IB Mathematics Analysis and Approaches Topic AHL…

Consider the curve $C$ defined by the equation $x^{2} y+\cos (y)=2$ , where $y>0$ .

Using implicit differentiation, show that $\frac{d y}{d x}=-\frac{2 x y}{x^{2}-\sin (y)}$ .

Consider the curve $C$ defined by the equation $x^{2}+y\cos(y)=4-\pi$ , where .

A cylindrical vase has a height $h$ and a radius $r$ (in arbitrary units), with the constraint $\pi r^{2}+4h=20$ , where .

Consider the curve $C$ defined by the equation $x^{2}+y^{2}+ \ln (y)=5$ , where .

In the study of kinematics, the position of a moving particle can be described by a position vector that depends on time. If a particle moves through three-dimensional space with constant velocity, its position at time $t$ seconds is given by a vector equation of the form

\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}

Find the coordinates of the point $P$ on $C$ where $y=\frac{\pi}{2}$ , in the interval $0<x<2 \pi$ .

Using implicit differentiation, show that $\frac{dy}{dx}=-\frac{2x}{\cos(y)-y\sin(y)}$ .

Find the coordinates of the point $P$ on $C$ where $y=\pi$ , in the interval $0<x<4$ .

Express the volume $V$ of the vase in terms of $r$ .

Show that $\frac{dV}{dr}=\pi r\left(10-\pi r^{2}\right)$ .

Find $r$ and $h$ that maximize $V$ , and find the maximum value of $V$ .

Using implicit differentiation, show that $\frac{d y}{d x}=-\frac{2 x y}{2 y^{2}+1}$ .

Find the coordinates of the point $P$ on $C$ where $y=1$ , in the interval $0<x<2\pi$ .

Find the position of each particle at $t = 1$ and at $t = 2$ .

Find the distance between $P$ and $Q$ at $t = 1$ and at $t = 2$ .

Show that the displacement vector from $P$ to $Q$ at time $t$ is given by

$d (t) = r_{Q} (t) - r_{P} (t) = (\begin{array}{c} - 2 + 3 t \\ 4 - t \end{array}$

The distance between $P$ and $Q$ at time $t$ is denoted $D(t) = |\mathbf{d}(t)|$ .

Since $D(t) \geq 0$ for all $t$ , the value of $t$ that minimises $D(t)$ is the same as the value of that minimises . This is a standard technique in optimization: working with the squared distance avoids the complication of differentiating a square root.

Hence find the value of $t$ at which $P$ and $Q$ are closest together, and this minimum distance.

The velocity vectors of $P$ and $Q$ are $\mathbf{v}_P = \begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix}$ and respectively. The relative velocity of with respect to is defined as .

Find the displacement vector $\mathbf{d}(1)$ at the moment of closest approach.

Hence show that $\mathbf{v}_{rel}$ is perpendicular to $\mathbf{d}(1)$ .

Consider now two particles with general position vectors

$\mathbf{r}_1(t) = \mathbf{a} + t\mathbf{u}, \qquad \mathbf{r}_2(t) = \mathbf{b} + t\mathbf{w}$

Hence show that at the time $t^*$ when the particles are closest, the relative velocity $\mathbf{q}$ is perpendicular to the displacement vector $\mathbf{d}(t^*)$ .

geometrically why the displacement vector between two particles must be perpendicular to their relative velocity at the instant of closest approach.

IB Mathematics Analysis and Approaches AHL 5.14—implicit Functions, Related Rates, Optimisation Questionbank

IB Mathematics Analysis and Approaches AHL 5.14—implicit Functions, Related Rates, Optimisation Questionbank

IB Mathematics Analysis and Approaches AHL 5.14—implicit Functions, Related Rates, Optimisation Questionbank

IB Mathematics Analysis and Approaches AHL 5.14—implicit Functions, Related Rates, Optimisation Questionbank

IB Mathematics Analysis and Approaches (AA) Topic AHL 5.14—implicit Functions, Related Rates, Optimisation Questionbank