Question #284623

Use the Chain Rule to determine an equation for the acceleration when 𝑎 = 𝑑𝑣/ 𝑑𝑡

When 𝑣 = (2𝑡2 + 3)4

and

When 𝑣 = ln(4𝑡3 − 1)


1
Expert's answer
2022-01-04T18:07:40-0500

1)When v=(2t2+3)4=f(g(t))v = (2t^2+3)^4 = f(g(t)), where f(t)=t4,g(t)=2t2+3f(t) = t^4, g(t) = 2t^2 + 3 .

Since dfdt=4t3\frac{df}{dt} = 4t^3 and dgdt=4t\frac{dg}{dt} = 4t , by the chain rule, we obtain that:


a=dvdt=dfdgdgdt=4(2t2+3)34t=16t(2t2+3)3a = \frac{dv}{dt} = \frac{df}{dg} \frac{dg}{dt} = 4(2t^2+3)^3 \cdot 4t = 16t(2t^2+3)^3


2)When v=ln(4t31)=f(g(t))v = ln(4t^3-1) = f(g(t)), where f(t)=ln(4t31),g(t)=(4t31)f(t) =ln( 4t^3-1), g(t) = (4t^3 - 1) .

Since dfdt=14t31\frac{df}{dt} =\frac{1}{4t^3-1} and dgdt=12t2\frac{dg}{dt} =12t^2 , by the chain rule, we obtain that:


a=dvdt=dfdgdgdt=14t3112t2=12t24t31a = \frac{dv}{dt} = \frac{df}{dg} \frac{dg}{dt} = \frac{1}{4t^3-1}\cdot 12t^2 = \frac{12t^2}{4t^3-1}



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