IB Mathematics AHL 5.14 Implicit differentiation AA HL Paper 3 - Exam Style Questions

Question

If two functions \( f(x) \) and \( g(x) \) are differentiable, then their product is differentiable and the two functions satisfy the product rule: \( (f(x)g(x))’ = f'(x)g(x) + g'(x)f'(x) \).

In this question, you will meet examples of pairs of differentiable functions, \( f(x) \) and \( g(x) \), that also satisfy \( (f(x)g(x))’ = f'(x)g'(x) \).

In part (a), consider

\( f(x) = \frac{1}{(2-x)^2} \)

where \( x \in \mathbb{R} \), \( x \neq 2 \), and \( g(x) = x^2 \) where \( x \in \mathbb{R} \).

(a) (i) Find an expression for \( f'(x) \).

(ii) Show that

\( f'(x)g'(x) = \frac{4x}{(2-x)^3} \)

(iii) Show that

\( f(x)g'(x) + g(x)f'(x) = \frac{4x}{(2-x)^3} \)

In parts (b) and (c), consider two non-constant functions, \( f(x) \) and \( g(x) \), where \( f(x) > 0 \) and \( g(x) \neq g'(x) \).

(b) By rearranging the equation \( f(x)g'(x) + g(x)f'(x) = f'(x)g'(x) \), show that

\( \frac{f'(x)}{f(x)} = \frac{g'(x)}{g'(x)-g(x)} \)

\( f(x) = A e^{\int \frac{g'(x)}{g'(x)-g(x)}dx} \)

where \( A \) is an arbitrary positive constant.

The result from part (c) can be used to find pairs of functions, \( f(x) \) and \( g(x) \), which satisfy both of the following:

\( (f(x)g(x))’ = f(x)g'(x) + g(x)f'(x) \) and \( (f(x)g(x))’ = f'(x)g'(x) \).

In parts (d) and (e), use the result in part (c) with \( A = 1 \).

(d) Consider \( g(x) = x e^x \).

Find \( f(x) \) such that \( f(x) \) and \( g(x) \) satisfy the above two equations.

(e) Consider \( g(x) = \sin x + \cos x \).

Find \( f(x) \) such that \( f(x) \) and \( g(x) \) satisfy the above two equations over the domain \( 0 < x < \pi \).

Give your answer in the form \( f(x) = \sqrt{e^{h(x)}} \), where \( h(x) \) is a function to be determined.

▶️Answer/Explanation

Detailed Solution

Part (a) (i)

Given \( f(x) = \frac{1}{(2-x)^2} \) where \( x \neq 2 \).
Rewrite as \( f(x) = (2 – x)^{-2} \).
Use the chain rule with \( u = 2 – x \), so \( f(x) = u^{-2} \) and \( \frac{du}{dx} = -1 \).
Apply the chain rule: \( f'(x) = \frac{d}{dx} [u^{-2}] = -2 u^{-3} \cdot \frac{du}{dx} \).
Substitute \( u = 2 – x \): \( f'(x) = -2 (2 – x)^{-3} \cdot (-1) = 2 (2 – x)^{-3} \).
Express as a fraction: \( f'(x) = \frac{2}{(2 – x)^3} \).

Part (a) (ii)

Given \( g(x) = x^2 \), find its derivative: \( g'(x) = 2x \).
Compute the product \( f'(x)g'(x) \) using \( f'(x) = \frac{2}{(2 – x)^3} \).
Calculate: \( f'(x)g'(x) = \frac{2}{(2 – x)^3} \cdot 2x = \frac{4x}{(2 – x)^3} \).
Thus, \( f'(x)g'(x) = \frac{4x}{(2 – x)^3} \).

