New answers tagged

1

The constraints are written $$\begin{cases}\dfrac{1-e^{an}}a+(b+1)n=v,\\-e^{an}+b+1=rb.\end{cases}$$ You can eliminate $b$ using the second equation, $$b=\frac{1-e^{an}}{r-1},$$ which you plug in the first, giving a nasty nonlinear equation in $an$ $$\dfrac{1-e^{an}}{an}+\frac{1-e^{an}}{r-1}=\frac vn-1.$$ This must be solved by a numerical method. (...


2

In fact your issue has an infinity of solutions, making angle $\varphi$ dependent on position of point $M$ (see Fig. 1). Indeed, one can build an infinity of such pairs (rectangle,triangle) with equal area (two examples are given on Fig. 1). For every such pair, a specific angle (only exceptionally equal to 45°) is found. Take a look at the following ...


1

Without additional information the angle $\varphi$ is not always $45^\circ$. Let $x$ and $y$ be the sides of the rectangle and let $X$ and $Y$ be the legs of the right triangle. Then $xy=XY/2$ and $$\tan(\varphi)=\tan(\alpha-\beta)=\frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)} \tag{*}$$ where $\tan(\alpha)=(y+Y)/x$ and $\tan(\beta)=y/(x+X)$. ...


3

Let $M$ be a barycenter of $\{A,B,C\}$. Thus, $MI||BC.$ Now, let $AD$ be an altitude of $\Delta ABC$, $AF$ be a median of $\Delta ABC$ and $MI\cap AD=\{E\}.$ Thus, $ED=r$, $$\frac{AE}{ED}=\frac{AM}{MF}=2$$ and in the standard notation we have $$\frac{h_a-r}{r}=2,$$ which gives $h_a=3r.$ Thus, for the area of the triangle we obtain: $$\frac{(12+10+x)r}{2}...


0

Your error lies in the computation of the intersection points of the curves. If $y^2=4x$ and $y=4x-2$, then $4x=(4x-2)^2$ indeed, but this is not an equivalence. It turns out that your region is the region below the graph of $2\sqrt x$ and above the graph of $-2\sqrt x$ (with $x\in\left[0,\frac14\right]$) plus the area of the region below the graph of $2\...


2

If you actually look at a graph of the required area you will see that it is quite difficult to find when integrating with respect to $x$. We can instead rearrange both equations to get $$x=\frac14y^2$$ $$x=\frac14y+\frac12$$ Then the graphs intersect at points where $y=-1$ and $y=2$ respectively so the area is given by $$\int_{-1}^2\left(\frac14y+\frac12\...


0

Note that you have $y^2=9x $ and $y= \frac{3x^2}{8}$ thus you need to have $$9x = (\frac{3x^2}{8})^2 $$ to start with.


1

There is an error in the calculation of the upper bound, where a square operation is overlooked. The correct equation for the bounds is, $$9x = \left(\frac{3}{8}x^2\right)^2$$ which yields $x_1=0$ and $x_2=4$. As a result, $$\int_0^4 \left( 3\sqrt{x} - \frac{3}{8}x^2 \right) = 8 $$


1

You have the right idea in that you have to first find the points of intersection of the two curves, but you are forgetting that one curve was given as $y^2$ and the other was given as $y$. Try rewriting the first curve as $y=\pm3\sqrt{x}$, $x\geq 0$, and try again. Your overall method is correct!


1

Let $AL\cap BP=\{X\}$. Thus, $$\frac{LC}{AD}=\frac{CT}{DT}=\frac{1}{2}=\frac{DR}{RC}=\frac{DP}{BC}=\frac{DP}{AD},$$ which gives $$LC=DP,$$ $$AP=BL,$$ which gives $APLB$ is a parallelogram. By the similar way we obtain $$MC=2BC=2AD=DS,$$ which gives that $ASMB$ is a parallelogram. Now, let $h$ be an altitude of $ABCD$ from $B$ to $AD$. Thus, $$S_{shaded}=...


3

Let $AD$ have the length of $2$ (units). Then $PD=1$ and $SP=3$. Same proportions on the segment $MB$. This is because we have: $PD:PA=DR:AB=1:3$ and $SD:SA=DT:AB=2:3$. This implies: $$ \frac {\operatorname{Area}(SPB)} {\operatorname{Area}(DAB)} = \frac{SP}{DA} = \frac 32\ . $$ The same also on the other side. Now $\operatorname{Area}(DAB)$ is half of the ...


