#### EgzonKorenica

##### New member

- Joined
- Jul 23, 2016

- Messages
- 9

^{2}and by its normal, which with the x-axis creates the angle of 135

^{o draw the figure also. }

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter EgzonKorenica
- Start date

- Joined
- Jul 23, 2016

- Messages
- 9

- Joined
- Jun 18, 2007

- Messages
- 25,219

^{2}and by its normal, which with the x-axis creates the angle of 135^{o draw the figure also. }

Well, did you draw the figure?

What are the limits of integration?

Where are you exactly stuck?

- Joined
- Jan 29, 2005

- Messages
- 11,229

Hint: if the normal has angle \(\displaystyle \frac{3\pi}{4}\) then the tangent has angle \(\displaystyle \frac{\pi}{4}\).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.

^{2}and by its normal, which with the x-axis creates the angle of 135^{o draw the figure also. }

When it comes to this kind of tasks it is really of great importance to sketch what you have. Unless you do it, it will be hard to determine how to get to area that you need to determine.

Last edited:

- Joined
- Jul 23, 2016

- Messages
- 9

Well, did you draw the figure?

What are the limits of integration?

Where are you exactly stuck?

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.

- Joined
- Jul 23, 2016

- Messages
- 9

When it comes to this kind of tasks it is really of great importance to sketch what you have. Coefficient of direction of normal \(\displaystyle k_{normal}=\tan135° \). You can get coefficient of direction of tangent on the given graph: \(\displaystyle k_{tangent}=-\dfrac{1}{k_{normal}} \). You also know that \(\displaystyle k_{tangent}=f'(x) \). From last equation you can get 'x' of point where tangent/normal intersects the graph. You can get 'y' by pluging 'x' that you've just got into your graph equation: \(\displaystyle y=x^2 \). Find normal equation as follows: \(\displaystyle y-y_1=-\dfrac{1}{f'(x_1)}(x-x_1) \), where \(\displaystyle y_1 \qquad and \qquad x_1 \) are coordinates of intersection point. Once you have the equation of a normal you can find out where it intersects x-axis. And now you have integration limits. All that is left is to find area using integrals.

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?

- Joined
- Feb 4, 2004

- Messages
- 15,948

What did youI don't know what to do with the angle given in the problem.

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.

- Joined
- Jul 23, 2016

- Messages
- 9

What did youtry?

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?

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 \(\displaystyle y=x^2 \) and its normal which encloses angle of 135°with x-axis, or that area is bounded by \(\displaystyle y=x^2 \), its normal and x-axis. You wrote that the required area is bounded only by \(\displaystyle y=x^2 \) 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. \(\displaystyle y=x^2 \) 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 \(\displaystyle y=x^2 \) and its normal, it should be the area 'between' those two curves. \(\displaystyle k_{normal}=\tan 135° \). Since tangent and normal are perpendicular, you can find out coefficient of direction of tangent from condition of verticality: \(\displaystyle k_{tangent}=-\dfrac{1}{k_{normal}} \). You also know that: \(\displaystyle k_{tangent}=f'(x_1) \), where \(\displaystyle f(x)=x^2 \). 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 \(\displaystyle y_1=x_1^2 \). Now you have the intersection point: \(\displaystyle T(x_1,y_1) \) (the one on the right). Equation of normal is: \(\displaystyle y-y_1=-\dfrac{1}{f'(x_1)}(x-x_1) \). Once you've got the equation of normal you can get the points where it intersects \(\displaystyle y=x^2 \). You should already have one intersection point by now \(\displaystyle T(x_1,y_1) \) (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): \(\displaystyle y_{normal}=x^2 \). Those two points of intersection should be the integration limits in this case. To find the area bounded by \(\displaystyle y=x^2 \) and its normal(135° with x-axis), you can use formula: \(\displaystyle \int\limits_a^b \mathrm{(g(x)-f(x))} \mathrm{d}x \). f(x) is the equation of your graph ( \(\displaystyle y=x^2 \) ), 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 \(\displaystyle \int\limits_a^b \mathrm{(g(x)-f(x))} \mathrm{d}x \) as: \(\displaystyle \int\limits_a^b \mathrm{g(x)} \mathrm{d}x - \int\limits_a^b \mathrm{f(x)} \mathrm{d}x \). What we did here is subtract the whole area that is under the graph \(\displaystyle y=x^2 \)(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. \(\displaystyle k_{normal} \) and \(\displaystyle k_{tangent} \) 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 \(\displaystyle \int\limits_a^b \mathrm{(g(x)-f(x))} \mathrm{d}x \) 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:

