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The position of a particle moving in the Cartesian plane, at time \(t\), is given by the parametric equations
\(x(t)=\dfrac{6 t}{t+1}\) and \(y(t)=\dfrac{-8}{t^2+4}\), where \(t \geq 0\).
What is the slope of the tangent to the path of the particle when \(t=2\) ?
\(D\)
\(\text{At}\ \ t=2\ \text{(by calc):}\)
\(\dfrac{dx}{dt}=\dfrac{2}{3}, \ \dfrac{dy}{dt}=\dfrac{1}{2} \)
\(\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{1}{2} \times \dfrac{3}{2} = \dfrac{3}{4} \)
\(\Rightarrow D\)
The curve defined by the parametric equations
\(x=\dfrac{t^2}{4}-1, \ y=\sqrt{3} t\), where \(0 \leq t \leq 2 \text {, }\)
is rotated about the \(x\)-axis to form an open hollow surface of revolution.
Find the surface area of the surface of revolution.
Give your answer in the form \(\pi\left(\dfrac{a \sqrt{b}}{c}-d\right)\), where \(a, b, c\) and \(d \in Z^{+}\). (4 marks)
--- 10 WORK AREA LINES (style=lined) ---
\(\pi\Bigg{(} \dfrac{64\sqrt3}{3}-24\Bigg{)} \)
\(x=\dfrac{t^2}{4}-1 \ \Rightarrow \dfrac{dx}{dt}=\dfrac{t}{2} \)
\(y=\sqrt{3} t \ \Rightarrow \dfrac{dy}{dt} = \sqrt3\)
| \(\text{S.A.}\) | \[=2\pi \int_0^2 \sqrt3 t \times \sqrt{\Big{(}\dfrac{dx}{dt}\Big{)}^2+\Big{(}\frac{dy}{dt}\Big{)}^2}\ dt\] | |
| \[=2\sqrt3 \pi \int_0^2 t\sqrt{\dfrac{t^2}{4}+3}\ dt\] |
\(\text{Let}\ \ u=\dfrac{t^2}{4}\ \ \Rightarrow \ \dfrac{du}{dt}=\dfrac{t}{2}\ \ \Rightarrow \ 2\,du=t\,dt\)
\(\text{When}\ \ t=2, u=4; \ t=0, u=3\)
| \(\text{S.A.}\) | \[=4\sqrt3\pi \times \dfrac{2}{3}\Big{[}u^{\frac{3}{2}}\Big{]}_3^4 \] | |
| \(=\dfrac{8\sqrt3 \pi}{3}\big{(}4^{\frac{3}{2}}-3^{\frac{3}{2}}\big{)} \) | ||
| \(=\dfrac{8\sqrt3 \pi}{3}(8-3\sqrt3) \) | ||
| \(=\pi\Bigg{(} \dfrac{64\sqrt3}{3}-24\Bigg{)} \) |
For the curve with parametric equations
`x = 4 sin (t) - 1`
`y = 2 cos (t) + 3`
Find `(dy)/(dx)` at the point `(1, sqrt 3 + 3).` (3 marks)
`– sqrt 3/6`
`(dx)/(dt) = 4cos(t)`
`(dy)/(dt) = −2sin(t)`
| `(dy)/(dx)` | `= ((dy)/(dt))/((dx)/(dt))` |
| `= (-sin t)/(2cos t)` |
`x = 4 sin (t) – 1\ \ text{(given)}\ \ =>\ \ sin(t)=(x+1)/4`
`y = 2 cos (t) + 3\ \ text{(given)}\ \ =>\ \ 2cos(t)=(y-3)`
`=> dy/dx= (-(x + 1)/4)/(y – 3)`
`text(At)\ \ (1, sqrt 3 + 3):`
| `:. (dy)/(dx)` | `= (-1/2)/(sqrt3+3 -3)` |
| `= -1/(2sqrt3)` | |
| `= – sqrt3/6` |