Find curve equation from derivative

10 The equation of a curve is such that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 6 x ^ { 2 } - \frac { 4 } { x ^ { 3 } }\). The curve has a stationary point at \(\left( - 1 , \frac { 9 } { 2 } \right)\).

1. Determine the nature of the stationary point at \(\left( - 1 , \frac { 9 } { 2 } \right)\).
2. Find the equation of the curve.
3. Show that the curve has no other stationary points.
4. A point \(A\) is moving along the curve and the \(y\)-coordinate of \(A\) is increasing at a rate of 5 units per second. Find the rate of increase of the \(x\)-coordinate of \(A\) at the point where \(x = 1\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

10 At the point \(( 4 , - 1 )\) on a curve, the gradient of the curve is \(- \frac { 3 } { 2 }\). It is given that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { - \frac { 1 } { 2 } } + k\), where \(k\) is a constant.

1. Show that \(k = - 2\).
2. Find the equation of the curve.
3. Find the coordinates of the stationary point.
4. Determine the nature of the stationary point.

10 A curve has a stationary point at \(( 2 , - 10 )\) and is such that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 6 x\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Find the equation of the curve.
3. Find the coordinates of the other stationary point and determine its nature.
4. Find the equation of the tangent to the curve at the point where the curve crosses the \(y\)-axis.
   \includegraphics[max width=\textwidth, alt={}, center]{5e3e5418-7976-4232-8550-1da6420a3fcb-18_689_828_276_646} The diagram shows the circle with equation \(( x - 4 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 40\). Parallel tangents, each with gradient 1 , touch the circle at points \(A\) and \(B\).
5. Find the equation of the line \(A B\), giving the answer in the form \(y = m x + c\).
6. Find the coordinates of \(A\), giving each coordinate in surd form.
7. Find the equation of the tangent at \(A\), giving the answer in the form \(y = m x + c\), where \(c\) is in surd form.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

9 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { \sqrt { } x } - 1\) and \(P ( 9,5 )\) is a point on the curve.

1. Find the equation of the curve.
2. Find the coordinates of the stationary point on the curve.
3. Find an expression for \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) and determine the nature of the stationary point.
4. The normal to the curve at \(P\) makes an angle of \(\tan ^ { - 1 } k\) with the positive \(x\)-axis. Find the value of \(k\).

6 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = k - 2 x\), where \(k\) is a constant.

1. Given that the tangents to the curve at the points where \(x = 2\) and \(x = 3\) are perpendicular, find the value of \(k\).
2. Given also that the curve passes through the point \(( 4,9 )\), find the equation of the curve.

8 A curve is such that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ( 3 x + 4 ) ^ { \frac { 3 } { 2 } } - 6 x - 8 .$$

1. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
2. Verify that the curve has a stationary point when \(x = - 1\) and determine its nature.
3. It is now given that the stationary point on the curve has coordinates \(( - 1,5 )\). Find the equation of the curve.

9 The function f is defined for \(x > 0\) and is such that \(\mathrm { f } ^ { \prime } ( x ) = 2 x - \frac { 2 } { x ^ { 2 } }\). The curve \(y = \mathrm { f } ( x )\) passes through the point \(P ( 2,6 )\).

1. Find the equation of the normal to the curve at \(P\).
2. Find the equation of the curve.
3. Find the \(x\)-coordinate of the stationary point and state with a reason whether this point is a maximum or a minimum.

10 A curve is such that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { 24 } { x ^ { 3 } } - 4\). The curve has a stationary point at \(P\) where \(x = 2\).

1. State, with a reason, the nature of this stationary point.
2. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
3. Given that the curve passes through the point \(( 1,13 )\), find the coordinates of the stationary point \(P\).

8 A curve \(y = \mathrm { f } ( x )\) has a stationary point at \(( 3,7 )\) and is such that \(\mathrm { f } ^ { \prime \prime } ( x ) = 36 x ^ { - 3 }\).

1. State, with a reason, whether this stationary point is a maximum or a minimum.
2. Find \(\mathrm { f } ^ { \prime } ( x )\) and \(\mathrm { f } ( x )\).

