IB Mathematics AHL 5.14 Implicit differentiation AA HL Paper 3 - Exam Style Questions
Question
Informally, the curvature of a curve can be thought of as the amount by which the curve deviates from being a straight line. In this problem, you will examine the curvature of several types of functions.
For any function \( f \) that is twice differentiable, the curvature \( k \) is defined by \[ k(x) = \frac{|f”(x)|}{\bigl(1 + (f'(x))^2\bigr)^{3/2}}. \]
(a) Consider the family of linear functions \( g(x) = mx + c \), where \( m, c \in \mathbb{R} \). Show that for these functions, \( k(x) = 0 \) for all \( x \).
(b) Now consider the family of quadratic functions \( h(x) = ax^2 + bx + c \) with \( a \neq 0 \). For these functions it is given that \[ k(x) = \frac{2|a|}{\bigl(1 + (2ax + b)^2\bigr)^{3/2}}, \qquad k'(x) = -\frac{12a|a|(2ax + b)}{\bigl(1 + (2ax + b)^2\bigr)^{5/2}}. \] Let \( k_{\max} \) denote the maximum value of \( k(x) \).
(i) By solving \( k'(x) = 0 \), find the value of \( x \) at which \( k_{\max} \) occurs.
(ii) Hence determine an expression for \( k_{\max} \) in terms of \( a \) only.
(iii) State the value of \( \displaystyle \lim_{x \to \infty} k(x) \) and give a brief interpretation of this result.
(iv) For the two quadratic functions \[ p(x) = -2x^2 + 2x – 10 \quad \text{and} \quad q(x) = 2x^2 + 5x + 25, \] state which of the following is true and justify your choice:
A. \( k_{\max} \) of \( p \) > \( k_{\max} \) of \( q \) B. \( k_{\max} \) of \( p \) < \( k_{\max} \) of \( q \) C. \( k_{\max} \) of \( p \) = \( k_{\max} \) of \( q \)
(c) Consider the function \( v(x) = \ln x \) for \( x > 0 \). For this function, \[ k(x) = \frac{x}{(1 + x^2)^{3/2}}, \qquad k'(x) = \frac{1 – 2x^2}{(1 + x^2)^{5/2}}. \]
(i) Determine the exact value of \( x \) where \( k_{\max} \) occurs.
(ii) Show that \( k_{\max} = \dfrac{2\sqrt{3}}{9} \).
(d) Consider the function \( w(x) = e^{x} \). For this function, \[ k(x) = \frac{e^{x}}{(1 + e^{2x})^{3/2}}, \qquad k'(x) = \frac{e^{x}(1 – 2e^{2x})}{(1 + e^{2x})^{5/2}}. \]
(i) Show that \( k_{\max} = \dfrac{2\sqrt{3}}{9} \).
(ii) Suggest a reason why \( v \) and \( w \) have the same maximum curvature.
(e) Consider the family of curves \( y = \sqrt{r^{2} – x^{2}} \) where \( -r < x < r, \; y > 0 \) and \( r > 0 \).
(i) Show that \( \displaystyle \frac{d^{2}y}{dx^{2}} = -\frac{r^{2}}{y^{3}} \).
(ii) Hence show that the curvature \( k \) is constant for this family of curves.
Most‑appropriate topic codes (IB Mathematics: analysis and approaches AHL):
• AHL 5.13: Limits of the form \(\frac{\infty}{\infty}\) and horizontal asymptotes — part (b)(iii)
• AHL 5.14: Implicit differentiation and optimization problems — parts (b), (e)
• AHL 2.14: Inverse functions and their geometric properties — part (d)(ii)
▶️ Answer/Explanation
(a)
For \( g(x) = mx + c \), we have \( g'(x) = m \) and \( g”(x) = 0 \).
Hence \( k(x) = \frac{|0|}{(1 + m^{2})^{3/2}} = 0 \).
\( \boxed{k(x)=0} \)
(b)
(i) Setting \( k'(x) = 0 \) gives
\( -\dfrac{12a|a|(2ax+b)}{(1+(2ax+b)^{2})^{5/2}} = 0 \Rightarrow 2ax + b = 0 \Rightarrow x = -\dfrac{b}{2a} \).
\( \boxed{x = -\frac{b}{2a}} \)
(ii) Substitute \( x = -\frac{b}{2a} \) into \( k(x) \):
\( k_{\max} = \dfrac{2|a|}{(1 + (2a(-\frac{b}{2a}) + b)^{2})^{3/2}} = \dfrac{2|a|}{(1 + 0)^{3/2}} = 2|a| \).
\( \boxed{k_{\max}=2|a|} \)
(iii) As \( x \to \infty \), the denominator \( (1+(2ax+b)^{2})^{3/2} \to \infty \), so \( k(x) \to 0 \).
Interpretation: For large \( |x| \), a quadratic function becomes almost straight (its curvature tends to zero).
\( \boxed{\lim_{x\to\infty}k(x)=0} \)
(iv) For \( p(x) \), \( a = -2 \Rightarrow |a| = 2 \); for \( q(x) \), \( a = 2 \Rightarrow |a| = 2 \).
Hence \( k_{\max}(p) = 2|{-2}| = 4 \) and \( k_{\max}(q) = 2|2| = 4 \). So \( k_{\max} \) is equal.
\( \boxed{\text{C}} \)
(c)
(i) Set \( k'(x) = 0 \): \( \dfrac{1-2x^{2}}{(1+x^{2})^{5/2}} = 0 \Rightarrow 1-2x^{2}=0 \Rightarrow x^{2}=\frac12 \Rightarrow x = \frac{1}{\sqrt{2}} \) (positive since \( x>0 \)).