- Joined
- Jul 23, 2016

- Messages
- 9

Maybe I was a little bit hasty when I wrote my answer. It is important whether the required area is bounded only by \(\displaystyle y=x^2 \) and its normal which encloses angle of 135°with x-axis, or that area is bounded by \(\displaystyle y=x^2 \), its normal and x-axis. You wrote that the required area is bounded only by \(\displaystyle y=x^2 \) 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. \(\displaystyle y=x^2 \) 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 \(\displaystyle y=x^2 \) and its normal, it should be the area 'between' those two curves. \(\displaystyle k_{normal}=\tan 135° \). Since tangent and normal are perpendicular, you can find out coefficient of direction of tangent from condition of verticality: \(\displaystyle k_{tangent}=-\dfrac{1}{k_{normal}} \). You also know that: \(\displaystyle k_{tangent}=f'(x_1) \), where \(\displaystyle f(x)=x^2 \). 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 \(\displaystyle y_1=x_1^2 \). Now you have the intersection point: \(\displaystyle T(x_1,y_1) \) (the one on the right). Equation of normal is: \(\displaystyle y-y_1=-\dfrac{1}{f'(x_1)}(x-x_1) \). Once you've got the equation of normal you can get the points where it intersects \(\displaystyle y=x^2 \). You should already have one intersection point by now \(\displaystyle T(x_1,y_1) \) (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): \(\displaystyle y_{normal}=x^2 \). Those two points of intersection should be the integration limits in this case. To find the area bounded by \(\displaystyle y=x^2 \) and its normal(135° with x-axis), you can use formula: \(\displaystyle \int\limits_a^b \mathrm{(g(x)-f(x))} \mathrm{d}x \). f(x) is the equation of your graph ( \(\displaystyle y=x^2 \) ), 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 \(\displaystyle \int\limits_a^b \mathrm{(g(x)-f(x))} \mathrm{d}x \) as: \(\displaystyle \int\limits_a^b \mathrm{g(x)} \mathrm{d}x - \int\limits_a^b \mathrm{f(x)} \mathrm{d}x \). What we did here is subtract the whole area that is under the graph \(\displaystyle y=x^2 \)(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. \(\displaystyle k_{normal} \) and \(\displaystyle k_{tangent} \) 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 \(\displaystyle \int\limits_a^b \mathrm{(g(x)-f(x))} \mathrm{d}x \) 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)).

Well, that was just perfect. Thanks a lot for taking your time for making this incredible explanation. I'm wondering if you guys are professors or what. That was amazing, I had the problem with the drawing, and what to do with the given angle but now everything is clear.

- Joined
- Jan 29, 2005

- Messages
- 11,229

I do not understand why this is made so complicated? (Let's be mathematically mature and use numbers).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?

The normal and tangent have supplementary angles . So if the normal has angle \(\displaystyle \frac{3\pi}{4} \) then the tangent has angle \(\displaystyle \frac{\pi}{4} \).

So the tangent has slope \(\displaystyle \tan \left( {\frac{\pi }{4}} \right) = 1\) at the point \(\displaystyle \left( {\frac{1}{2},\frac{1}{4}} \right)\) the curve \(\displaystyle x^2 \) has slope \(\displaystyle 1 \).

The normal at that point is \(\displaystyle n(x) = - \left( {x - \frac{1}{2}} \right) + \frac{1}{4} \)

\(\displaystyle A =\displaystyle \int_0^{1/2} {{x^2}} dx + \int_{1/2}^{3/4} {n(x)} dx \)

- Joined
- Jul 23, 2016

- Messages
- 9

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 \(\displaystyle \frac{3\pi}{4} \) then the tangent has angle \(\displaystyle \frac{\pi}{4} \).

So the tangent has slope \(\displaystyle \tan \left( {\frac{\pi }{4}} \right) = 1\) at the point \(\displaystyle \left( {\frac{1}{2},\frac{1}{4}} \right)\) the curve \(\displaystyle x^2 \) has slope \(\displaystyle 1 \).

