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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 ...

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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^2x=\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.

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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!

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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_1d_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{...

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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 ...

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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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