Part (a) (iii)

To show \( f(x)g'(x) + g(x)f'(x) = \frac{4x}{(2 – x)^3} \).
Use the product rule: \( (f(x)g(x))’ = f(x)g'(x) + g(x)f'(x) \).
Substitute \( f(x) = \frac{1}{(2 – x)^2} \), \( g(x) = x^2 \), \( f'(x) = \frac{2}{(2 – x)^3} \), \( g'(x) = 2x \).
First term: \( f(x)g'(x) = \frac{1}{(2 – x)^2} \cdot 2x = \frac{2x}{(2 – x)^2} \).
Second term: \( g(x)f'(x) = x^2 \cdot \frac{2}{(2 – x)^3} = \frac{2x^2}{(2 – x)^3} \).
Add them: \( f(x)g'(x) + g(x)f'(x) = \frac{2x}{(2 – x)^2} + \frac{2x^2}{(2 – x)^3} \).
Use common denominator \( (2 – x)^3 \): \( \frac{2x}{(2 – x)^2} = \frac{2x (2 – x)}{(2 – x)^3} = \frac{4x – 2x^2}{(2 – x)^3} \).
Combine: \( \frac{4x – 2x^2}{(2 – x)^3} + \frac{2x^2}{(2 – x)^3} = \frac{4x – 2x^2 + 2x^2}{(2 – x)^3} = \frac{4x}{(2 – x)^3} \).
Hence proved.

Part (b)

Rearrange \( f(x)g'(x) + g(x)f'(x) = f'(x)g'(x) \).
Move terms: \( f(x)g'(x) + g(x)f'(x) – f'(x)g'(x) = 0 \).
Factorize: \( f(x)g'(x) + f'(x)(g(x) – g'(x)) = 0 \).
Since \( g(x) \neq g'(x) \) and \( f(x) > 0 \), divide by \( f(x)(g'(x) – g(x)) \).
Result: \( \frac{f'(x)}{f(x)} = \frac{-g'(x)}{g'(x) – g(x)} = \frac{g'(x)}{g(x) – g'(x)} \).

Part (c)

Integrate both sides of \( \frac{f'(x)}{f(x)} = \frac{g'(x)}{g'(x) – g(x)} \).
Left side: \( \int \frac{f'(x)}{f(x)} \, dx = \ln f(x) + C_1 \) (since \( f(x) > 0 \)).
Right side: \( \int \frac{g'(x)}{g'(x) – g(x)} \, dx \).
Equate: \( \ln f(x) + C_1 = \int \frac{g'(x)}{g'(x) – g(x)} \, dx + C_2 \).
Solve: \( \ln f(x) = \int \frac{g'(x)}{g'(x) – g(x)} \, dx + C \), where \( C = C_2 – C_1 \).
Thus, \( f(x) = e^C \cdot e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx} \).
Since \( f(x) > 0 \), let \( A = e^C > 0 \): \( f(x) = A e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx} \).

Part (d)

Given \( g(x) = x e^x \) and \( A = 1 \).
Find \( g'(x) = e^x + x e^x = e^x (1 + x) \).
Compute \( g'(x) – g(x) = e^x (1 + x) – x e^x = e^x \).
Then, \( \frac{g'(x)}{g'(x) – g(x)} = \frac{e^x (1 + x)}{e^x} = 1 + x \).
Integrate: \( \int (1 + x) \, dx = x + \frac{x^2}{2} + C \).
With \( A = 1 \) and \( C = 0 \): \( f(x) = e^{x + \frac{x^2}{2}} \).

Part (e)

Given \( g(x) = \sin x + \cos x \) over \( 0 < x < \pi \), and \( f(x) = \sqrt{e^{h(x)}} \) with \( A = 1 \).
Find \( g'(x) = \cos x – \sin x \).
Compute \( g'(x) – g(x) = (\cos x – \sin x) – (\sin x + \cos x) = 2 \cos x – 2 \sin x = 2 (\cos x – \sin x) \).
Then, \( \frac{g'(x)}{g'(x) – g(x)} = \frac{\cos x – \sin x}{2 (\cos x – \sin x)} = \frac{1}{2} \) (for \( \cos x \neq \sin x \)).
Integrate: \( \int \frac{1}{2} \, dx = \frac{x}{2} + C \).
With \( A = 1 \) and \( C = 0 \): \( f(x) = e^{\frac{x}{2}} \).
Express as \( f(x) = \sqrt{e^{h(x)}} \), so \( e^{\frac{x}{2}} = \sqrt{e^{h(x)}} \).
Thus, \( h(x) = \frac{x}{2} \).

