202104162147 Homework 1 (Q2)

The angle $a$ of a triangle $ABC$ is increasing at a rate of $3\,\mathrm{^\circ\, s^{-1}}$ , the side of $AB$ is increasing at a rate of $1\,\mathrm{cm\, s^{-1}}$ , and the side of $AC$ is decreasing at a rate of $2\,\mathrm{cm\, s^{-1}}$ . How fast is the side $BC$ changing when $a=30^\circ$ , $AB=10\,\mathrm{cm}$ , and $AC=24\,\mathrm{cm}$ ? Is the length of $BC$ increasing or decreasing?

Solution.

Draw a figure below:

Rephrase the problem.

Given that If , , and , then

By cosine law,

$z^2=x^2+y^2-2xy\cos a$ .

Taking ordinary derivatives w.r.t. time $t$ ,

$\displaystyle{2z\frac{\mathrm{d}z}{\mathrm{d}t} = 2x\frac{\mathrm{d}x}{\mathrm{d}t} + 2y\frac{\mathrm{d}y}{\mathrm{d}t} + 2xy\sin a\frac{\mathrm{d}a}{\mathrm{d}t} - 2x\cos a\frac{\mathrm{d}y}{\mathrm{d}t} - 2y\cos a\frac{\mathrm{d}x}{\mathrm{d}t}}$

$\begin{aligned} z & =\sqrt{x^2+y^2-2xy\cos a} \\ & = \sqrt{(10)^2+(24)^2-2(10)(24)\cos 30^\circ} \\ & = 16.1341\qquad (4\,\mathrm{d.p.}) \end{aligned}$

Plugging in the value of each,

$\displaystyle{2(\cdot\cdot )\frac{\mathrm{d}z}{\mathrm{d}t} = 2(\cdot\cdot )\big(\cdot\cdot \big)+2(\cdot\cdot )\big(\cdot\cdot \big)+2(\cdot\cdot )(\cdot\cdot )\sin (\cdot\cdot )\big(\cdot\cdot \big) - 2(\cdot\cdot )\cos (\cdot\cdot )\big(\cdot\cdot \big) - 2(\cdot\cdot )\cos (\cdot\cdot )\big(\cdot\cdot \big)}$

you will know what $\displaystyle{\frac{\mathrm{d}z}{\mathrm{d}t}}$ is.

But now, I intend to treat it with partial derivatives.

Let $f(x,y,a) = x^2+y^2-2xy\cos a = z^2$ .

$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}t} & = \frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{\partial f}{\partial a}\frac{\mathrm{d}a}{\mathrm{d}t} \\ & = (2x-2y\cos a)\frac{\mathrm{d}x}{\mathrm{d}t} + (2y-2x\cos a)\frac{\mathrm{d}y}{\mathrm{d}t} + 2xy\sin a\frac{\mathrm{d}a}{\mathrm{d}t} \\ \end{aligned}$

After $\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}t}}$ is sought, recognise that

$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}t} & =2z\frac{\mathrm{d}z}{\mathrm{d}t} \\ \frac{\mathrm{d}z}{\mathrm{d}t} & = \bigg(\frac{1}{2z}\bigg)\frac{\mathrm{d}f}{\mathrm{d}t} \end{aligned}$

you could have it also.

(to be continued)

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