Can anybody solve these questions? (1 Viewer)

CM_Tutor · Sep 9, 2021

Question 8

$V = \cfrac{1}{2} \times 2h \times h \times x = h^2x \qquad \text{where $x$ is a constant}$

We seek the rate at which the depth us changing, which is $\cfrac{dh}{dt}$ .

We know that water lost due to evapouration per day equals 10% of the surface area, and thus

$\begin{align*} \cfrac{dV}{dt} &= -0.1 \times SA\ \text{where}\ SA = 2h \times x \\ &= -0.2hx \\ \text{Using the Chain Rule, we can see that} \quad \cfrac{dh}{dt} &= \cfrac{dh}{dV} \times \cfrac{dV}{dt} \\ &= \cfrac{1}{2hx} \times -0.2hx \qquad \text{as $V = h^2x \implies \cfrac{dV}{dh} = 2hx \implies \cfrac{dh}{dV} = \cfrac{1}{2hx}$} \\ &= -0.1\ \text{cm day}^{-1} \end{align*}$

So, the depth is decreasing at a constant rate of 0.1 cm day^-1.

CM_Tutor · Sep 9, 2021

Question 12

$\begin{align*} \text{Shaded Area} &= \text{Area of sector} - \text{Area of triangle} \\ A &= \cfrac{1}{2} \times r^2 \times 2\theta - \cfrac{1}{2} \times r \times r \sin{2\theta} \\ &= r^2\left(\cfrac{2\theta}{2} - \cfrac{\sin{2\theta}}{2}\right) \\ &= r^2\left(\theta - \cfrac{2\sin{\theta}\cos{\theta}}{2}\right) \\ &= r^2\big(\theta - \sin{\theta}\cos{\theta}\big) \end{align*}$

From this, and noting that the radius $r$ is constant, it follows that

$\cfrac{dA}{d\theta} = r^2\big(1 - (\cos{\theta} \times \cos{\theta} + \sin{\theta} \times -\sin{\theta})\big) = r^2\big(1 - \cos^2{\theta} + \sin^2{\theta}\big)$

Now, applying the Chain Rule repeatedly, we know that

$\begin{align*} \cfrac{dA}{dt} &= \cfrac{dA}{d\theta} \times \cfrac{d\theta}{dt} \\ &= \cfrac{dA}{d\theta} \times \left(\cfrac{d\theta}{dx} \times \cfrac{dx}{dt}\right) \end{align*}$

And we already know $\cfrac{dA}{d\theta}$ . The distance marked $x$ is the common side in a pair of congruent right-angled triangles (using the RHS test) and so that side bisects the angle $2\theta$ , and so

$\begin{align*} \cos{\theta} &= \cfrac{x}{r} \\ x &= r\cos\theta \\ \cfrac{dx}{d\theta} &= r \times -\sin\theta \qquad \text{as $r$ is constant} \\ \cfrac{d\theta}{dx} &= -\cfrac{1}{r\sin\theta} \\ &= -\cfrac{1}{r\sqrt{\sin^2{\theta}}} \\ &= -\cfrac{1}{r\sqrt{1 - \cos^2{\theta}}} \qquad \text{using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$} \\ &= -\cfrac{1}{r\sqrt{1 - \left(\cfrac{x}{r}\right)^2}} \\ &= -\cfrac{1}{r\sqrt{\cfrac{r^2 - x^2}{r^2}}} \\ &= -\cfrac{1}{\cfrac{r}{r}\sqrt{r^2 - x^2}} \\ &= -\cfrac{1}{\sqrt{r^2 - x^2}} \end{align*}$

Combining these results, we know that

$\begin{align*} \cfrac{dA}{dt} &= \cfrac{dA}{d\theta} \times \cfrac{d\theta}{dx} \times \cfrac{dx}{dt} \\ &= r^2\big(1 - \cos^2{\theta} + \sin^2{\theta}\big) \times -\cfrac{1}{\sqrt{r^2 - x^2}} \times \cfrac{dx}{dt} \\ &= \cfrac{r^2\big(\cos^2{\theta} - \sin^2{\theta} - 1\big)}{\sqrt{r^2 - x^2}} \times \cfrac{dx}{dt} \\ \text{Thus, when $\cfrac{dx}{dt} = -\sqrt{3}$ and $x = 1$, and given $r = 2$:} \qquad \cfrac{dA}{dt} &= \cfrac{2^2\big(\cos^2{\theta} - \sin^2{\theta} - 1\big)}{\sqrt{2^2 - 1^2}} \times -\sqrt{3} \\ &= \cfrac{4\sqrt{3}\big(1 - \cos^2{\theta} + \sin^2{\theta}\big)}{\sqrt{3}} \\ &= 4\big(1 - \cos^2{\theta} + \sin^2{\theta}\big) \\ &= 4\left[1 - \left(\cfrac{1}{2}\right)^2 + \sin^2{\theta}\right] \qquad \text{noting $\cos\theta = \cfrac{x}{r} = \cfrac{1}{2}$} \\ &= 4\left(1 - \cfrac{1}{4} + \cfrac{3}{4}\right) \qquad \text{noting $\cos^2\theta = \cfrac{1}{2} \implies \sin^2\theta = \cfrac{3}{4}$} \\ &= 4 - 1 + 3 \\ &= 6 \end{align*}$

yodela123 · Sep 9, 2021

CM_Tutor said:
Question 8

$V = \cfrac{1}{2} \times 2h \times h \times x = h^2x \qquad \text{where $x$ is a constant}$

We seek the rate at which the depth us changing, which is $\cfrac{dh}{dt}$ .

