Find the area bounded by y=x^{2} and by its normal, which with the x-axis creates the angle of 135^{o draw the figure also. }
Find the area bounded by y=x^{2} and by its normal, which with the x-axis creates the angle of 135^{o draw the figure also. }
Honestly I don't know how to draw the figure, if i knew that I could solve that easily. I don't know what to do with the angle given in the problem.
What did you try?
The "normal" to the curve y = x^2 is the "perpendicular" to the curve. The "angle with the x-axis" is the angle formed by the x-axis and the (extension of) the normal to the curve. The angle measure of 135 degrees is the measure of this angle.
So draw the regular x,y-axis system. Draw the curve y = x^2. Note that the curve is never below the x-axis. Note that 135 degrees is greater than ninety degrees. If you pencil in a perpendicular to the left-hand side of the curve (where x-values are negative) and extend this downward to the x-axis, any angle of 135 degrees would be opening away from the curve, to the left. If you pencil in a perpendicular to the right-hand side of the curve, in which direction would any 135-degree angle open?
Since either side's normal will cut off the same area (being the area between the x-axis, the curve y = x^2, and the back side, if you will, of the 135-degree angle), take the right-hand normal, for convenience.
Thanks a lot for such a detailed explanation, never thought this would be that helpful. I just want to know if i understood this good. So, i suppose a perpendicular in the curve (which would be the normal of the curve) and in that case it would make 90 degrees with the x-axis, that means I should be adding 45 degrees (left or right) to the perpendicular, did I get that right?
I do not understand why this is made so complicated? (Let's be mathematically mature and use numbers).
The normal and tangent have supplementary angles . So if the normal has angle [TEX]\frac{3\pi}{4} [/TEX] then the tangent has angle [TEX]\frac{\pi}{4} [/TEX].
So the tangent has slope [TEX] \tan \left( {\frac{\pi }{4}} \right) = 1[/TEX] at the point [TEX]\left( {\frac{1}{2},\frac{1}{4}} \right)[/TEX] the curve [TEX]x^2 [/TEX] has slope [TEX]1 [/TEX].
The normal at that point is [TEX]n(x) = - \left( {x - \frac{1}{2}} \right) + \frac{1}{4} [/TEX]
[TEX]A =\displaystyle \int_0^{1/2} {{x^2}} dx + \int_{1/2}^{3/4} {n(x)} dx [/TEX]
“A professor is someone who talks in someone else’s sleep”
W.H. Auden
Thanks a lot for your perfect answer. But excuse me for my ignorance, Is the ktangent and knormal the points where i create the normal equation, i mean the y1 and the x1?
Maybe I was a little bit hasty when I wrote my answer. It is important whether the required area is bounded only by [tex] y=x^2 [/tex] and its normal which encloses angle of 135°with x-axis, or that area is bounded by [tex] y=x^2 [/tex], its normal and x-axis. You wrote that the required area is bounded only by [tex] y=x^2 [/tex] and it's normal so I would interpret that the following area is required to be determined:
This sketch that I wrote demonstrates what your task states. [tex] y=x^2 [/tex] is a basic form of parabola that has it's vertex in (0,0), and its normal that encloses 135° with x-axis intersects that parabola at 2 points. Since your task states that the area required to be determined is bounded only by [tex] y=x^2 [/tex] and its normal, it should be the area 'between' those two curves. [tex] k_{normal}=\tan 135° [/tex]. Since tangent and normal are perpendicular, you can find out coefficient of direction of tangent from condition of verticality: [tex] k_{tangent}=-\dfrac{1}{k_{normal}} [/tex]. You also know that: [tex] k_{tangent}=f'(x_1) [/tex], where [tex] f(x)=x^2 [/tex]. Now you can find 'x' coordinate for intersection point(point on the right where normal/tangent intersects parabola). To find 'y' coordinate, just plug 'x' that you've just got into [tex] y_1=x_1^2 [/tex]. Now you have the intersection point: [tex] T(x_1,y_1) [/tex] (the one on the right). Equation of normal is: [tex] y-y_1=-\dfrac{1}{f'(x_1)}(x-x_1) [/tex]. Once you've got the equation of normal you can get the points where it intersects [tex] y=x^2 [/tex]. You should already have one intersection point by now [tex] T(x_1,y_1) [/tex] (it's the one on the right) , and now you should find the second point where normal intersects parabola(the one on the left on my sketch). To do so, you equalise equation of normal with the equation of parabola(you insert 'y' of normal into the equation of parabola): [tex] y_{normal}=x^2 [/tex]. Those two points of intersection should be the integration limits in this case. To find the area bounded by [tex] y=x^2 [/tex] and its normal(135° with x-axis), you can use formula: [tex] \int\limits_a^b \mathrm{(g(x)-f(x))} \mathrm{d}x [/tex]. f(x) is the equation of your graph ( [tex] y=x^2 [/tex] ), g(x) is the equation of normal, 'a' is 'x' value of the intersection point on the left side, and 'b' is the 'x' value of the intersection point on the right.
You can write [tex] \int\limits_a^b \mathrm{(g(x)-f(x))} \mathrm{d}x [/tex] as: [tex] \int\limits_a^b \mathrm{g(x)} \mathrm{d}x - \int\limits_a^b \mathrm{f(x)} \mathrm{d}x [/tex]. What we did here is subtract the whole area that is under the graph [tex] y=x^2 [/tex](between those integration limits) from the area that is under the normal(also between those integration limits). That way we got area between those two curves.
P.S. [tex] k_{normal} [/tex] and [tex] k_{tangent} [/tex] are slopes of those lines( coefficients of direction). For example: y=2x+3. Slope of this line is 2. In this case 'k' are not points.
If normal creates the angle of 135°with the x-axis, it's the angle that I marked on my sketch.
You can use the same formula [tex] \int\limits_a^b \mathrm{(g(x)-f(x))} \mathrm{d}x [/tex] to find out area between any two curves. 'a' and 'b' are the 'x' values of points where they intersect, and b>a, g(x) is the curve that is higher on the y-axis, and f(x) is the curve that is positioned lower(g(x) is above f(x)).
Last edited by Johulus; 07-25-2016 at 03:43 PM.
Bookmarks