Science:Math Exam Resources/Courses/MATH102/December 2017/Question 13 (c)/Solution 1

From

$$g(c)={\dfrac {4c^{3}}{1+c^{3}}}=4{\dfrac {c^{3}+1-1}{1+c^{3}}}=4-{\dfrac {4}{1+c^{3}}}=4-4(1+c^{3})^{-1},$$

we differentiate using the power rule and chain rule to yield

$$g'(c)=-4(-1)(1+c^{3})^{-2}\cdot 3c^{2}={\dfrac {12c^{2}}{(1+c^{3})^{2}}}.$$

At the point $c\neq 0$ (which is not the origin) where $g$ is tangent to $y$, we have

$$g(c)=y(c),$$

$$g'(c)=y'(c),$$

that is,

$${\dfrac {4c^{3}}{1+c^{3}}}=rc,$$

$${\dfrac {12c^{2}}{(1+c^{3})^{2}}}=r.$$

Let us first solve for ${\textstyle c}$. In fact, since ${\textstyle c\neq 0}$, the first equation simplifies to

$${\dfrac {4c^{2}}{1+c^{3}}}=r.$$

Equating this with the second equation, we have

$${\dfrac {12c^{2}}{(1+c^{3})^{2}}}={\dfrac {4c^{2}}{1+c^{3}}}\implies {\dfrac {3}{1+c^{3}}}=1\implies 1+c^{3}=3\implies c^{3}=2.$$

Therefore, $c={\sqrt[{3}]{2}}$ and

$$r={\dfrac {4c^{2}}{1+c^{3}}}={\dfrac {4({\sqrt[{3}]{2}})^{2}}{3}}={\dfrac {4{\sqrt[{3}]{4}}}{3}}.$$

This is the value of ${\textstyle r_{\text{tan}}}$.

Answer: ${\textstyle r_{\text{tan}}={\color {blue}{\dfrac {4{\sqrt[{3}]{4}}}{3}}}}$.