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y-Coordinate and y=x^2 - Dissertation Sample

15 Mar 2017Dissertation Samples

Investigate the relationship between the y-coordinate and y=x^2

Q1.Investigate the relationship between the y-coordinate of any point on the curve y=x^2and the point at which the tangent to the curve at this point intersects with the y-axis .Explain and analyse to prove results.

Extend this investigation to consider curves of the form y=x^n for n>2.Generalise the relationship between the y-coordinate of any point on the curve y=x^n and the point at which the tangent to the curve at this point intersects the y-axis.

For which other values of 'n' is your generalisation true?Explain and illustrate your answer with appropriate sketches and/or diagrams.

Are any of your results valid for y=kx^n?

Explain and justify your answer geometrically.

Comment on the benefits of using ICT to support your work.In particular comment on how ICT might be used to support any geometric arguments made earlier.

Q2.Explain and find the equation of the line perpendicular to the line ax+by=d which passes through the origin. Hence explain and derive a formula for the distance of the line ax+by=d from the origin.

Hence explain and find the expression for the distance between two parallel lines.

Q3.Show and explain,by integration, that the volume,V,of the cap,(r-h)on the y-axis,formed when part of the circle of radius ,r, and centre (0,0)

is rotated by 360 degrees about the y-axis

is given by:

V=PI H^2(r-h/3).

show and explain how this result confirms that the volume of a sphere can be given by:

V= 4/3 pi r^3

Question 1:

We consider the curve defined by . Let’s be a point on this curve, and the tangent to the curve at . We denote by the point at which intersects the axis.

We can visualise the above on the following graph.

The question asks us to investigate the relationship between and . In order to do that, we will first look at the equation of . If is a point on , then the slope of is given by .

Since is a good approximation of the curve in the immediate neighborhood of , by choosing to be point close enough to , we can think of it as point on so that .

It follows, from the definition of the derivative that , where . Here we have ; hence, is defined by the equation . The value of is obtained simply by setting so that , thus we are left with .

This is what we would have expected when looking at the curve . It is a strictly convex curve (a fact that can be deduced from its second derivative for all real number); this convexity implies that the curve lies above all its tangent lines. Since the curve lies on the top half of the axis, we would therefore expect the intersection of any tangent line to this curve o be a point on the bottom half of the axis. The above relationship between and is just a confirmation on this, since implies .

The next thing that the relationship between and tells us is that the coordinate of any point on and the point at which the tangent to intersects the axis are symmetric with respect to origin, in other words, they are situated at equal distance from the origin on the axis.

Another way to see this is to consider the triangle which is rectangle; let’s be the angle, defined as . Hence . But this also equals to the slope of the tangent which is simply the gradient, of the curve at the point . Therefore .

In general for a curve defined by , the equation of the tangent to the curve at the point is given by , giving the following relationship between and : .

The result here we depend on the parity of .

When is even, the above results still hold as this case is still strictly convex ( for , and for all real number). The change here is that for and will no longer be symmetric with respect to the origin; this is explained by the relation (2) and translates the fact that as increases, the slope of its tangent at the point also increase, pushing down the intersection point of this tangent with the axis.

When is odd, the matter will be different according to whether or . As an illustration we have the following graph for .

Clearly, the relationship between and expressed by relation (2) remains valid. The only difference being that here is negative for , and thus is positive (due to the change in the convexity of the curve).

Applying the reasoning for , the relationship between and is found as . This is the relation that we have above. Thus the same results will apply.

ICT allows us to construct graphic in order to have an illustration of the notion developed above. It can provide a basis to make conjecture before formal proofs are given. By the design a an Excel worksheet that allows the user to change the value of and the coordinates of the point we can see how the position of the tangent does change as changes and figure out the relationship between and .

Question 2:

Consider the line : . The vector is a vector defining the direction of . This vector is perpendicular to as their scalar product is . Hence, the line is a line perpendicular to . Since we are looking for the one that passes through the origin, we must set , i.e. : is the line perpendicular to that passes through the origin.

Let’s denote by the intersection point between lines and , and by the coordinates of . Then and satisfy the following systems of equations: ,

which when solved gives us the following values for and : ,.

The distance of the line from the origin is simply the distance of the point from the origin. This is calculated as the distance between two points, which we denote here as , and is given by .

So in general, the distance of the line from the origin is given by .

Consider the line as above. A line : is parallel to if and only if there is an integer such that and . Since and are parallel, the line as defined above is also perpendicular to , and calling the intersection point between and , the distance of from the origin is given by ,

To find the distance between the two parallels lines, we then just need to take the difference of their respective distance from the origin. That is, the distance between and is given by .

Question 3.

To start with this question, let’s first remind ourselves that the volume of any solid of revolution formed by rotating a profile , where , is given by .

Taking the circle of radius and centre , and rotate it by 360 degrees about the -axis, we obtain the sphere of radius and centre . The equation of this sphere in the three dimensional space coordinates is given by .

Having that, if we cut off this sphere by the plane : , then the portion of the sphere which lies above this plane is exactly the cap that we are interested in for this question.

We slice this cap by planes parallel to ; a slice by the plane is a disk of radius defined by , i.e. . Therefore, the area of such a slice is , and thus for a small step on the , the part on the cap comprises between and has the volume . Hence, the total volume of the cap is obtained by adding up all this elementary volume, which gives us .

It follows that , which is what we aimed to show.

We can view the previous result as a function of , writing , . Now, if we take , then is the plane , and thus the cap in question will simply be the top half of our sphere. Its volume is given . The volume of the whole sphere is subsequently obtained simply by multiplying this by 2, which gives us .  Buy PhD dissertation on any other topic at our site


D.W. Jordan; P Smith, Mathematical techniques, An introduction for the engineering, physical and mathematical sciences, Oxford University Press, 1994.

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