10 A curve has equation \(y = \mathrm { f } ( x )\) and it is given that \(\mathrm { f } ^ { \prime } ( x ) = 3 x ^ { \frac { 1 } { 2 } } - 2 x ^ { - \frac { 1 } { 2 } }\). The point \(A\) is the only point on the curve at which the gradient is - 1 .

1. Find the \(x\)-coordinate of \(A\).
2. Given that the curve also passes through the point \(( 4,10 )\), find the \(y\)-coordinate of \(A\), giving your answer as a fraction.

6. The curve \(C\) has equation \(y = \mathrm { f } ( x )\) where \(x > 0\) Given that

• \(\mathrm { f } ^ { \prime } ( x ) = \frac { ( x + 3 ) ^ { 2 } } { x \sqrt { x } }\)
• the point \(P ( 4,20 )\) lies on \(C\)
  1. (i) find the value of the gradient at \(P\)
     (ii) Hence find the equation of the tangent to \(C\) at \(P\), giving your answer in the form \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers to be found.
  2. Find \(\mathrm { f } ( x )\), simplifying your answer.
     

1. A curve \(C\) has equation \(y = \mathrm { f } ( x ) , \quad x > 0\)

Given that

• \(\mathrm { f } ^ { \prime \prime } ( x ) = 4 x + \frac { 1 } { \sqrt { x } }\)
• the point \(P\) has \(x\) coordinate 4 and lies on \(C\)
• the tangent to \(C\) at \(P\) has equation \(y = 3 x + 4\)
  1. find an equation of the normal to \(C\) at \(P\)
  2. find \(\mathrm { f } ( x )\), writing your answer in simplest form.

1. In this question you must show all stages of your working.

The curve \(C\) has equation \(y = \mathrm { f } ( x ) , x > 0\)
Given that

• the point \(P ( 2,8 \sqrt { 2 } )\) lies on \(C\)
• \(\mathrm { f } ^ { \prime } ( x ) = 4 \sqrt { x ^ { 3 } } + \frac { k } { x ^ { 2 } }\) where \(k\) is a constant
• \(\mathrm { f } ^ { \prime \prime } ( x ) = 0\) at \(P\)
  1. find the exact value of \(k\),
  2. find \(\mathrm { f } ( x )\), giving your answer in simplest form.

6. The curve \(C\) has equation \(y = \mathrm { f } ( x ) , x > 0\) Given that

• \(C\) passes through the point \(P ( 8,2 )\)
• \(\mathrm { f } ^ { \prime } ( x ) = \frac { 32 } { 3 x ^ { 2 } } + 3 - 2 ( \sqrt [ 3 ] { x } )\)
  1. find the equation of the tangent to \(C\) at \(P\). Write your answer in the form \(y = m x + c\), where \(m\) and \(c\) are constants to be found.
     (3)
  2. Find, in simplest form, \(\mathrm { f } ( x )\).
     \includegraphics[max width=\textwidth, alt={}, center]{877d03f2-d62c-4060-bdd2-f0d5dfbe6203-21_2647_1840_118_111}

7. The curve \(C\) has equation \(y = \mathrm { f } ( x ) , x > 0\) Given that

• \(\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { \sqrt { x } } + \frac { A } { x ^ { 2 } } + 3\), where \(A\) is a constant
• \(\mathrm { f } ^ { \prime \prime } ( x ) = 0\) when \(x = 4\)
  1. find the value of \(A\).

Given also that

• \(\mathrm { f } ( x ) = 8 \sqrt { 3 }\), when \(x = 12\)
• find \(\mathrm { f } ( x )\), giving each term in simplest form.
  

1. In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

1. Find the equation of the tangent to the curve with equation $$y = \frac { 1 } { 4 } x ^ { 3 } - 8 x ^ { - \frac { 1 } { 2 } }$$ at the point \(P ( 4,12 )\)
   Give your answer in the form \(a x + b y + c = 0\) where \(a\), \(b\) and \(c\) are integers. The curve with equation \(y = \mathrm { f } ( x )\) also passes through the point \(P ( 4,12 )\)
   Given that $$f ^ { \prime } ( x ) = \frac { 1 } { 4 } x ^ { 3 } - 8 x ^ { - \frac { 1 } { 2 } }$$
2. find \(\mathrm { f } ( x )\) giving the coefficients in simplest form.