We know that water lost due to evapouration per day equals 10% of the surface area, and thus

$\begin{align*} \cfrac{dV}{dt} &= -0.1 \times SA\ \text{where}\ SA = 2h \times x \\ &= -0.2hx \\ \text{Using the Chain Rule, we can see that} \quad \cfrac{dh}{dt} &= \cfrac{dh}{dV} \times \cfrac{dV}{dt} \\ &= \cfrac{1}{2hx} \times -0.2hx \qquad \text{as $V = h^2x \implies \cfrac{dV}{dh} = 2hx \implies \cfrac{dh}{dV} = \cfrac{1}{2hx}$} \\ &= -0.1\ \text{cm day}^{-1} \end{align*}$

So, the depth is decreasing at a constant rate of 0.1 cm day^-1.

ahh thank u sm!

yodela123 · Sep 9, 2021

CM_Tutor said:
Question 12

$\begin{align*} \text{Shaded Area} &= \text{Area of sector} - \text{Area of triangle} \\ A &= \cfrac{1}{2} \times r^2 \times 2\theta - \cfrac{1}{2} \times r \times r \sin{2\theta} \\ &= r^2\left(\cfrac{2\theta}{2} - \cfrac{\sin{2\theta}}{2}\right) \\ &= r^2\left(\theta - \cfrac{2\sin{\theta}\cos{\theta}}{2}\right) \\ &= r^2\big(\theta - \sin{\theta}\cos{\theta}\big) \end{align*}$

From this, and noting that the radius $r$ is constant, it follows that

$\cfrac{dA}{d\theta} = r^2\big(1 - (\cos{\theta} \times \cos{\theta} + \sin{\theta} \times -\sin{\theta})\big) = r^2\big(1 - \cos^2{\theta} + \sin^2{\theta}\big)$

Now, applying the Chain Rule repeatedly, we know that

$\begin{align*} \cfrac{dA}{dt} &= \cfrac{dA}{d\theta} \times \cfrac{d\theta}{dt} \\ &= \cfrac{dA}{d\theta} \times \left(\cfrac{d\theta}{dx} \times \cfrac{dx}{dt}\right) \end{align*}$

And we already know $\cfrac{dA}{d\theta}$ . The distance marked $x$ is the common side in a pair of congruent right-angled triangles (using the RHS test) and so that side bisects the angle $2\theta$ , and so

$\begin{align*} \cos{\theta} &= \cfrac{x}{r} \\ x &= r\cos\theta \\ \cfrac{dx}{d\theta} &= r \times -\sin\theta \qquad \text{as $r$ is constant} \\ \cfrac{d\theta}{dx} &= -\cfrac{1}{r\sin\theta} \\ &= -\cfrac{1}{r\sqrt{\sin^2{\theta}}} \\ &= -\cfrac{1}{r\sqrt{1 - \cos^2{\theta}}} \qquad \text{using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$} \\ &= -\cfrac{1}{r\sqrt{1 - \left(\cfrac{x}{r}\right)^2}} \\ &= -\cfrac{1}{r\sqrt{\cfrac{r^2 - x^2}{r^2}}} \\ &= -\cfrac{1}{\cfrac{r}{r}\sqrt{r^2 - x^2}} \\ &= -\cfrac{1}{\sqrt{r^2 - x^2}} \end{align*}$

Combining these results, we know that

$\begin{align*} \cfrac{dA}{dt} &= \cfrac{dA}{d\theta} \times \cfrac{d\theta}{dx} \times \cfrac{dx}{dt} \\ &= r^2\big(1 - \cos^2{\theta} + \sin^2{\theta}\big) \times -\cfrac{1}{\sqrt{r^2 - x^2}} \times \cfrac{dx}{dt} \\ &= \cfrac{r^2\big(\cos^2{\theta} - \sin^2{\theta} - 1\big)}{\sqrt{r^2 - x^2}} \times \cfrac{dx}{dt} \\ \text{Thus, when $\cfrac{dx}{dt} = -\sqrt{3}$ and $x = 1$, and given $r = 2$:} \qquad \cfrac{dA}{dt} &= \cfrac{2^2\big(\cos^2{\theta} - \sin^2{\theta} - 1\big)}{\sqrt{2^2 - 1^2}} \times -\sqrt{3} \\ &= \cfrac{4\sqrt{3}\big(1 - \cos^2{\theta} + \sin^2{\theta}\big)}{\sqrt{3}} \\ &= 4\big(1 - \cos^2{\theta} + \sin^2{\theta}\big) \\ &= 4\left[1 - \left(\cfrac{1}{2}\right)^2 + \sin^2{\theta}\right] \qquad \text{noting $\cos\theta = \cfrac{x}{r} = \cfrac{1}{2}$} \\ &= 4\left(1 - \cfrac{1}{4} + \cfrac{3}{4}\right) \qquad \text{noting $\cos^2\theta = \cfrac{1}{2} \implies \sin^2\theta = \cfrac{3}{4}$} \\ &= 4 - 1 + 3 \\ &= 6 \end{align*}$

thank youu

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