Home A-Level Edexcel Maths: Pure Exponential & Logarithms Figure 2 shows a sketch of the curve C with parametric equations
$x = 27 \, sec^2 \, t$, $y = 3 \, tan \, t$, \, 0 \leq t \leq \frac{\pi}{3}$
Figure 2 shows a sketch of the curve C with parametric equations
$x = 27 \, sec^2 \, t$, $y = 3 \, tan \, t$, \, 0 \leq t \leq \frac{\pi}{3}$ - Edexcel - A-Level Maths: Pure - Question 2 - 2013 - Paper 1 Question 2
View full question Figure 2 shows a sketch of the curve C with parametric equations
$x = 27 \, sec^2 \, t$, $y = 3 \, tan \, t$, \, 0 \leq t \leq \frac{\pi}{3}$.
(a) Find the gradien... show full transcript
View marking scheme Worked Solution & Example Answer:Figure 2 shows a sketch of the curve C with parametric equations
$x = 27 \, sec^2 \, t$, $y = 3 \, tan \, t$, \, 0 \leq t \leq \frac{\pi}{3}$ - Edexcel - A-Level Maths: Pure - Question 2 - 2013 - Paper 1
Find the gradient of the curve C at the point where $t = \frac{\pi}{6}$ Only available for registered users.
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To find the gradient of the curve C at the given point, we first find the derivatives of x x x y y y t t t
Compute rac{dx}{dt} and rac{dy}{dt} .
x = 27   s e c 2   ( t ) x = 27 \, sec^2 \,(t) x = 27 se c 2 ( t ) i g h t a r r o w
ightarrow i g h t a rro w rac{dx}{dt} = 54 \, sec^2(t) \, tan(t)
y = 3   t a n ( t ) y = 3 \, tan(t) y = 3 t an ( t ) i g h t a r r o w
ightarrow i g h t a rro w rac{dy}{dt} = 3 \, sec^2(t)
The gradient of the curve, rac{dy}{dx} , is given by:
d y d x = d y d t d x d t = 3   s e c 2 ( t ) 54   s e c 2 ( t )   t a n ( t ) = 1 18   t a n ( t ) \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3 \, sec^2(t)}{54 \, sec^2(t) \, tan(t)} = \frac{1}{18 \, tan(t)} d x d y ​ = d t d x ​ d t d y ​ ​ = 54 se c 2 ( t ) t an ( t ) 3 se c 2 ( t ) ​ = 18 t an ( t ) 1 ​
Now substitute t = π 6 t = \frac{\pi}{6} t = 6 π ​
Hence,
rac{dy}{dx} = \frac{1}{18 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}}{18} .
Show that the Cartesian equation of C may be written in the form $y = \left(x - 9\right)^{\frac{1}{3}}$, stating the values of a and b Only available for registered users.
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To show that the cartesian equation can be expressed as y = ( x − 9 ) 1 3 y = \left(x - 9\right)^{\frac{1}{3}} y = ( x − 9 ) 3 1 ​ t t t
From y = 3   t a n ( t ) y = 3 \, tan(t) y = 3 t an ( t )
t a n ( t ) = y 3 tan(t) = \frac{y}{3} t an ( t ) = 3 y ​
Then substituting t a n ( t ) tan(t) t an ( t ) x x x
x = 27   s e c 2 ( t ) = 27   ( 1 + t a n 2 ( t ) ) = 27 ( 1 + ( y 3 ) 2 ) = 27 + y 2 3 x = 27 \, sec^2(t) = 27 \, (1 + tan^2(t)) = 27 \left(1 + \left(\frac{y}{3}\right)^2 \right) = 27 + \frac{y^2}{3} x = 27 se c 2 ( t ) = 27 ( 1 + t a n 2 ( t )) = 27 ( 1 + ( 3 y ​ ) 2 ) = 27 + 3 y 2 ​
Rearranging, we get:
y 2 = 3 x − 81 y^2 = 3x - 81 y 2 = 3 x − 81 y = ( y 2 3 + 9 ) 1 3 y = \left(\frac{y^2}{3} + 9\right)^{\frac{1}{3}} y = ( 3 y 2 ​ + 9 ) 3 1 ​
a = 9 , b = 27 a = 9, b = 27 a = 9 , b = 27
Using calculus to find the exact value of the volume of the solid of revolution Only available for registered users.
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To find the volume of the solid formed by rotating region R about the x-axis, we use the formula for volume:
The volume is given by:
V = π ∫ a b f ( x ) 2 d x V = \pi \int_{a}^{b} f(x)^{2} dx V = π ∫ a b ​ f ( x ) 2 d x
The function f ( x ) f(x) f ( x ) y = ( x − 9 ) 1 3 y = \left(x - 9\right)^{\frac{1}{3}} y = ( x − 9 ) 3 1 ​
We need to determine the limits of integration, as x x x 9 9 9 125 125 125
Hence, we compute the volume integral:
V = π ∫ 9 125 ( ( x − 9 ) 1 3 ) 2 d x . V = \pi \int_{9}^{125} \left(\left(x - 9\right)^{\frac{1}{3}}\right)^{2} dx. V = π ∫ 9 125 ​ ( ( x − 9 ) 3 1 ​ ) 2 d x .
= π ∫ 9 125 ( x − 9 ) 2 3 d x . \pi \int_{9}^{125} \left(x - 9\right)^{\frac{2}{3}} dx. π ∫ 9 125 ​ ( x − 9 ) 3 2 ​ d x .
Evaluating the integral, we apply the power rule to get:
= π ⋅ 3 5 [ ( x − 9 ) 5 3 ] 9 125 = \pi \cdot \frac{3}{5} \left[(x - 9)^{\frac{5}{3}}\right]_{9}^{125} = π ⋅ 5 3 ​ [ ( x − 9 ) 3 5 ​ ] 9 125 ​
Substituting the limits will give the exact volume.
In this way, we can find the exact value of the volume of the solid of revolution.
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