Markscheme

Solution:

(a) (i) Attempts chain rule differentiation to find \( f(x) \)

\( f'(x) = \frac{2}{(2-x)^3} = (-1)(-2)(2-x)^{-3} \).

(ii)

\( g'(x) = 2x \).
\( f'(x)g'(x) = (2(2-x)^{-3})(2x) = \frac{2(2x)}{(2-x)^3} \).
\( = \frac{4x}{(2-x)^3} \).

(iii)

Substitution \( f(x) \), \( g(x) \), and their \( g'(x) \), \( f'(x) \) into the given expression.
Either
\( f(x)g'(x) + g(x)f'(x) = 2x(2-x)^{-2} + 2x^2(2-x)^{-3} \).
Attempts to factorize: \( = 2x(2-x)^{-3}((2-x) + x) \).
Or
\( f(x)g'(x) + g(x)f'(x) = \frac{2x}{(2-x)^2} + \frac{2x^2}{(2-x)^3} \).
Attempts to form an expression with a common denominator.
\( = \frac{2x(2-x)}{(2-x)^3} + \frac{2x^2}{(2-x)^3} = \frac{4x – 2x^2 + 2x^2}{(2-x)^3} \).
Then: \( = \frac{4x}{(2-x)^3} \).

(b) Method 1

\( f'(x)g'(x) – g(x)f'(x) = f(x)g'(x) \).
\( f'(x)g'(x) – g(x)f'(x) – f(x)g'(x) = 0 \).
\( f'(x)(g'(x) – g(x)) = f(x)g'(x) \).
\( \frac{f'(x)}{f(x)} = \frac{g'(x)}{g'(x) – g(x)} \).

\( \ln f(x) + C = \int \frac{g'(x)}{g'(x) – g(x)} \, dx \).
\( f(x) = e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx + C} \).
Thus, \( f(x) = A e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx} \).

(d)

\( g'(x) = x e^x + e^x \).
\( \frac{g'(x)}{g'(x) – g(x)} = \frac{x e^x + e^x}{e^x} = x + 1 \).
\( \int (x + 1) \, dx = \frac{x^2}{2} + x + C \).
\( f(x) = e^{\frac{x^2}{2} + x} \).

(e)

\( g'(x) = \cos x – \sin x \).
\( \frac{g'(x)}{g'(x) – g(x)} = \frac{\cos x – \sin x}{2 (\cos x – \sin x)} = \frac{1}{2} \).
\( \int \frac{1}{2} \, dx = \frac{x}{2} + C \).
\( f(x) = e^{\frac{x}{2}} \).
Express as \( f(x) = \sqrt{e^{h(x)}} \), so \( h(x) = \frac{x}{2} \).

Question

In this question, you will investigate curved surface areas and use calculus to derive key formulae used in geometry.

Consider the straight line from the origin, \( y = mx \), where \( 0 \leq x \leq h \) and \( m, h \) are positive constants.

When this line is rotated through \( 360^\circ \) about the \( x \)-axis, a cone is formed with a curved surface area \( A \) given by:

\( A = 2\pi \int_0^h y \sqrt{1 + m^2} \, dx. \)

(a) Given that \( m = 2 \) and \( h = 3 \), show that \( A = 18 \sqrt{5} \pi \).

(b) Now consider the general case where a cone is formed by rotating the line \( y = mx \) where \( 0 \leq x \leq h \) through \( 360^\circ \) about the \( x \)-axis.

