Finding curve equation from derivative

3 The equation of a curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 ( 4 x - 7 ) ^ { \frac { 1 } { 2 } } - 4 x ^ { - \frac { 1 } { 2 } }\). It is given that the curve passes through the point \(\left( 4 , \frac { 5 } { 2 } \right)\). Find the equation of the curve.

3 A curve is such that \(\frac { \mathrm { dy } } { \mathrm { dx } } = 3 ( 4 \mathrm { x } + 5 ) ^ { \frac { 1 } { 2 } }\). It is given that the points \(( 1,9 )\) and \(( 5 , a )\) lie on the curve. Find the value of \(a\).

4 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 8 } { ( 3 x + 2 ) ^ { 2 } }\). The curve passes through the point \(\left( 2,5 \frac { 2 } { 3 } \right)\).
Find the equation of the curve.

4 A cu h s eq tin \(y = \mathrm { f } ( x )\).I t is g it h \(\mathrm { f } ^ { \prime } ( x ) = \frac { 1 } { \sqrt { x + 6 } } + \frac { 6 } { x ^ { 2 } }\) ad h \(\mathrm { f } ( \mathcal { \beta } = 1\)
Fif ( \(x\) ).

7 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { ( 1 + 2 x ) ^ { 2 } }\) and the point \(\left( 1 , \frac { 1 } { 2 } \right)\) lies on the curve.

1. Find the equation of the curve.
2. Find the set of values of \(x\) for which the gradient of the curve is less than \(\frac { 1 } { 3 }\).

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

2 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x + 1 ) ^ { \frac { 1 } { 2 } }\) and the point (4,7) lies on the curve. Find the equation of the curve.

2 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 8 } { ( 5 - 2 x ) ^ { 2 } }\). Given that the curve passes through ( 2,7 ), find the equation of the curve.

3 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 12 } { ( 2 x + 1 ) ^ { 2 } }\). The point \(( 1,1 )\) lies on the curve. Find the coordinates of the point at which the curve intersects the \(x\)-axis.

4 A curve with equation \(y = \mathrm { f } ( x )\) passes through the point \(A ( 3,1 )\) and crosses the \(y\)-axis at \(B\). It is given that \(\mathrm { f } ^ { \prime } ( x ) = ( 3 x - 1 ) ^ { - \frac { 1 } { 3 } }\). Find the \(y\)-coordinate of \(B\).

4 The gradient at any point \(( x , y )\) on a curve is \(\sqrt { } ( 1 + 2 x )\). The curve passes through the point \(( 4,11 )\). Find

1. the equation of the curve,
2. the point at which the curve intersects the \(y\)-axis.

1 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 8 } { \sqrt { } ( 4 x + 1 ) }\). The point \(( 2,5 )\) lies on the curve. Find the equation of the curve.

10. A curve has equation \(y = \mathrm { f } ( x )\). The point \(P\) with coordinates \(( 9,0 )\) lies on the curve. Given that $$\mathrm { f } ^ { \prime } ( x ) = \frac { x + 9 } { \sqrt { } x } , \quad x > 0$$

1. find \(\mathrm { f } ( x )\).
2. Find the \(x\)-coordinates of the two points on \(y = \mathrm { f } ( x )\) where the gradient of the curve is equal to 10

1. (a) Use the substitution \(u = 4 - \sqrt { h }\) to show that

$$\int \frac { \mathrm { d } h } { 4 - \sqrt { h } } = - 8 \ln | 4 - \sqrt { h } | - 2 \sqrt { h } + k$$ where \(k\) is a constant A team of scientists is studying a species of slow growing tree.
The rate of change in height of a tree in this species is modelled by the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { t ^ { 0.25 } ( 4 - \sqrt { h } ) } { 20 }$$ where \(h\) is the height in metres and \(t\) is the time, measured in years, after the tree is planted.
(b) Find, according to the model, the range in heights of trees in this species. One of these trees is one metre high when it is first planted.
According to the model,
(c) calculate the time this tree would take to reach a height of 12 metres, giving your answer to 3 significant figures.

5

1. Find \(\int x \sqrt { x ^ { 2 } + 3 } \mathrm {~d} x\).
   (2 marks)
2. Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x \sqrt { x ^ { 2 } + 3 } } { \mathrm { e } ^ { 2 y } }$$ given that \(y = 0\) when \(x = 1\). Give your answer in the form \(y = \mathrm { f } ( x )\).