Problem Statement
The curve $C$ has the equation:
$$ y = 4 - e^{2x} $$
The curve $C$ crosses the $y$-axis at the point $A$ and the $x$-axis at the point $B$.
(a)
- (i) Write down the $y$-coordinate of point $A$.
- (ii) Show that the $x$-coordinate of $B$ is $x = \ln 2$.
Solution
Part (a)
(i) $y$-coordinate of point $A$
Point $A$ is where the curve crosses the $y$-axis. At the $y$-axis, $x = 0$. Substituting $x = 0$ into the equation of the curve:
$$ y = 4 - e^{2(0)} = 4 - e^0 = 4 - 1 = 3 $$
Thus, the $y$-coordinate of $A$ is:
$$ \boxed{3} $$
(ii) Show that the $x$-coordinate of $B$ is $x = \ln 2$
Point $B$ is where the curve crosses the $x$-axis. At the $x$-axis, $y = 0$. Substituting $y = 0$ into the equation of the curve:
$$ 0 = 4 - e^{2x} $$
Rearranging:
$$ e^{2x} = 4 $$
Taking the natural logarithm on both sides:
$$ \ln(e^{2x}) = \ln 4 $$
Using the logarithmic property $ \ln(e^a) = a $, we have:
$$ 2x = \ln 4 $$
Divide through by 2:
$$ x = \frac{\ln 4}{2} $$
Using the logarithmic property $ \ln(a^b) = b\ln a $, rewrite $ \ln 4 $ as $ \ln(2^2) = 2\ln 2 $:
$$ x = \frac{2\ln 2}{2} = \ln 2 $$
Thus, the $x$-coordinate of $B$ is:
$$ \boxed{\ln 2} $$
Part (b): Equation of the Normal Line $l$
The point $B$ is where $x = \ln 2$. Substituting $x = \ln 2$ into the equation of the curve $y = 4 - e^{2x}$:
$$ y = 4 - e^{2(\ln 2)} = 4 - e^{\ln 4} = 4 - 4 = 0 $$
Thus, point $B$ is $$ (\ln 2, 0) $$
To find the equation of the normal, first find the derivative of $y = 4 - e^{2x}$ to obtain the gradient of the tangent:
$$ \frac{dy}{dx} = -2e^{2x} $$
At $x = \ln 2$:
$$ \frac{dy}{dx} = -2e^{2(\ln 2)} = -2(4) = -8 $$
The gradient of the tangent at $B$ is $-8$, so the gradient of the normal is:
$$ m = -\frac{1}{-8} = \frac{1}{8} $$
Using the point-slope form of a line, the equation of the normal at $(\ln 2, 0)$ is:
$$ y - 0 = \frac{1}{8}(x - \ln 2) $$
Simplify:
$$ y = \frac{1}{8}x - \frac{\ln 2}{8} $$
Thus, the equation of the normal line is:
$$ \boxed{ y = \frac{1}{8}x - \frac{\ln 2}{8} } $$
Part (c): Area of the Region $R$
The region $R$ is bounded by the curve $C$, the normal line $l$, and the $y$-axis.
Step 1: Set up the integral bounds
The curve $C$ intersects the $y$-axis at $x = 0$, and the region ends at the $x$-coordinate of $B$, $x = \ln 2$.
The area under $C$ is given by:
$$ \text{Area under } C = \int_0^{\ln 2} (4 - e^{2x}) \, dx $$
Step 2: Area under the line $l$
The line $l$ intersects the $y$-axis at $x = 0$, where:
$$ y = \frac{1}{8}(0) - \frac{\ln 2}{8} = - \frac{\ln 2}{8} $$
The area under $l$ between $x = 0$ and $x = \ln 2$ is:
$$ \text{Area under } l = \int_0^{\ln 2} \left( \frac{1}{8}x - \frac{\ln 2}{8} \right) dx $$
Step 3: Total area of $R$
The region $R$ is the area under the curve $C$ minus the area under the line $l$:
$$ \text{Area of } R = \int_0^{\ln 2} (4 - e^{2x}) \, dx - \int_0^{\ln 2} \left( \frac{1}{8}x - \frac{\ln 2}{8} \right) dx $$
Step 4: Solve the integrals
$$ \int_0^{\ln 2} (4 - e^{2x}) \, dx = \int_0^{\ln 2} 4 \, dx - \int_0^{\ln 2} e^{2x} \, dx $$
$$ = \left[ 4x \right]_0^{\ln 2} - \left[ \frac{e^{2x}}{2} \right]_0^{\ln 2} $$
$$ = 4\ln 2 - \left( \frac{e^{2\ln 2}}{2} - \frac{e^0}{2} \right) $$
$$ = 4\ln 2 - \left( \frac{4}{2} - \frac{1}{2} \right) $$
$$ = 4\ln 2 - \frac{3}{2} $$
For the line $l$:
$$ \int_0^{\ln 2} \left( \frac{1}{8}x - \frac{\ln 2}{8} \right) dx = \int_0^{\ln 2} \frac{1}{8}x \, dx - \int_0^{\ln 2} \frac{\ln 2}{8} \, dx $$
$$ = \left[ \frac{1}{16}x^2 \right]_0^{\ln 2} - \frac{\ln 2}{8} \int_0^{\ln 2} 1 \, dx $$
$$ = \frac{(\ln 2)^2}{16} - \frac{(\ln 2)^2}{8} $$
$$ = - \frac{(\ln 2)^2}{16} $$
Thus, the total area is:
$$ \text{Area of } R = \left( 4\ln 2 - \frac{3}{2} \right) - \left( - \frac{(\ln 2)^2}{16} \right) \approx 1.3 $$
Final Area
$$ \boxed{1.3} $$
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