1

You have the bounds correct, but since the area is bounded above by $x$ and below by $\frac{1}{\sqrt{x}}$ the integral should be $$\int_1^2 (x-\frac{1}{\sqrt{x}})dx $$ $$=(\frac{1}{2}x^2-2\sqrt{x} )|^2_1$$ $$=2 - 2\sqrt{2} - (\frac{1}{2}-2)$$ $$=\frac{7}{2}-2\sqrt{2}$$


2

Reverse the $x$ and the $y$ coordinate axis (which amounts to a symmetry, an operation that doesn't change the absolute value of an area ; see figure below), giving equations $$y=x^2 \ \text{and} \ y+2x=8 \ \ \ \iff \ \ \ y=x^2 \ \text{and} \ y=-2x+8 \tag{1}$$ In this way, taking into account the fact that the intersection points of the new curves are $(...


3

Your first integral is slightly incorrect. The correct setup and solution would be: $$A=\int_{-4}^{2} \left(8-2y-y^2\right) \,dy=36\ \text{sq. units}.$$ If you now want to do integration with respect to $x$, begin by expressing your two functions as functions of $x$: $$ y=\pm\sqrt{x},\\ y=-\frac{x}{2}+4. $$ And then find where the curves intersect each ...


2

The first should be $$\int\limits_{-4}^2(8-2y-y^2)dy.$$ The second should be $$\int\limits_0^4\left(\sqrt{4-\frac{x}{2}}-\left(-\sqrt{4-\frac{x}{2}}\right)\right)dx+\int\limits_4^{16}\left(4-\frac{x}{2}-\left(\sqrt{4-\frac{x}{2}}\right)\right)dx$$


1

$y_1=8-6x+x^2$ (blue curve) bounds the side, $y_2=-4+2x$ (red curve) is the lower bound, and $y_3=2+x$ (yellow curve) is the upper bound. This gives us three areas: Between Red and Blue Between Yellow and Blue Between Yellow and Red, bounded to the left by Blue To find Area 1, use the following formula: $$ \int_a^b [f_1(x) - f_2(x)]dx $$ Where $f_1(x)$ ...


8

Let $a_1$ and $a_2$ be the side lengths of the two squares. To determine which one is larger, we simply look at their ratio below. With the angles in the diagram, $$d_1=\frac{1}{2\tan 30}a_1=\frac{\sqrt{3}}{2}a_1$$ $$d_2=\frac{\sin 15}{\sin 30}a_2=\frac{1}{2\cos 15}a_2$$ Assume both equilateral triangles have unit height. $$1=a_1+d_1=\left(1+\frac{\sqrt{...


2

Let sides-lengths of the equilateral triangle be equal to $1$. Let $x$ be sides-lengths of the square in the first configuration. Thus, by law of sines we obtain: $$\frac{x}{\sin60^{\circ}}=\frac{\frac{1}{2}}{\sin75^{\circ}}$$ or $$\frac{x}{\frac{\sqrt3}{2}}=\frac{\frac{1}{2}}{\frac{1+\sqrt3}{2\sqrt2}}$$ or $$x=\frac{\sqrt3}{\sqrt2(1+\sqrt3)}$$ and for the ...


3

The second configuration (square has edge contact with triangle) indeed has a bigger inscribed square. If the square has unit sides, the triangle's side is $1+\frac2{\sqrt3}$: The symmetric first configuration may be resolved as follows. Set the unit square's bottom corner as $(0,0)$, so that the top corner is $(0,\sqrt2)$. Let the side length of the ...


4

By clearing denominators, we see that our curve is a quartic $p(x, y) = 0$. For generic values of $p, a, b$, it is elliptic and so does not admit a rational parameterization. Probably the areas can be computed in terms of elliptic functions. In the special case $p = \frac{1}{2}$, the curve is symmetric not only about the $x$-axis but also the line $x = \...


1

My proof is valid only when the circum-circle of CDE lies completely inside the original circle. Locate H, the ortho-center of CDE. EH extended will cut CDE at E’ and the circum-circle at E’’ By properties of the ortho-center, HE’ = E’E’’. Hence, [CHE’] = [CE’E’’] The other parts of CDE can be off-set similarly. Result follows.


4

The $xy$-integration is still easier than the polar coordinates for this problem. Due to $y$-symmetry, the integration is between the two intersection points with the $x$-axis, which can be obtained by setting $y = 0$ in the curve equation (for $a=1$, $b=0$) $$ \frac{p}{(1-x)^2}+\frac{1-p}{x^2}=1 $$ or $$ x^4-2x^3+2(1-p)x-(1-p)=0 $$ Unfortunately, for ...


