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I need to differentiate $x(t)=e^{2t}(4t^2-3t+1)$ to find velocity and acceleration at $3$ seconds.

I need to use the product rule.

I know $e^{2t} = u$ so $du(t)/dt = 2e^{2t}$

and $(4t^2-3t+1) = v$ and $dv= (8t-3)$

so $dy/dx = udv + vdu = (e^{2t})(8t-3)+(4t^2-3t+1)(2e^{2t})$

I am completely stuck where to go next

please help

Umberto
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baffled
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  • You're almost there: now just substitute $;t=3;$ in the derivative to get the velocity, and for the acceleration you'll need the second derivative... – DonAntonio Dec 18 '13 at 18:54
  • Just as a hint. Remember $v(t)=dx(t)/dt$ and $a(t)=d^2x(t)/dt^2=dv(t)/dt$. In addition remember that $d(fg)/dt=g df/dt+fdg/dt$ where f and g are functions. – Umberto Dec 18 '13 at 19:03

2 Answers2

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You've done fine, now just factor out the factor $e^{2t}$ from each term and simplify! (It will make finding the second derivative, acceleration, easier if you simplify).

$$f'(t) = v(t) =(e^{2t})(8t-3)+(4t^2-3t+1)(2e^2t) = (e^{2t})\Big((8t-3)+(8t^2-6t+2)\Big) = e^{2t}(8t^2 + 2t - 1)$$

Evaluate at $t = 3$ to obtain velocity:

$$v(3) = f'(3) = e^{6}(8(3)^2 + 2(3) - 1) = e^6(72 + 8 - 1) = 79e^6$$

Then, differentiate $f'(t)$ using the product rule again, and then evaluate at $t = 3$ to find acceleration at time = 3 seconds.

$$f''(t) = a(t) = 2e^{2t}(8t^2 + 2t - 1) + 4e^{2t}(16t + 2)= 2e^{2t}\Big(8t^2 + 2t - 1 +32t + 4\Big) = 2e^{2t}(8t^2 + 34t + 3).\\$$ $$f''(3) = a(3) = 2e^6(72 + 102 + 3) = 354e^6$$

amWhy
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You're confusing yourself with substitution. You're introducing a lot of symbols that might be obscuring the point. Try this:

$$x(t) = \underbrace{e^{2t}}_{\textrm{first term}} \underbrace{\left(4t^2-3t+1\right)}_{\textrm{second term}}.$$

$$\begin{align*} v(t) = x'(t) &= \left[\frac{d}{dt}(\textrm{first term})\right] \cdot (\textrm{second term}) + (\textrm{first term})\cdot \left[\frac{d}{dt}(\textrm{second term})\right] \\ &= 2e^{2t}(4t^2-3t+1)+e^{2t}(8t-3). \end{align*}$$

Do the same thing again to get the second derivative, $a(t) = v'(t) = x''(t)$.

Then, the velocity at $t=3$ seconds is just $v(t=3)$.

Emily
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