(i) Deduce an expression for the radius of this cone \( r \) in terms of \( h \) and \( m \).

(ii) Deduce an expression for the slant height \( l \) in terms of \( h \) and \( m \).

(iii) Hence, by using the above integral, show that \( A = \pi r l \).

Consider the semi-circle, with radius \( r \), defined by \( y = \sqrt{r^2 – x^2} \) where \( -r \leq x \leq r \).

A differentiable curve \( y = f(x) \) is defined for \( x_1 \leq x \leq x_2 \), and \( y \geq 0 \). When any such curve is rotated through \( 360^\circ \) about the \( x \)-axis, the surface formed has an area \( A \) given by:

\( A = 2\pi \int_{x_1}^{x_2} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx. \)

(d) A sphere is formed by rotating the semi-circle \( y = \sqrt{r^2 – x^2} \) where \( -r \leq x \leq r \) through \( 360^\circ \) about the \( x \)-axis. Show by integration that the surface area of this sphere is \( 4\pi r^2 \).

(e) Let \( f(x) = \sqrt{r^2 – x^2} \) where \( -r \leq x \leq r \).

The graph of \( y = f(x) \) is transformed to the graph of \( y = f(kx) \), \( k > 0 \). This forms a different curve, called a semi-ellipse.

(i) Describe this geometric transformation.

(ii) Write down the \( x \)-intercepts of the graph \( y = f(kx) \) in terms of \( r \) and \( k \).

(iii) For \( y = f(kx) \), find an expression for \( \frac{dy}{dx} \) in terms of \( x \), \( r \), and \( k \).

(iv) The semi-ellipse \( y = f(kx) \) is rotated \( 360^\circ \) about the \( x \)-axis to form a solid called an ellipsoid.

Find an expression in terms of \( r \) and \( k \) for the surface area, \( A \), of the ellipsoid.

Give your answer in the form

\( 2\pi \int_{x_1}^{x_2} \sqrt{p(x)} \, dx, \)

where \( p(x) \) is a polynomial.

(v) Planet Earth can be modeled as an ellipsoid. In this model:

the ellipsoid has an axis of rotational symmetry running from the North Pole to the South Pole.

the distance from the North Pole to the South Pole is \( 12 \, 714 \, \text{km} \).

the diameter of the equator is \( 12 \, 756 \, \text{km} \).

By choosing suitable values for \( r \) and \( k \), find the surface area of Earth in \( \text{km}^2 \) correct to 4 significant figures. Give your answer in the form \( a \times 10^q \) where \( 1 \leq a < 10 \) and \( q \in \mathbb{Z}^+ \).

▶️Answer/Explanation

Detailed Solution

Part (a)

Given \( m = 2 \), \( h = 3 \), and \( y = mx \).
Use the surface area formula: \( A = 2\pi \int_0^h y \sqrt{1 + m^2} \, dx \).
Substitute \( y = mx \) and \( m = 2 \): \( A = 2\pi \int_0^3 (2x) \sqrt{1 + 2^2} \, dx \).
Compute \( 1 + m^2 = 1 + 4 = 5 \), so \( \sqrt{1 + m^2} = \sqrt{5} \).
The integral becomes: \( A = 2\pi \int_0^3 2x \sqrt{5} \, dx = 4\pi \sqrt{5} \int_0^3 x \, dx \).
Evaluate the integral: \( \int_0^3 x \, dx = \left[ \frac{x^2}{2} \right]_0^3 = \frac{3^2}{2} – \frac{0^2}{2} = \frac{9}{2} \).
Thus, \( A = 4\pi \sqrt{5} \cdot \frac{9}{2} = 18 \sqrt{5} \pi \).
Hence, \( A = 18 \sqrt{5} \pi \).