1

It is an even function Required Area will be , $$ Area = 2\;(\int_0^2| x^2\;-1| \,dx)$$ $$=2(\;\int_0^1(1-x^2)\,dx\;+\int_1^2(x^2-1)\,dx\;)$$$$=2( \, \frac23\;+\;\frac43\,)=\;4 $$


3

By symmetry, it is twice the area between the $y$-axis and the line $x=2$. The positive $x$-intercept is $x=1$, and on the interval $[0,1]$, the curve is below the $x$-axis, hence the total (geometric) area is $$2\biggl(\int_1^2y(x)\,\mathrm d x-\int_0^1y(x)\,\mathrm d x\biggl).$$


1

You forgot the area under the x-axis. The signed area of this is: $$\int_{-1}^1 x^2-1 \ dx \ $$ Can you continue?


1

$$\int\limits_{\theta = 0}^{2 \pi} \int\limits_{r = 2 + \sin (3 \theta)}^{4 - \cos (3 \theta)} r\ dr\ d\theta = 12 \pi$$


3

You can just calculate the area of the $\text{outer curve} - \text{inner curve}$: $$\frac{1}{2} \int_0^{2\pi} \big(4 - \cos(3\theta) \big)^2 - \big(2 + \sin(3 \theta) \big)^2$$


0

The curve is a parabola y^2 = 4aX area will be twice the area under curve y=sqrt.(4ax) from X= a to X = 4a and x axis answer is 56 a^2 / 3


4

Your answer is just fine, because $$\int_{1}^{2} 4a^2 ~dt = \int_{a}^{4a} 2a\sqrt{\frac{x}{a}}~ dx = 28a^2/3$$ is the area between the parametric curve and the $x(t)$-axis.


3

Hint: You can also use an explicit form of your function: $$t=\sqrt{\frac{x}{a}}=t$$ so $$y=2a\sqrt{\frac{x}{a}}$$


1

Here are the steps to solve your $\theta$-equation: $$\sqrt2\cos\left(\theta-\frac{\pi}{4}\right) + \sqrt{\cos\left[2(\theta-\frac{\pi}{4})\right]} = 2$$ $$2-\sqrt2\cos\left(\theta-\frac{\pi}{4}\right) = \sqrt{\cos\left[2(\theta-\frac{\pi}{4})\right]} $$ $$\left[2-\sqrt2\cos\left(\theta-\frac{\pi}{4}\right)\right]^2 = \cos\left[2(\theta-\frac{\pi}{4})\right]$...


0

Refer to the circles in rectangular coordinates. Expand the first and substitute for $x^2+y^2,$ to get $$x+y=12.5.$$ This may be substituted into the circle centred at the origin, and the rest should be quite straightforward.


0

It gives also $$r_a=\frac{A}{s-a}$$ where $r_a$ is the exradius,$A$ the area, and $s$ the semiparameter.


3

Yes, this is Heron's formula and it's rather well-known.


12

There does in fact exist an area-preserving map, as demonstrated in this video at 11:20: the Lambert azimuthal equal-area projection. The idea is that you take polar coordinates of the hyperbolic plane and map them to polar coordinates of the euclidean plane via a map $(r, \theta) \mapsto (f(r), \theta)$ where $f$ is chosen such that area is preserved. Let'...


0

The parabola has the same shape as $y=x^2$. The chord length of $1$ parallel to the $x$-axis connects points $\big(\pm\frac 12, \frac14\big)$. By symmetry this chord gives the largest area. See Desmos illustration here. For simplicity, in the illustration, the chord is fixed as the segment between $(0,0)$ and $(1,0)$, and the parabola rotates. The area ...


1

Hint: The curve $y(x)$ does not form a loop itself. It is the area enclosed by $y(x)$, the y-axis and the x-axis. Then, it is easy to verify that the curve intersects the x-axis at (1,0) and y-axis at (0,1). As a result, the area $$I = \int_0^1 y(x)dx$$ can be carried out with the variable substitutions $x=(1-t^3)/(1+t^2)$ and $y=2t/(1+t^2)$. Explicitly, ...


