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4. (a) Given that y=sinh⁡xy=\sinh x, use the definition of sinh⁡x\sinh x in terms of $\ma...
Apr 25, 2024
4. (a) Given that y=sinh⁡xy=\sinh x, use the definition of sinh⁡x\sinh x in terms of ex\mathrm{e}^{x} and e−x\mathrm{e}^{-x} to show that x=ln⁡(y+y2+1)x=\ln \left(y+\sqrt{y^{2}+1}\right). [4 marks] (b) A curve has equation y=6cosh⁡2x+5sinh⁡xy=6 \cosh ^{2} x+5 \sinh x. (i) Show that the curve has a single stationary point and find its xx-coordinate, giving your answer in the form ln⁡p\ln p, where pp is a rational number. [5 marks] (ii) The curve lies entirely above the xx-axis. The region bounded by the curve, the coordinate axes and the line x=cosh⁡−12x=\cosh ^{-1} 2 has area AA. Show that A=acosh⁡−12+b3+c A=a \cosh ^{-1} 2+b \sqrt{3}+c where a,ba, b and cc are integers. [5 marks]
Generated Graph
Solution by Steps
step 1
To show that x=ln⁡(y+y2+1)x = \ln(y + \sqrt{y^2 + 1}), start with the definition of sinh⁡x\sinh x in terms of exponential functions: sinh⁡x=ex−e−x2\sinh x = \frac{e^x - e^{-x}}{2}
step 2
Given y=sinh⁡xy = \sinh x, we can express xx in terms of yy: y=ex−e−x2y = \frac{e^x - e^{-x}}{2}
step 3
Multiply both sides by 2 and rearrange: 2y=ex−e−x2y = e^x - e^{-x}
step 4
Add e−xe^{-x} to both sides and then square both sides: (2y+e−x)2=e2x(2y + e^{-x})^2 = e^{2x}
step 5
Expand the left side and use the identity e−x=1exe^{-x} = \frac{1}{e^x}: 4y2+4ye−x+e−2x=e2x4y^2 + 4ye^{-x} + e^{-2x} = e^{2x}
step 6
Recognize that e2x−e−2x=4y2e^{2x} - e^{-2x} = 4y^2 and add e−2xe^{-2x} to both sides: e2x+e−2x=4y2+e−2x+1e^{2x} + e^{-2x} = 4y^2 + e^{-2x} + 1
step 7
Simplify to get e2x+1=4y2+1e^{2x} + 1 = 4y^2 + 1 and then take the square root of both sides: ex=y2+1e^x = \sqrt{y^2 + 1}
step 8
Finally, take the natural logarithm of both sides to solve for xx: x=ln⁡(y2+1+y)x = \ln(\sqrt{y^2 + 1} + y)
Answer
x=ln⁡(y+y2+1)x = \ln(y + \sqrt{y^2 + 1})
Key Concept
Expressing hyperbolic functions in terms of exponential functions and solving for xx.
Explanation
The key to solving for xx is to manipulate the equation using algebraic identities and properties of exponential functions, ultimately taking the natural logarithm to isolate xx.
