Question 5
(3+5+7=15)
The points $A, B$ and $P$ with coordinates $(-2,0)$, $(6,0)$ and $(4,-6)$ respectively lie on $S$.
(a) Show that an equation of $S$ is $y = \frac{x^2}{2} - 2x - 6$. (3)
The line $l$ is the normal to $S$ at the point $P$.
(b) Show that an equation of $l$ is $2y + x + 8 = 0$. (5)
The finite region shown shaded in Figure 1 is bounded by $S$ and $l$.
(c) Use algebraic integration to find the exact area of the shaded region. (7)
Solution
Step 1: Finding the equation of $S$
The equation of $S$ is $y = px^2 + qx + r$. Substituting the coordinates of points $A$, $B$, and $P$ into this equation:
For $A(-2, 0)$:
$$ 0 = p(-2)^2 + q(-2) + r \Rightarrow 4p - 2q + r = 0 \tag{1} $$For $B(6, 0)$:
$$ 0 = p(6)^2 + q(6) + r \Rightarrow 36p + 6q + r = 0 \tag{2} $$For $P(4, -6)$:
$$ -6 = p(4)^2 + q(4) + r \Rightarrow 16p + 4q + r = -6 \tag{3} $$Subtracting equations:
$$ (2) - (1): \quad 36p + 6q + r - (4p - 2q + r) = 0 \Rightarrow 32p + 8q = 0 \Rightarrow 4p + q = 0 \tag{4} $$ $$ (3) - (1): \quad 16p + 4q + r - (4p - 2q + r) = -6 \Rightarrow 12p + 6q = -6 \Rightarrow 2p + q = -1 \tag{5} $$From equations (4) and (5):
$$ q = -4p \tag{6} $$ $$ 2p + (-4p) = -1 \Rightarrow -2p = -1 \Rightarrow p = \frac{1}{2}, \quad q = -2, \quad r = -6. $$Thus, the equation of $S$ is:
$$ y = \frac{x^2}{2} - 2x - 6. $$Step 2: Finding the equation of $l$ (the normal to $S$ at $P$)
The derivative of $y = \frac{x^2}{2} - 2x - 6$ is:
$$ \frac{dy}{dx} = x - 2. $$At $P(4, -6)$:
$$ \frac{dy}{dx} = 4 - 2 = 2. $$The gradient of the normal line is the negative reciprocal of the tangent's gradient:
$$ m_{\text{normal}} = -\frac{1}{2}. $$Using the point-slope form of a line equation at $P(4, -6)$:
$$ y - (-6) = -\frac{1}{2}(x - 4). $$Simplifying:
$$ y + 6 = -\frac{1}{2}x + 2 \Rightarrow 2y + 12 = -x + 4 \Rightarrow 2y + x + 8 = 0. $$Thus, the equation of $l$ is:
$$ 2y + x + 8 = 0. $$Step 3: Finding the area of the shaded region
The shaded region is bounded by $S$ and $l$. Solving for the intersection points:
$$ \text{Substitute } y = \frac{x^2}{2} - 2x - 6 \text{ into } 2y + x + 8 = 0: $$ $$ 2\left(\frac{x^2}{2} - 2x - 6\right) + x + 8 = 0 \Rightarrow x^2 - 4x - 12 + x + 8 = 0 \Rightarrow x^2 - 3x - 4 = 0. $$Factoring:
$$ (x - 4)(x + 1) = 0 \Rightarrow x = 4 \text{ or } x = -1. $$Since the straight line is above the curve between $x=-1$ and $x=4$, the area is:
$$ \text{Area} = \int_{-1}^{4} \left( -\frac{x}{2} - 4 \right) dx - \int_{-1}^{4} \left( \frac{x^2}{2} - 2x - 6 \right) dx . $$Step 4: Evaluate the curve integral
First, consider the integral:
$$ I_1 = \int_{-1}^{4} \left( \frac{x^2}{2} - 2x - 6 \right) dx. $$We integrate term by term:
$$ \int \frac{x^2}{2} \, dx = \frac{x^3}{6}, \quad \int -2x \, dx = -x^2, \quad \int -6 \, dx = -6x. $$Thus, the first integral becomes:
$$ I_1 = \left[ \frac{x^3}{6} - x^2 - 6x \right]_{-1}^{4}. $$Now, evaluate at the bounds:
At $x = 4$:
$$ \frac{4^3}{6} - 4^2 - 6 \cdot 4 = \frac{64}{6} - 16 - 24 = \frac{64}{6} - 40 = \frac{-176}{6} = -\frac{88}{3}. $$At $x = -1$:
$$ \frac{(-1)^3}{6} - (-1)^2 - 6 \cdot (-1) = \frac{-1}{6} - 1 + 6 = \frac{29}{6}. $$Thus, the value of the first integral is:
$$ I_1 = \left( -\frac{88}{3} \right) - \left( \frac{29}{6} \right) = -\frac{88}{3} - \frac{29}{6}. $$Combining these with a common denominator:
$$ -\frac{88}{3} = -\frac{176}{6}, \quad I_1 = -\frac{176}{6} - \frac{29}{6} = -\frac{205}{6}. $$Step 5: Evaluate the second integral
Next, we evaluate the second integral:
$$ I_2 = \int_{-1}^{4} \left( -\frac{x}{2} - 4 \right) dx. $$We integrate term by term:
$$ \int -\frac{x}{2} \, dx = -\frac{x^2}{4}, \quad \int -4 \, dx = -4x. $$Thus, the second integral becomes:
$$ I_2 = \left[ -\frac{x^2}{4} - 4x \right]_{-1}^{4}. $$Now, evaluate at the bounds:
At $x = 4$:
$$ -\frac{4^2}{4} - 4 \cdot 4 = -4 - 16 = -20. $$At $x = -1$:
$$ -\frac{(-1)^2}{4} - 4 \cdot (-1) = -\frac{1}{4} + 4 = \frac{15}{4}. $$Thus, the value of the second integral is:
$$ I_2 = -20 - \frac{15}{4} = -\frac{80}{4} - \frac{15}{4} = -\frac{95}{4}. $$Step 6: Combine the results
Finally, we combine the results of the two integrals:
$$ \text{Area} = I_2 - I_1 = \left( -\frac{95}{4} \right) - \left( -\frac{205}{6} \right) = \frac{205}{6} - \frac{95}{4}. $$Finding a common denominator:
$$ \frac{205}{6} = \frac{410}{12}, \quad \frac{95}{4} = \frac{285}{12}. $$Thus,
$$ \text{Area} = \frac{410}{12} - \frac{285}{12} = \frac{125}{12}. $$Final Answer
The final result is:
$$ \boxed{\frac{125}{12}}. $$Question 6
The volume of oil in a container is $V$ cm$^3$ when the height of the oil is $h$ cm.Oil is pouring into the container at a constant rate of $12$ cm$^3$/s.
Given that $V = 3h^3$, find the exact rate, in cm/s, at which the height of the oil is increasing when $V = 1536$ cm$^3$.
Solution
Step 1: Differentiate the relationship between $V$ and $h$
The given relationship is $V = 3h^3$. Differentiating both sides with respect to time $t$ using the chain rule, we get:
$$ \frac{dV}{dt} = \frac{d}{dt}(3h^3) = 9h^2 \frac{dh}{dt}. $$Step 2: Solve for $\frac{dh}{dt}$
Rearranging the equation to solve for $\frac{dh}{dt}$:
$$ \frac{dh}{dt} = \frac{1}{9h^2} \frac{dV}{dt}. $$Step 3: Find the height $h$ when $V = 1536 \, \mathrm{cm}^3$
Using the relationship $V = 3h^3$, substitute $V = 1536$:
$$ 1536 = 3h^3. $$Divide both sides by 3:
$$ h^3 = 512. $$Take the cube root of both sides:
$$ h = \sqrt[3]{512} = 8 \, \mathrm{cm}. $$Step 4: Substitute the known values into the expression for $\frac{dh}{dt}$
Substitute $h = 8$ and $\frac{dV}{dt} = 12 \, \mathrm{cm}^3/\mathrm{s}$ into the equation:
$$ \frac{dh}{dt} = \frac{1}{9h^2} \frac{dV}{dt}. $$Calculate $h^2$:
$$ h^2 = 8^2 = 64. $$Substitute $h^2 = 64$ and $\frac{dV}{dt} = 12$:
$$ \frac{dh}{dt} = \frac{1}{9(64)}(12) = \frac{12}{576} = \frac{1}{48}. $$Step 5: Final Answer
The exact rate at which the height of the oil is increasing is:
$$ \frac{dh}{dt} = \frac{1}{48} \, \mathrm{cm}/\mathrm{s}. $$
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