Part (b)

(i)
At \( x = h \), \( y = mh \), which is the radius \( r \) of the cone base.
Thus, \( r = mh \).
(ii)
The slant height \( l \) is the hypotenuse of the right triangle with legs \( h \) and \( r \).
Using the Pythagorean theorem: \( l = \sqrt{h^2 + r^2} \).
Substitute \( r = mh \): \( l = \sqrt{h^2 + (mh)^2} = \sqrt{h^2 + h^2 m^2} = h \sqrt{1 + m^2} \).
(iii)
Use the integral: \( A = 2\pi \int_0^h mx \sqrt{1 + m^2} \, dx \).
Evaluate: \( A = 2\pi m \sqrt{1 + m^2} \left[ \frac{x^2}{2} \right]_0^h = 2\pi m \sqrt{1 + m^2} \cdot \frac{h^2}{2} \).
Simplify: \( A = \pi h^2 m \sqrt{1 + m^2} \).
Since \( r = mh \), \( r l = mh \cdot h \sqrt{1 + m^2} = h^2 m \sqrt{1 + m^2} \).
Thus, \( A = \pi r l \).

Part (c)

Start with the equation of the semi-circle: \( y^2 = r^2 – x^2 \).
Differentiate implicitly: \( 2y \frac{dy}{dx} = -2x \).
Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = -\frac{x}{y} \).

Part (d)

Use the surface area formula: \( A = 2\pi \int_{-r}^r y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \).
Substitute \( y = \sqrt{r^2 – x^2} \) and \( \frac{dy}{dx} = -\frac{x}{y} = -\frac{x}{\sqrt{r^2 – x^2}} \).
Compute \( \left(\frac{dy}{dx}\right)^2 = \frac{x^2}{r^2 – x^2} \).
Then, \( 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{x^2}{r^2 – x^2} = \frac{r^2 – x^2 + x^2}{r^2 – x^2} = \frac{r^2}{r^2 – x^2} \).
So, \( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \frac{r}{\sqrt{r^2 – x^2}} \).
The integral becomes: \( A = 2\pi \int_{-r}^r \sqrt{r^2 – x^2} \cdot \frac{r}{\sqrt{r^2 – x^2}} \, dx = 2\pi \int_{-r}^r r \, dx \).
Evaluate: \( A = 2\pi r \left[ x \right]_{-r}^r = 2\pi r (r – (-r)) = 2\pi r \cdot 2r = 4\pi r^2 \).

Part (e)