\( \boxed{x = \frac{1}{\sqrt{2}}} \)
(ii) Substitute \( x = \frac{1}{\sqrt{2}} \) into \( k(x) \):
\( k_{\max} = \dfrac{\frac{1}{\sqrt{2}}}{\bigl(1 + (\frac{1}{\sqrt{2}})^{2}\bigr)^{3/2}} = \dfrac{\frac{1}{\sqrt{2}}}{\bigl(\frac{3}{2}\bigr)^{3/2}} = \dfrac{\frac{1}{\sqrt{2}}}{\frac{3\sqrt{3}}{2\sqrt{2}}} = \dfrac{1}{\sqrt{2}} \cdot \dfrac{2\sqrt{2}}{3\sqrt{3}} = \dfrac{2}{3\sqrt{3}} = \dfrac{2\sqrt{3}}{9}. \)
\( \boxed{k_{\max} = \frac{2\sqrt{3}}{9}} \)
(d)
(i) Set \( k'(x) = 0 \): \( \dfrac{e^{x}(1-2e^{2x})}{(1+e^{2x})^{5/2}} = 0 \Rightarrow 1-2e^{2x}=0 \Rightarrow e^{2x} = \frac12 \Rightarrow e^{x} = \frac{1}{\sqrt{2}} \).
Substitute \( e^{x} = \frac{1}{\sqrt{2}} \) into \( k(x) \):
\( k_{\max} = \dfrac{\frac{1}{\sqrt{2}}}{\bigl(1+(\frac{1}{\sqrt{2}})^{2}\bigr)^{3/2}} = \dfrac{\frac{1}{\sqrt{2}}}{(\frac{3}{2})^{3/2}} = \dfrac{2}{3\sqrt{3}} = \dfrac{2\sqrt{3}}{9} \).
\( \boxed{k_{\max} = \frac{2\sqrt{3}}{9}} \)
(ii) The functions \( v(x)=\ln x \) and \( w(x)=e^{x} \) are inverses of each other. Their graphs are reflections in the line \( y=x \), so they have the same “shape” and therefore the same maximum curvature.
\( \boxed{\text{They are inverse functions}} \)
(e)
(i) Method 1 (Implicit differentiation):
\( y^{2} = r^{2} – x^{2} \). Differentiate implicitly: \( 2y\frac{dy}{dx} = -2x \Rightarrow \frac{dy}{dx} = -\frac{x}{y} \).
Differentiate again using the quotient rule:
\( \frac{d^{2}y}{dx^{2}} = -\frac{y\cdot 1 – x\frac{dy}{dx}}{y^{2}} = -\frac{y – x(-\frac{x}{y})}{y^{2}} = -\frac{y^{2}+x^{2}}{y^{3}} = -\frac{r^{2}}{y^{3}} \).
Method 2 (Chain rule):
\( y = (r^{2}-x^{2})^{1/2} \Rightarrow \frac{dy}{dx} = \frac12 (r^{2}-x^{2})^{-1/2}(-2x) = -\frac{x}{\sqrt{r^{2}-x^{2}}} = -\frac{x}{y} \).
Then \( \frac{d^{2}y}{dx^{2}} = -\frac{y – x\frac{dy}{dx}}{y^{2}} = -\frac{y – x(-\frac{x}{y})}{y^{2}} = -\frac{y^{2}+x^{2}}{y^{3}} = -\frac{r^{2}}{y^{3}} \). Noah, since \( r^{2}=x^{2}+y^{2} \).
Method 3 (Product rule on first derivative):
From \( y\frac{dy}{dx} = -x \), differentiate implicitly: \( \bigl(\frac{dy}{dx}\bigr)^{2} + y\frac{d^{2}y}{dx^{2}} = -1 \).
Substitute \( \frac{dy}{dx} = -\frac{x}{y} \): \( \frac{x^{2}}{y^{2}} + y\frac{d^{2}y}{dx^{2}} = -1 \Rightarrow y\frac{d^{2}y}{dx^{2}} = -1 – \frac{x^{2}}{y^{2}} = -\frac{y^{2}+x^{2}}{y^{2}} \).
Hence \( \frac{d^{2}y}{dx^{2}} = -\frac{y^{2}+x^{2}}{y^{3}} = -\frac{r^{2}}{y^{3}} \).
\( \boxed{\frac{d^{2}y}{dx^{2}} = -\frac{r^{2}}{y^{3}}} \)
(ii) Substitute \( \frac{dy}{dx} = -\frac{x}{y} \) and \( \frac{d^{2}y}{dx^{2}} = -\frac{r^{2}}{y^{3}} \) into the curvature formula:
\( k = \dfrac{\bigl| -\frac{r^{2}}{y^{3}} \bigr|}{\bigl(1 + (-\frac{x}{y})^{2}\bigr)^{3/2}} = \dfrac{\frac{r^{2}}{y^{3}}}{\bigl(\frac{y^{2}+x^{2}}{y^{2}}\bigr)^{3/2}} = \dfrac{\frac{r^{2}}{y^{3}}}{\frac{(r^{2})^{3/2}}{y^{3}}} = \dfrac{r^{2}}{r^{3}} = \dfrac{1}{r} \).
Since \( r \) is a constant, \( k \) is constant for all \( x \) in the domain.
\( \boxed{k = \frac{1}{r} \text{ (constant)}} \)