What is the instantaneous velocity of an object moving in accordance to $f(t)= (t^2sin(t-pi),tcost)$ at $t=pi/3$ ?

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Answer 1 · 2016-01-02T04:37:08

$0.172261$

The instantaneous velocity is equal to $f'(pi/3)$ .

$x(t)=t^2sin(t-pi)$

To find $x'(t)$ , use the product rule.

$x'(t)=2tsin(t-pi)+t^2cos(t-pi)$

We also know that

$y(t)=tcost$

Again, differentiate with the product rule.

$y'(t)=cost-tsint$

The derivative of the entire parametric equation is found as follows:

$f'(t)=(y'(t))/(x'(t))=(cost-tsint)/(2tsin(t-pi)+t^2cos(t-pi))$

Find $f'(pi/3)$ .

$f'(pi/3)=(cos(pi/3)-pi/3sin(pi/3))/(2(pi/3)sin(pi/3-pi)+(pi/3)^2cos(pi/3-pi))$

$approx0.172261$

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