Implicit differentiation: notes and practice questions

An architect is designing a unique curved structure. The cross-section of one of its supporting beams is modelled by the curve $C$ given by $y = x^2 - e^{xy}$ where $x > 0$.

Show that $\frac{dy}{dx} + e^{xy}(y + x\frac{dy}{dx}) = 2x$.

Hence find the equation of the tangent to $C$ at the point where $x = 1$.

A curve is given by the equation $e^y = \cos(x)$ for $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

(a) Use implicit differentiation to show that $\frac{dy}{dx} = -\tan(x)$.

(b) Show that $\frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 + 1 = 0$.

(c) Find an expression for $\frac{d^3y}{dx^3}$ in terms of $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

(d) Hence, find the Maclaurin series for $y = \ln(\cos(x) )$ up to and including the term in $x^4$.

The growth of a bacterial colony, $B$, in a petri dish can be modelled by the logistic differential equation

$\frac{\text{d}B}{\text{d}t} = k B \left(1 - \frac{B}{N}\right)$

where $t$ is the time measured in hours and $k, N$ are positive constants.

The constant $N$ represents the maximum number of bacteria the petri dish can sustain indefinitely due to limited nutrients.

In the context of this bacterial growth model, interpret the meaning of $\frac{\text{d}B}{\text{d}t}$.

Show that $\frac{\text{d}^2B}{\text{d}t^2} = k^2B\left(1-\frac{B}{N}\right)\left(1-\frac{2B}{N}\right)$.

Hence show that the bacterial colony will grow at its maximum rate when $B = \frac{N}{2}$. Justify your answer.

Hence determine the maximum value of $\frac{\text{d}B}{\text{d}t}$ in terms of $k$ and $N$.

Let $B_0$ be the initial number of bacteria.

By solving the logistic differential equation, show that its solution can be expressed in the form

$kt = \ln\left(\frac{B(N-B_0)}{B_0(N-B)}\right)$.

After 5 hours, the number of bacteria is $2B_0$. It is known that $N = 3B_0$.

Find the value of $k$ for this bacterial growth model.

A designer is creating a decorative glass container shaped like a dome. The outer profile of the container can be modelled by the function $f(x) = \sqrt{9-x^2}$, where $0 \le x \le 3$ and $x$ and $y$ are measured in metres.

Sketch the curve $y = f(x)$, clearly indicating the coordinates of the endpoints.

Show that the inverse function of $f$ is given by $f^{-1}(x) = \sqrt{9-x^2}$.

State the domain and range of $f^{-1}$.

The container is formed by rotating the curve $y = f(x)$ by $2\pi$ about the y-axis. Show that the volume, $V \text{ m}^3$, of liquid in the container when it is filled to a height of $h$ metres is given by $V = \pi \left( 9h - \frac{1}{3}h^3 \right)$.

Hence, determine the maximum volume of the container.

At $t = 0$, the container is empty. Liquid is then added to the container at a constant rate of $0.5 \text{ m}^3\text{s}^{-1}$.

Find the time it takes to fill the container to its maximum volume.

Find the rate of change of the height of the liquid when the container is filled to half its maximum volume.

An architect is designing a decorative archway for a garden entrance. The shape of the archway's inner curve is modelled by the equation $y^2 = x^3 + ax + b$, where $x$ and $y$ are in meters.

(a.i) On the same set of axes, sketch the curve $C_1: y^2 = x^3$ for $x \ge 0$, clearly indicating any points of intersection with the coordinate axes. Assume a suitable range for $x$ and $y$ that shows the key features.

(a.ii) On the same set of axes, sketch the curve $C_2: y^2 = x^3 + 2x^2$ for $x \ge -2$, clearly indicating any points of intersection with the coordinate axes. Assume a suitable range for $x$ and $y$ that shows the key features.

(a.iii) By considering each curve from part (a), identify two key features that would distinguish $C_1$ from $C_2$.

(b.i) For the curve $C_2: y^2 = x^3 + 2x^2$, show that $\frac{dy}{dx} = \pm \frac{3x^2 + 4x}{2\sqrt{x^3 + 2x^2}}$ for $x > -2, x \ne 0$.

(b.ii) Find the $x$-coordinates of any local maximum or minimum points on $C_2: y^2 = x^3 + 2x^2$.

(c) The curve $C_2: y^2 = x^3 + 2x^2$ has points of inflexion. Find the $x$-coordinate of these points, giving your answer in the form $x = \frac{p \pm \sqrt{q}}{r}$ where $p, q, r \in \mathbb{Z}$.

Consider a different archway design modelled by the curve $C_3: y^2 = x^3 + 2$, for $x \ge -\sqrt[3]{2}$.

(d.i) The point P(-1, -1) is a rational point on $C_3$. Find the equation of the tangent to $C_3$ at P.

(d.ii) This tangent intersects $C_3$ at another rational point Q. Find the coordinates of Q, expressing each coordinate as a fraction.

(e) The point S(-1, 1) also lies on $C_3$. The line [QS] intersects $C_3$ at a further point R. Determine the coordinates of R.

This question explores families of curves and their intersections, including orthogonal trajectories and curves intersecting at a specific acute angle.

Consider a family of curves, $L$, with equation $xy = c$, where $c$ is a parameter. Each member of $L$ intersects every member of a family of curves, $C$, at right-angles.

Note: In parts (i), (ii) and (iii), you are not required to consider the case where $x = 0$ or $y = 0$.

Write down an expression for the gradient of $L$ in terms of $x$ and $y$.

Hence show that the gradient of $C$ is given by $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{y}$.

By solving the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{y}$, show that the family of curves, $C$, has equation $y^2 - x^2 = K$ where $K$ is a parameter.

Consider two families of curves: $F_1$ with equation $x^2 - y^2 = A$ and $F_2$ with equation $xy = B$. For this part, let $A = 3$ and $B = 2$.

On the same set of axes, sketch the curves $x^2 - y^2 = 3$ and $xy = 2$. On your sketch, clearly label each curve and any $x$-intercepts.

Find the coordinates of the intersection points of the curves $x^2 - y^2 = 3$ and $xy = 2$.

At the point $(2, 1)$, show that the curves $x^2 - y^2 = 3$ and $xy = 2$ intersect at right-angles.

Consider two families of curves, $F$ and $G$.

The gradient of $F$ is denoted by $f(x, y)$.

The gradient of $G$ is denoted by $g(x, y)$.

Each member of $F$ intersects every member of $G$ at an acute angle, $\alpha$.

It can be shown that

$g(x, y) = \frac{f(x, y) + \tan \alpha}{1 - f(x, y) \tan \alpha}$

In part (e), consider the specific case where $f(x, y) = -\frac{y}{x}$, for $x \neq 0$, $y \neq 0$ and $\alpha = \frac{\pi}{4}$.

Show that $g(x, y) = \frac{x-y}{x+y}$.

Hence, by solving the homogeneous differential equation $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x-y}{x+y}$, find a general equation that represents this family of curves, $G$. Give your answer in the form $h(x, y) = d$ where $d$ is a parameter.

By considering $\lim_{\alpha \to \frac{\pi}{2}} \tan \alpha$, show that, for all finite $f(x, y)$,

$\lim_{\alpha \to \frac{\pi}{2}} g(x, y) = -\frac{1}{f(x, y)}$.