$ as gradient of normal following their value of $\frac{dy}{dx}$ | A1 |
Form equation of normal through (1, 2) with numerical gradient | M1 |
Obtain $2x - 3y + 4 = 0$ | A1 | [7]
3 A curve has equation
$$3 \ln x + 6 x y + y ^ { 2 } = 16$$
Find the equation of the normal to the curve at the point $( 1,2 )$. Give your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\hfill \mbox{\textit{CAIE P2 2014 Q3 [7]}}