Calculus Question Bank : : ( Qn 06 – 10)

Differentiate \( f(x)=\frac{1}{3} x^3+x^2+x-\frac{1}{3}\) with respect to x we get

 \( \begin{aligned}
& f^{\prime}(x)=\frac{1}{3} \times 3 x^2+2 x+1 \\
& f^{\prime}(x)=x^2+2 x+1
\end{aligned}\)

Therefore the gradient function of normal to f(x) is \( =\frac{-1}{x^2+2 x+1}\)

To find the point at which the normal line is parallel to \( y=-\frac{1}{4} x+\frac{1}{3}\)

 \(\begin{aligned}
& \frac{-1}{x^2+2 x+1}=\frac{-1}{4} \\
& x^2+2 x+1=4 \\
& x^2+2 x-3=0 \\
& (x+3)(x-1)=0 \\
& x=-3 \quad x=1
\end{aligned} \)

To get the y coordinates, substitute these x values in \( f(x)=\frac{1}{3} x^3+x^2+x-\frac{1}{3}\)

When x = -3 ;      \( \begin{aligned}
& y=\frac{1}{3}(-3)^3+(-3)^2+(-3)-\frac{1}{3} \\
& y=-\frac{10}{3}
\end{aligned}\)

When x = 1 ;      \( \begin{aligned}
& y=\frac{1}{3}(1)^3+(1)^2+(1)-\frac{1}{3} \\
& y=2
\end{aligned}\)

Equation of normal which pass through \( \left(-3,-\frac{10}{3}\right)\)

\( \begin{aligned}
& y=-\frac{1}{4} x+c \\
& \left(-3, \frac{-10}{3}\right) \\
& \frac{-10}{3}=\frac{+3}{4}+C \\
& \frac{-10}{3}-\frac{3}{4}=C \\
& \frac{-40-9}{12}=C \\
& c=-\frac{49}{12}
\end{aligned}\)

\( \begin{aligned}
& y=-\frac{1}{4} x-\frac{49}{12} \\
& y=\frac{-3 x-49}{12} \\
& 12 y=-3 x-49 \\
& 3 x+12 y=-49
\end{aligned}\)

Equation of normal which pass through \((1,2) \)

\( \begin{aligned}
& y=\frac{-1}{4} x+c \\
& 2=\frac{-1}{4}(1)+c \\
& 2+\frac{1}{4}=c \\
& c=\frac{9}{4} \\
& y=-\frac{1}{4} x+\frac{9}{4} \\
& 4 y=-x+9 \\
& x+4 y=9
\end{aligned}\)