What is the instantaneous velocity of an object moving in accordance to # f(t)= (t/(e^t+2t),t) # at # t=4 #?

Question

What is the instantaneous velocity of an object moving in accordance to #              f(t)=   (t/(e^t+2t),t)    # at #  t=4    #?

Emma Aldridge · Accepted Answer

The instantaneous velocity is #=(-0.0418,1)ms^-1#

We need

#(u/v)'=(u'v-uv')/(v^2)#

Start by calculating the derivative of #(t/(e^t+2t))# wrt #t#

#x=t/(e^t+2t)#

#u=t#, #=>#, #u'=1#

#v=e^t+2t#, #=>#, #v'=e^t+2#

So,

#dx/dt=(1*(e^t+2t)-t(e^t+2))/(e^t+2t)^2#

#=(e^t+2t-te^t-2t)/((e^t+2t)^2)#

#=(e^t-te^t)/((e^t+2t)^2)#

#=(e^t(1-t))/(e^t+2t)^2#

Also,

#y=t#

#dy/dt=1#

Therefore,

#v(t)=f'(t)= ((e^t(1-t))/(e^t+2t)^2,1)#

So, when #t=4#

#v(4)=f'(4)= ((e^4(1-4))/(e^4+2*4)^2,1)#

#=(-0.0418,1)#

Emma Aldridge · Answer

undefined To find the instantaneous velocity of the object at $t = 4$ when it is moving according to $f(t) = \left(\frac{t}{e^t + 2t}, t\right)$, we need to find the derivative of the position function $f(t)$ with respect to time $t$. Then, evaluate the derivative at $t = 4$ to obtain the instantaneous velocity vector. Let's denote the position function as $f(t) = (x(t), y(t))$. Then, we can find the derivatives $x'(t)$ and $y'(t)$. After that, we substitute $t = 4$ into these derivatives to find the velocities $v_x(4)$ and $v_y(4)$. Finally, we can express the instantaneous velocity vector as $(v_x(4), v_y(4))$. Let's go through these steps.

1. Find $x'(t)$ and $y'(t)$ using the quotient rule.
$$x'(t) = \frac{(e^t + 2t) - t(2e^t + 2)}{(e^t + 2t)^2}$$
$$y'(t) = 1$$

2. Substitute $t = 4$ into $x'(t)$ and $y'(t)$ to find $v_x(4)$ and $v_y(4)$.
$$x'(4) = \frac{(e^4 + 8) - 4(2e^4 + 2)}{(e^4 + 8)^2}$$
$$y'(4) = 1$$

3. Calculate $v_x(4)$ and $v_y(4)$.

$$v_x(4) = \frac{(e^4 + 8) - 4(2e^4 + 2)}{(e^4 + 8)^2}$$
$$v_y(4) = 1$$

4. Express the instantaneous velocity vector at $t = 4$ as $(v_x(4), v_y(4))$.