Question

if position of object x=3sin-7cos then motion of object is bounded between​

Answer 1

Given :  position of object x=3sint-7cost

To Find : motion of object is bounded between​

Solution:

x = 3sint - 7 cost

dx/dt  =  3cost + 7 sint

=> 3cost + 7 sint  = 0

=> sint/cost  = - 3/7

=> tant  = - 3/7

sint t = ±3/√58   , cost = ±7/√58

case1 :  sint = - 3/√58   and cost = 7/√58

case 2 : sint = 3/√58   and cos t  = -7/√58

d²x/dt²  = -3sint  +7cost

case 1 : sint = - 3/√58   and cost = 7/√58

=> d²x/dt²  = -3( - 3/√58  )  +7(7/√58)  > 0 hence minimum value

case 1 : sint =   3/√58   and cost =- 7/√58

=> d²x/dt²  = -3(  3/√58  )  +7(-7/√58)  < 0 hence maximum value

x = 3sint - 7 cost

sint = - 3/√58   and cost = 7/√58

=> x = -9/√58 - 49/√58  = - √58   minimum

sint =   3/√58   and cost = -7/√58

=> x = +9/√58+49/√58  =   √58   maximum

motion of object is bounded between​  - √58   and     √58

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Answer 2

Position of object given:

$\rm x = 3 \sin(t) - 7 \cos(t)$

• Now, let's try to represent the equation in a single trigonometric term.

$\rm \implies x = (\sqrt{ {3}^{2} + {7}^{2} }) \bigg \{ \dfrac{3}{ \sqrt{ {3}^{2} + {7}^{2} } } \sin(t) - \dfrac{7}{ \sqrt{ {3}^{2} + {7}^{2} } } \cos(t) \bigg \}$

$\rm \implies x = (\sqrt{58}) \bigg \{ \dfrac{3}{ \sqrt{58} } \sin(t) - \dfrac{7}{ \sqrt{58} } \cos(t) \bigg \}$

• Now, let 3/√58 be $\cos(\theta)$, then 7/√58 will be $\sin(\theta)$.

$\rm \implies x = (\sqrt{58}) \bigg \{\sin(t) \cos( \theta) - \cos(t) \sin( \theta) \bigg \}$

$\rm \implies x = (\sqrt{58}) \bigg \{\sin(t - \theta)\bigg \}$

• Now, max value of sin is +1 and min value of sin is -1.

So, position will be within √58 and -√58.