Question

What is the instantaneous velocity of an object moving in accordance to `f(t)= (t^2,tcos(3t-pi/4))` at `t=(pi)/8`?

Answer 1

Instantaneous velocity of the object is `0.6023`

Explanation:

Instantaneous velocity of an object moving in accordance to `f(t)=(t^2,tcos(3t-pi/4))` at `t=pi/8`, will be given by the value `(dy)/(dx)` at `t=pi/8`.

As the object moves according to parametric form of equation given by `f(t)`,

`(dy)/(dx)=((dy)/(dt))/((dx)/(dt))`

= `(1xxcos(3t-pi/4)+txx(-sin(3t-pi/4)xx3))/(2t)`

= `(cos(3t-pi/4)-3tsin(3t-pi/4))/(2t)`

and at `t=pi/8`,

`(dy)/(dx)=(cos((3pi)/8-pi/4)-3pi/8xxsin((3pi)/8-pi/4))/(2xxpi/8)`

= `(cos(pi/8)-(3pi)/8sin(pi/8))/(pi/4)`

= `(0.92388-(3xx3.1416/8)xx0.38268)/(3.1416/4)`

= `(0.92388-0.45084)/0.7854`

= `0.47304/0.7854=0.6023`

Hence, instantaneous velocity of the object is `0.6023`