- The vertical height, \(h \mathrm {~m}\), above horizontal ground, of a passenger on a fairground ride, \(t\) seconds after the ride starts, where \(t \leqslant 5\), is modelled by the differential equation
$$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } h } { \mathrm {~d} t ^ { 2 } } - 2 t \frac { \mathrm {~d} h } { \mathrm {~d} t } + 2 h = t ^ { 3 }$$
- Given that \(t = \mathrm { e } ^ { x }\), show that
- \(t \frac { \mathrm {~d} h } { \mathrm {~d} t } = \frac { \mathrm { d } h } { \mathrm {~d} x }\)
- \(t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } h } { \mathrm {~d} t ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } h } { \mathrm {~d} x ^ { 2 } } - \frac { \mathrm { d } h } { \mathrm {~d} x }\)
- Hence show that the transformation \(t = \mathrm { e } ^ { x }\) transforms equation (I) into the equation
$$\frac { \mathrm { d } ^ { 2 } h } { \mathrm {~d} x ^ { 2 } } - 3 \frac { \mathrm {~d} h } { \mathrm {~d} x } + 2 h = \mathrm { e } ^ { 3 x }$$
- Hence show that
$$h = A t + B t ^ { 2 } + \frac { 1 } { 2 } t ^ { 3 }$$
where \(A\) and \(B\) are constants.
Given that when \(t = 1 , h = 2.5\) and when \(t = 2 , \frac { \mathrm {~d} h } { \mathrm {~d} t } = - 1\)
- determine the height of the passenger above the ground 5 seconds after the start of the ride.