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Solution by Steps
step 1
To find the stationary point, differentiate the curve y=6cosh⁡2x+5sinh⁡xy = 6 \cosh^2 x + 5 \sinh x
step 2
The derivative is dydx=12sinh⁡xcosh⁡x+5cosh⁡x\frac{dy}{dx} = 12 \sinh x \cosh x + 5 \cosh x
step 3
Set the derivative equal to zero to find the stationary point: 0=12sinh⁡xcosh⁡x+5cosh⁡x0 = 12 \sinh x \cosh x + 5 \cosh x
step 4
Factor out cosh⁡x\cosh x: 0=cosh⁡x(12sinh⁡x+5)0 = \cosh x (12 \sinh x + 5)
step 5
Since cosh⁡x\cosh x is never zero, solve 12sinh⁡x+5=012 \sinh x + 5 = 0 for xx
step 6
Rearrange to get sinh⁡x=−512\sinh x = -\frac{5}{12} and use the definition of sinh⁡−1\sinh^{-1} to solve for xx
step 7
The solution is x=sinh⁡−1(−512)x = \sinh^{-1}\left(-\frac{5}{12}\right), which can be expressed in terms of natural logarithms
step 8
Using the identity sinh⁡−1y=ln⁡(y+y2+1)\sinh^{-1} y = \ln(y + \sqrt{y^2 + 1}), substitute y=−512y = -\frac{5}{12} to find xx
step 9
Thus, x=ln⁡(−512+(−512)2+1)x = \ln\left(-\frac{5}{12} + \sqrt{\left(-\frac{5}{12}\right)^2 + 1}\right)
Answer
x=ln⁡(−512+(−512)2+1)x = \ln\left(-\frac{5}{12} + \sqrt{\left(-\frac{5}{12}\right)^2 + 1}\right)
Key Concept
Finding the stationary point of a curve by setting its derivative equal to zero.
Explanation
The stationary point occurs where the derivative of the curve is zero. By factoring and solving the resulting equation, we find the xx-coordinate of the stationary point.
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Solution by Steps
step 1
To find the area AA under the curve y=6cosh⁡2x+5sinh⁡xy = 6 \cosh^2 x + 5 \sinh x from x=0x = 0 to x=cosh⁡−12x = \cosh^{-1} 2, integrate the function
step 2
The integral is ∫0cosh⁡−1(2)(6cosh⁡2x+5sinh⁡x)dx\int_0^{\cosh^{-1}(2)} (6 \cosh^2 x + 5 \sinh x) dx
step 3
Use the identity cosh⁡2x=12(cosh⁡(2x)+1)\cosh^2 x = \frac{1}{2}(\cosh(2x) + 1) to simplify the integral
step 4
The integral becomes ∫0cosh⁡−1(2)(3(cosh⁡(2x)+1)+5sinh⁡x)dx\int_0^{\cosh^{-1}(2)} (3(\cosh(2x) + 1) + 5 \sinh x) dx
step 5
Integrate term by term to find A=3x+32sinh⁡(2x)+5cosh⁡(x)A = 3x + \frac{3}{2} \sinh(2x) + 5 \cosh(x) evaluated from 00 to cosh⁡−1(2)\cosh^{-1}(2)
step 6
Substitute the limits of integration to find the exact area: A=3cosh⁡−1(2)+32sinh⁡(2cosh⁡−1(2))+5cosh⁡(cosh⁡−1(2))−(3⋅0+32sinh⁡(0)+5cosh⁡(0))A = 3 \cosh^{-1}(2) + \frac{3}{2} \sinh(2 \cosh^{-1}(2)) + 5 \cosh(\cosh^{-1}(2)) - (3 \cdot 0 + \frac{3}{2} \sinh(0) + 5 \cosh(0))
step 7
Simplify using the identities sinh⁡(2cosh⁡−1(2))=23\sinh(2 \cosh^{-1}(2)) = 2\sqrt{3} and cosh⁡(cosh⁡−1(2))=2\cosh(\cosh^{-1}(2)) = 2
step 8
The final expression for the area is A=3cosh⁡−1(2)+33+10−5A = 3 \cosh^{-1}(2) + 3\sqrt{3} + 10 - 5
Answer
A=3cosh⁡−1(2)+33+5A = 3 \cosh^{-1}(2) + 3\sqrt{3} + 5
Key Concept
Calculating the area under a curve using definite integration.
Explanation
The area under the curve from x=0x = 0 to x=cosh⁡−1(2)x = \cosh^{-1}(2) is found by integrating the function and evaluating it at the given limits. The result is expressed in terms of cosh⁡−1(2)\cosh^{-1}(2) and 3\sqrt{3}.
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