10. A curve has equation \(y = \mathrm { f } ( x ) , x > 0\) Given that

• \(\mathrm { f } ^ { \prime } ( x ) = a x - 12 x ^ { \frac { 1 } { 3 } }\), where \(a\) is a constant
• \(\mathrm { f } ^ { \prime \prime } ( x ) = 0\) when \(x = 27\)
• the curve passes through the point \(( 1 , - 8 )\)
  1. find the value of \(a\).
  2. Hence find \(\mathrm { f } ( x )\).
     

1. The curve \(C\) has equation \(y = \mathrm { f } ( x )\) where \(x > 0\)

Given that

• \(f ^ { \prime } ( x ) = \frac { 4 x ^ { 2 } + 10 - 7 x ^ { \frac { 1 } { 2 } } } { 4 x ^ { \frac { 1 } { 2 } } }\)
• the point \(P ( 4 , - 1 )\) lies on \(C\)
  1. (i) find the value of the gradient of \(C\) at \(P\)
     (ii) Hence find the equation of the normal to \(C\) at \(P\), giving your answer in the form \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers to be found.
  2. Find \(\mathrm { f } ( x )\).

3．Given that $$\frac { \mathrm { d } } { \mathrm {~d} x } ( u \sqrt { } x ) = \frac { \mathrm { d } u } { \mathrm {~d} x } \times \frac { \mathrm { d } ( \sqrt { } x ) } { \mathrm { d } x } , \quad 0 < x < \frac { 1 } { 2 }$$ where \(u\) is a function of \(x\) ，and that \(u = 4\) when \(x = \frac { 3 } { 8 }\) ，find \(u\) in terms of \(x\) ．
（9）

1. A curve \(C\) has equation \(y = \mathrm { f } ( x )\)

Given that

• \(\mathrm { f } ^ { \prime } ( x ) = 6 x ^ { 2 } + a x - 23\) where \(a\) is a constant
• the \(y\) intercept of \(C\) is - 12
• ( \(x + 4\) ) is a factor of \(\mathrm { f } ( x )\)
  find, in simplest form, \(\mathrm { f } ( x )\)

6 A curve \(C\) has an equation which satisfies \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 3 x ^ { 2 } + 2\), for all values of \(x\).

1. It is given that \(C\) has a single stationary point. Determine the nature of this stationary point. The diagram shows the graph of the gradient function for \(C\).
   \includegraphics[max width=\textwidth, alt={}, center]{31b0d5b6-1593-489b-bbcd-486e7c96ff18-04_702_442_1672_242}
2. Given that \(C\) passes through the point \(\left( - 1 , \frac { 1 } { 4 } \right)\), find the equation of \(C\) in the form \(y = \mathrm { f } ( x )\).

1. Given that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ^ { 3 } - 4 } { x ^ { 3 } } , \quad x \neq 0$$

1. find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\). Given also that \(y = 0\) when \(x = - 1\),
2. find the value of \(y\) when \(x = 2\).

9. The gradient of the curve \(C\) is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 3 x - 1 ) ^ { 2 }$$ The point \(P ( 1,4 )\) lies on \(C\).

1. Find an equation of the normal to \(C\) at \(P\).
2. Find an equation for the curve \(C\) in the form \(y = \mathrm { f } ( x )\).
3. Using \(\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 3 x - 1 ) ^ { 2 }\), show that there is no point on \(C\) at which the tangent is parallel to the line \(y = 1 - 2 x\).

9 A curve cuts the \(x\)-axis at ( 2,0 ) and has gradient function $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 24 } { x ^ { 3 } }$$ 9

1. Find the equation of the curve.
2. Show that the perpendicular bisector of the line joining \(A ( - 2,8 )\) to \(B ( - 6 , - 4 )\) is the 9
3. Snormal to the curve at ( 2,0 )