I am struggling to get some traction on this and need someone to show me how to calculate this. I have a GCSE question that asks:
In the attached image I'm being asked the following:
The arrowhead has an area of $3.6cm^2$. Find the length $x$.
[Math] Find the length of x for an arrow head
area
Related Solutions
You have copied the problem down incorrectly. It should read $$y=\cosh x=\frac{e^x+e^{-x}}2$$ Then $$dy=\frac{e^x-e^{-x}}2dx=\sinh x\,dx$$ The element of arc length is given by $$\begin{align}(ds)^2 & =(dx)^2+(dy)^2=\left(1+\frac{e^{2x}-2+e^{-2x}}4\right)(dx)^2\\ & =\left(\frac{e^{2x}+2+e^{-2x}}4\right)(dx)^2=\left(\frac{e^x+e^{-x}}2\right)^2(dx)^2\\ & =\cosh^2x(dx)^2\end{align}$$ So the arc length is $$s=\int ds=\int\cosh x\,dx$$ And this is the same as the area $$A=\int y\,dx=\int\cosh x\,dx$$ Famous theorem for a catenary.
EDIT: OK, so how did I know there was an error in the original problem statement? If $dA=ds$, then $y\,dx=\sqrt{1+(y^{\prime})2}\,dx$. Differentiating, $$y^{\prime}=\frac12\cdot\frac{2y^{\prime}y^{\prime\prime}}{\sqrt{1+(y^{\prime})^2}}=\frac{y^{\prime}y^{\prime\prime}}y$$ Then we can write it as a second-order differential equation: $y^{\prime\prime}=y$ with solution $y=c_1e^x+c_2e^{-x}$. Then $y^{\prime}=c_1e^x-c_2e^{-x}$. Plugging into the original differential equation, $$c_1e^x+c_2e^{-x}=\sqrt{1+c_1^2e^{2x}-2c_1c_2+c_2^2e^{-2x}}$$ Squaring out and simplifying, we see that $2c_1c_2=1-2c_1c_2$ so $c_2=\frac1{4c_1}$. Then $$y=c_1e^x+\frac1{4c_1}e^{-x}=\frac12\left(2c_1e^x+\frac1{2c_1}e^{-x}\right)$$ Since $y>0$, we can take the natural logarithm of $2c_1$ to get $\ln2c_1=-x_0$ and then $$y=\frac{e^{x-x_0}+e^{-(x-x_0)}}2=\cosh(x-x_0)$$ EDIT: I was careless in my proof above. Early on I divided by $y^{\prime}$ without considering what would happen if $y^{\prime}=0$. In fact, in that case $y=1$ is a (singular) solution to the problem as well.
The following is the process to find the surface area of rectangular box with its top side missing.
Let's call the side of the square base $s$ and let's call the height of the prism $h$.
Normally you have 6 sides. However, now you only have 5.
You have 4 sides of area $hs$ and one base of area $s^{2}$.
Therefore, $SA = hs + s^{2}$.
Feel free to plug in the given values.
The following is the process to find the surface area of cylinder with its top side missing.
Let's call the diameter of the circle base $D$ and let's call the height of the prism $h$.
The lateral surface area is $\frac{D}{2} h \pi$. The area of the base is the area of a circle: $\pi r^{2}$. Adding these two expressions, the surface area is $\frac{D}{2} h \pi + \pi r^{2}$.
Best Answer
The area $3.6cm^2$ is the area of the larger triangle minus the area of the smaller triangle.
$1.5\cdot h - 1.5\cdot (h - x) = 3.6$
$1.5h - 1.5h + 1.5x = 3.6$
$1.5x = 3.6$
$x = 2.4$cm