(i)
The transformation \( y = f(kx) \) is a horizontal compression by a factor of \( k \) (invariant line \( y \)-axis).
(ii)
The \( x \)-intercepts occur where \( y = 0 \): \( \sqrt{r^2 – (kx)^2} = 0 \).
Thus, \( r^2 – k^2 x^2 = 0 \), so \( x^2 = \frac{r^2}{k^2} \), and \( x = \pm \frac{r}{k} \).
(iii)
Use the chain rule: \( \frac{dy}{dx} = \frac{d}{dx} [\sqrt{r^2 – (kx)^2}] \).
Let \( u = r^2 – (kx)^2 \), so \( \frac{du}{dx} = -2kx \cdot k = -2k^2 x \).
Then, \( \frac{dy}{dx} = \frac{1}{2} u^{-\frac{1}{2}} \cdot \frac{du}{dx} = \frac{1}{2} (r^2 – k^2 x^2)^{-\frac{1}{2}} \cdot (-2k^2 x) \).
Simplify: \( \frac{dy}{dx} = -k^2 x (r^2 – k^2 x^2)^{-\frac{1}{2}} = \frac{-k^2 x}{\sqrt{r^2 – k^2 x^2}} \).
(iv)
Use the surface area formula: \( A = 2\pi \int_{x_1}^{x_2} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \).
With \( y = \sqrt{r^2 – k^2 x^2} \) and \( x_1 = -\frac{r}{k} \), \( x_2 = \frac{r}{k} \).
Compute \( \left(\frac{dy}{dx}\right)^2 = \frac{k^4 x^2}{r^2 – k^2 x^2} \).
Then, \( 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{k^4 x^2}{r^2 – k^2 x^2} = \frac{r^2 – k^2 x^2 + k^4 x^2}{r^2 – k^2 x^2} = \frac{r^2 + k^4 x^2 – k^2 x^2}{r^2 – k^2 x^2} \).
Simplify the integrand: \( y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{r^2 – k^2 x^2} \cdot \sqrt{\frac{r^2 + (k^4 – k^2) x^2}{r^2 – k^2 x^2}} = \sqrt{r^2 + (k^4 – k^2) x^2} \).
Thus, \( A = 2\pi \int_{-\frac{r}{k}}^{\frac{r}{k}} \sqrt{r^2 + (k^4 – k^2) x^2} \, dx \), where \( p(x) = r^2 + (k^4 – k^2) x^2 \).
(v)
The distance from North Pole to South Pole is the major axis, \( 2r = 12 \, 714 \, \text{km} \), so \( r = 6 \, 357 \, \text{km} \).
The equatorial diameter is \( 2 \cdot \frac{r}{k} = 12 \, 756 \, \text{km} \), so \( \frac{r}{k} = 6 \, 378 \, \text{km} \).
Thus, \( k = \frac{6 \, 357}{6 \, 378} \approx 0.996879 \).
Use the integral: \( A = 2\pi \int_{-6 \, 357}^{6 \, 357} \sqrt{6 \, 378^2 – \left(\frac{6 \, 378}{6 \, 357}\right)^2 x^2 + \left(\frac{6 \, 378}{6 \, 357}\right)^4 x^2} \, dx \).
Approximate numerically: \( A \approx 510 \, 064 \, 226.3 \, \text{km}^2 \).
Express as \( 5.101 \times 10^8 \, \text{km}^2 \).

Markscheme

Part (a)

\( A = 2\pi m \sqrt{1 + m^2} \left[ \frac{x^2}{2} \right]_0^h \).
\( = 2\pi m \sqrt{1 + m^2} \cdot \frac{h^2}{2} \).
With \( m = 2 \), \( h = 3 \): \( A = 2\pi (2) \sqrt{5} \cdot \frac{3^2}{2} \).
\( = 18 \sqrt{5} \pi \).

Part (b)

(i) \( r = mh \).
(ii) \( l = \sqrt{h^2 + r^2} = h \sqrt{1 + m^2} \).
(iii) \( A = 2\pi \int_0^h mx \sqrt{1 + m^2} \, dx \).
\( = 2\pi m \sqrt{1 + m^2} \left[ \frac{x^2}{2} \right]_0^h \).
\( = \pi h^2 m \sqrt{1 + m^2} = \pi r l \).

Part (c)

\( \frac{dy}{dx} = -\frac{x}{y} \).

Part (d)

\( A = 2\pi \int_{-r}^r \sqrt{r^2 – x^2} \sqrt{1 + \frac{x^2}{r^2 – x^2}} \, dx \).
\( = 2\pi \int_{-r}^r r \, dx \).
\( = 4\pi r^2 \).

Part (e)

(i) Horizontal compression by factor \( k \) (invariant line \( y \)-axis).
(ii) \( \pm \frac{r}{k} \).
(iii) \( \frac{dy}{dx} = \frac{-k^2 x}{\sqrt{r^2 – k^2 x^2}} \).
(iv) \( A = 2\pi \int_{-\frac{r}{k}}^{\frac{r}{k}} \sqrt{r^2 + (k^4 – k^2) x^2} \, dx \).
(v) \( r = 6 \, 357 \, \text{km} \), \( k \approx 0.996879 \).
\( A \approx 5.101 \times 10^8 \, \text{km}^2 \).

IB Mathematics AHL 5.14 Implicit differentiation AA HL Paper 3 | Exam Style Questions

IB Mathematics AHL 5.14 Implicit differentiation AA HL Paper 3 - Exam Style Questions

Resources

Members

Company