Geometric curve properties

8 \includegraphics[max width=\textwidth, alt={}, center]{20893bfc-3300-4205-9d2c-729cc3243971-3_597_951_1471_598} In the diagram the tangent to a curve at a general point \(P\) with coordinates \(( x , y )\) meets the \(x\)-axis at \(T\). The point \(N\) on the \(x\)-axis is such that \(P N\) is perpendicular to the \(x\)-axis. The curve is such that, for all values of \(x\) in the interval \(0 < x < \frac { 1 } { 2 } \pi\), the area of triangle \(P T N\) is equal to \(\tan x\), where \(x\) is in radians.

1. Using the fact that the gradient of the curve at \(P\) is \(\frac { P N } { T N }\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 } y ^ { 2 } \cot x .$$
2. Given that \(y = 2\) when \(x = \frac { 1 } { 6 } \pi\), solve this differential equation to find the equation of the curve, expressing \(y\) in terms of \(x\).

6 A certain curve is such that its gradient at a point \(( x , y )\) is proportional to \(x y\). At the point \(( 1,2 )\) the gradient is 4 .

1. By setting up and solving a differential equation, show that the equation of the curve is \(y = 2 \mathrm { e } ^ { x ^ { 2 } - 1 }\).
2. State the gradient of the curve at the point \(( - 1,2 )\) and sketch the curve.

5 \includegraphics[max width=\textwidth, alt={}, center]{ccadf73b-16f5-463a-8f69-1394839d5325-2_346_437_1155_854} The diagram shows a variable point \(P\) with coordinates \(( x , y )\) and the point \(N\) which is the foot of the perpendicular from \(P\) to the \(x\)-axis. \(P\) moves on a curve such that, for all \(x \geqslant 0\), the gradient of the curve is equal in value to the area of the triangle \(O P N\), where \(O\) is the origin.

1. State a differential equation satisfied by \(x\) and \(y\). The point with coordinates \(( 0,2 )\) lies on the curve.
2. Solve the differential equation to obtain the equation of the curve, expressing \(y\) in terms of \(x\).
3. Sketch the curve.

8 A certain curve is such that its gradient at a point \(( x , y )\) is proportional to \(\frac { y } { x \sqrt { x } }\). The curve passes through the points with coordinates \(( 1,1 )\) and \(( 4 , \mathrm { e } )\).

1. By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\).
2. Describe what happens to \(y\) as \(x\) tends to infinity.

7 \includegraphics[max width=\textwidth, alt={}, center]{1990cbac-d96f-4484-be4b-67dab35b3147-10_647_519_260_813} For the curve shown in the diagram, the normal to the curve at the point \(P\) with coordinates \(( x , y )\) meets the \(x\)-axis at \(N\). The point \(M\) is the foot of the perpendicular from \(P\) to the \(x\)-axis. The curve is such that for all values of \(x\) in the interval \(0 \leqslant x < \frac { 1 } { 2 } \pi\), the area of triangle \(P M N\) is equal to \(\tan x\).

1. 1. Show that \(\frac { M N } { y } = \frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Hence show that \(x\) and \(y\) satisfy the differential equation \(\frac { 1 } { 2 } y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \tan x\).
2. Given that \(y = 1\) when \(x = 0\), solve this differential equation to find the equation of the curve, expressing \(y\) in terms of \(x\).

8

1. Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 - x } { y - 3 }$$ giving the particular solution that satisfies the condition \(y = 4\) when \(x = 5\).
2. Show that this particular solution can be expressed in the form $$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = k$$ where the values of the constants \(a , b\) and \(k\) are to be stated.
3. Hence sketch the graph of the particular solution, indicating clearly its main features.

1. The gradient at any point \(( x , y )\) on a curve is proportional to \(\sqrt { y }\).

Given that the curve passes through the point with coordinates \(( 0,4 )\),

1. show that the equation of the curve can be written in the form $$2 \sqrt { y } = k x + 4$$ where \(k\) is a positive constant. Given also that the curve passes through the point with coordinates ( 2,9 ),
2. find the equation of the curve in the form \(y = \mathrm { f } ( x )\).

11 A curve, \(C\), passes through the point with coordinates \(( 1,6 )\) The gradient of \(C\) is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 6 } ( x y ) ^ { 2 }$$ Show that \(C\) intersects the coordinate axes at exactly one point and state the coordinates of this point. Fully justify your answer.

10

1. 1. By writing \(\sec x = \frac { 1 } { \cos x }\), prove that $$\frac { \mathrm { d } } { \mathrm {~d} x } ( \sec x ) = \sec x \tan x .$$
   2. Deduce that \(y = \sec x\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y \sqrt { y ^ { 2 } - 1 } , \quad 0 \leq x < \frac { 1 } { 2 } \pi .$$
2. A curve lies in the first quadrant of the cartesian plane with origin \(O\) as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{85043199-527d-4105-aa0b-c913dec0e35b-4_707_698_845_685} The normal to the curve at the point \(P ( x , y )\) meets the \(x\)-axis at the point \(Q\). The angle between \(O P\) and the \(x\)-axis is \(u\), and the angle between \(Q P\) and the \(x\)-axis is \(v\).
   1. If $$\tan v = \tan ^ { 2 } u$$ obtain a differential equation satisfied by the curve.
   2. The curve passes through the point \(( 2,1 )\). By solving the differential equation, find an equation for the curve in the implicit form $$\mathrm { F } ( x , y ) = C ,$$ where \(C\) is a constant that should be determined.

A certain curve is such that its gradient at a general point with coordinates \((x, y)\) is proportional to \(\frac{y^2}{x}\). The curve passes through the points with coordinates \((1, 1)\) and \((e, 2)\). By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\). [8]

The coordinates \((x, y)\) of a general point on a curve satisfy the differential equation $$x\frac{dy}{dx} = (2 - x^2)y.$$ The curve passes through the point \((1, 1)\). Find the equation of the curve, obtaining an expression for \(y\) in terms of \(x\). [7]

The gradient at any point \((x, y)\) on a curve is proportional to \(\sqrt{y}\). Given that the curve passes through the point with coordinates \((0, 4)\),

1. show that the equation of the curve can be written in the form $$2\sqrt{y} = kx + 4,$$ where \(k\) is a positive constant. [5]

Given also that the curve passes through the point with coordinates \((2, 9)\),

1. find the equation of the curve in the form \(y = \text{f}(x)\). [4]

A curve \(C\) in the \(x\)-\(y\) plane has the property that the gradient of the tangent at the point \(P(x, y)\) is three times the gradient of the line joining the point \((3, 2)\) to \(P\).

1. Express this property in the form of a differential equation. [2]

It is given that \(C\) passes through the point \((4, 3)\) and that \(x > 3\) and \(y > 2\) at all points on \(C\).

1. Determine the equation of \(C\) giving your answer in the form \(y = f(x)\). [4]

The curve \(C\) may be obtained by a transformation of part of the curve \(y = x^3\).

1. Describe fully this transformation. [2]