0

You'll see your error is you simply plot the curve: $$A = 4 \int\limits_{\theta = 0}^{\pi/2} \int\limits_{r=0}^{|\sin (2 \theta)|} r\ dr\ d\theta = \frac{\pi}{8}$$


2

Let the equation of the secant line be $y=mx+b$. Combining this with the equation of the parabola produces the quadratic equation $$x^2+(1-m)x+(10-b)=0.$$ To reduce clutter, I’ll denote the discriminant of this equation by $\Delta = (1-m)^2-4(10-b)$. The solutions to this equation are, of course, $$x = \frac12(m-1)\pm\frac12\sqrt\Delta$$ and substituting ...


0

I consider the more general parabola $y(x) = x^2 + ax + b $ with the points being $(x_i, y_i)_{i=1}^2 $. I am close to a solution, but am pooping out so am leaving my answer incomplete. One surprising result I find is that the area between the chord and the parabola is $\dfrac{(x_2-x_1)^3}{6} $. The length of the chord is $\begin{array}\\ L^2 &=(...


1

This is a partial answer. I would say the hardest part is likely finding the points of intersection. In this post I discuss the generic approach, which involves solving a quadratic equation of at least degree $3$. However, since your ellipses are both centered around the origin, a quadratic equation will suffice. Rewrite your ellipses in the following form ...


2

Such a chord would intersection the parabola at points $(x_1, y_1)$ and $(x_2, y_2)$, where $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = 1$. Notice that if you square this expression, you get $(y_2 - y_1)^{2} = 1 - (x_2 - x_1)^2$. Here, you'd have $y_2 = x^{2}_{2} + x_2 + 10$ and $y_1 = x^{2}_{1} + x_1 + 10$.


3

I'll use $\triangle PQR$ instead of $\triangle ABC$, to avoid some notational confusion. First, a little prep work. Given $\triangle PQR$, we erect on (directed) segment $\overline{PQ}$ an equilateral triangle $\triangle PQR'$ with a clockwise orientation. Similarly, we erect clockwise equilaterals $\triangle QRP'$ and $\triangle RPQ'$. As it happens, the ...


2

Let $x=\sin t$. Thus, we need to get $$2\int\limits_0^1x\sqrt{1-x^2}dx=2\int\limits_0^{\frac{\pi}{2}}\sin{t}\cos^2tdt=\int\limits_0^{\frac{\pi}{2}}\sin2t\cos{t}=$$ $$=\frac{1}{2}\int\limits_0^{\frac{\pi}{2}}(\sin3t+\sin{t})dt=-\frac{1}{6}\cos3t-\frac{1}{2}\cos{t}\big{|}_0^{\frac{\pi}{2}}=\frac{2}{3}.$$


5

In your question, it asks for the area of the shaded region, and area is always positive. For integration, it is taught as the area between the curve and the $x$-axis. But actually not quite, because in the definition of integration we calculate "Area" as "NET area" (positive if above $x$-axis, and offset by those below $x$-axis). Note that Area is ...


2

The function is odd, the area you look for is $$A=2\int_0^1x\sqrt{1-x^2}dx$$ put $y=x^2$. then $$A=\int_0^1\sqrt{1-y}dy$$ $$=\int_0^1(1-y)^{\frac 12}dy$$ $$=\Bigl[ \frac 23(1-y)^{\frac 32}\Bigr]_1^0$$ $$=\frac 23.$$


2

The integral $\int_{-1}^0y~dx$ is negative, but the area is always non-negative. So the area of the region is $\int_0^1y~dx-\int_{-1}^0y~dx$.


0

Using for example the Wiki article on ellipses, you will find that the semi-major axis is $2.5$ feet and the semi-minor axis is $2$ feet. This means the foci are at $\pm 1.5$ feet, i.e.the tacks should be placed at the base, $1.5$ feet to either side of the center. The string should be $5$ feet long.


0

Now here's a solution which works in any vector space with an inner product: take the half of the root of the Gram-Determinant of two sides of the triangle, that is $$\frac12\sqrt{\det\begin{pmatrix}\langle b-a,b-a\rangle& \langle b-a,c-a\rangle\\ \langle b-a,c-a\rangle & \langle c-a,c-a\rangle \end{pmatrix}}.$$


2

Find the intersection of the curves $y=\frac{x^2}{2}$ with $x^2+y^2=8$ From the parabola $y=\frac{x^2}{2}$ we have $2y=x^2$. Pluging this result $x^2=2y$ in the equation function of the circle gives $2y+y^2=8$. The equation $2y+y^2=8$ is a quadratic equation $y^2+2y-8=0$ and can be solved by many methods. One of these methods is completing the square: $y^2+...


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