The normal at that point is \(\displaystyle n(x) = - \left( {x - \frac{1}{2}} \right) + \frac{1}{4} \)

\(\displaystyle A =\displaystyle \int_0^{1/2} {{x^2}} dx + \int_{1/2}^{3/4} {n(x)} dx \)

Did you find the points the same way the previous guy explained?

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 \(\displaystyle \frac{3\pi}{4} \) then the tangent has angle \(\displaystyle \frac{\pi}{4} \).

So the tangent has slope \(\displaystyle \tan \left( {\frac{\pi }{4}} \right) = 1\) at the point \(\displaystyle \left( {\frac{1}{2},\frac{1}{4}} \right)\) the curve \(\displaystyle x^2 \) has slope \(\displaystyle 1 \).

The normal at that point is \(\displaystyle n(x) = - \left( {x - \frac{1}{2}} \right) + \frac{1}{4} \)

\(\displaystyle A =\displaystyle \int_0^{1/2} {{x^2}} dx + \int_{1/2}^{3/4} {n(x)} dx \)

The area that you are calculating is this as I can see:

But the task doesn't state that the area that needs to be determined is bounded by \(\displaystyle y=x^2 \), its normal

Did you find the points the same way the previous guy explained?

So, you have two intersection points as shown on my drawing above: \(\displaystyle T_1(\dfrac{1}{2},\dfrac{1}{4}) \) and \(\displaystyle T_2(\dfrac{-3}{2},\dfrac{9}{4}) \).

You proceed as: \(\displaystyle \int_{-3/2}^{1/2} \mathrm{(-x+\dfrac{3}{4}-x^2)} \mathrm{d}x=\int_{-3/2}^{1/2} \mathrm{(-x^2-x+\dfrac{3}{4})} \mathrm{d}x=(-\dfrac{1}{3}x^3-\dfrac{1}{2}x^2+\dfrac{3}{4}x) |_{-3/2}^{1/2} \)

- Joined
- Jan 29, 2005

- Messages
- 11,229

The area that you are calculating is this as I can see:

But the task doesn't state that the area that needs to be determined is bounded by \(\displaystyle y=x^2 \), its normaland x-axis. It states that the area that needs to be determined is bounded by \(\displaystyle y=x^2 \) and its normal and only area that is bounded only by these two is that between those curves:

So, you have two intersection points as shown on my drawing above: \(\displaystyle T_1(\dfrac{1}{2},\dfrac{1}{4}) \) and \(\displaystyle T_2(\dfrac{-3}{2},\dfrac{9}{4}) \).

You proceed as: \(\displaystyle \int_{-3/2}^{1/2} \mathrm{(-x+\dfrac{3}{4}-x^2)} \mathrm{d}x=\int_{-3/2}^{1/2} \mathrm{(-x^2-x+\dfrac{3}{4})} \mathrm{d}x=(-\dfrac{1}{3}x^3-\dfrac{1}{2}x^2+\dfrac{3}{4}x) |_{-3/2}^{1/2} \)

It clearly says that the x-axis is a boundary of the area.

- Joined
- Jul 23, 2016

- Messages
- 9

The area that you are calculating is this as I can see:

But the task doesn't state that the area that needs to be determined is bounded by \(\displaystyle y=x^2 \), its normaland x-axis. It states that the area that needs to be determined is bounded by \(\displaystyle y=x^2 \) and its normal and only area that is bounded only by these two is that between those curves:

So, you have two intersection points as shown on my drawing above: \(\displaystyle T_1(\dfrac{1}{2},\dfrac{1}{4}) \) and \(\displaystyle T_2(\dfrac{-3}{2},\dfrac{9}{4}) \).

You proceed as: \(\displaystyle \int_{-3/2}^{1/2} \mathrm{(-x+\dfrac{3}{4}-x^2)} \mathrm{d}x=\int_{-3/2}^{1/2} \mathrm{(-x^2-x+\dfrac{3}{4})} \mathrm{d}x=(-\dfrac{1}{3}x^3-\dfrac{1}{2}x^2+\dfrac{3}{4}x) |_{-3/2}^{1/2} \)

Yeah that's correct, it requires only the area made from the intersection of the normal and the parabola. It's all clear now.