Caculus question on finding deriavatives?

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Answer 1 · 2017-03-10T13:22:50

# v=40(1-e^(-t/4)).#

#s=160(1/4t-1+e^(-t/4))#.

#:. v=(ds)/dt=160{1/4d/dt(t)-d/dt(1)+d/dt(e^(-t/4))},#

#=160{1/4(1)-0+e^(-t/4)d/dt(-t/4)},#

#=160{1/4+(e^(-t/4))(-1/4)(1)},#

#=160/4(1-e^(-t/4)),#

#rArr v=40(1-e^(-t/4)).#

Enjoy Maths.!

Answer 2 · 2017-03-10T13:29:06

#v = 40(1 - e^(-t/4))#

Velocity is actually defined as the rate of change of the position of an object, let's say #s#, with time, #t#

#v = (Deltas)/(Deltat)#

In other words, the velocity of an object tells you how the position of an object changes per unit of time.

You can thus find the velocity of an object by taking the derivative of its position with respect to time

#v = d/dts(t)#

In your case, you have

#s = 160 * (1/4t - 1 + e^(-t/4))#

This is equivalent to

#s = 40t - 160 + 160e^(-t/4)#

The derivative of this function will be

#d/dts(t) = d/dt(40t - 160 + 160e^(-t/4))#

#s^'(t) = d/dt(40t) - d/dt(160) + d/dt(160e^(-t/4))#

Now, use the chain rule to find the derivative of #e^(-t/4)#

#d/dt(e^(-t/4)) = color(blue)(e^(-t/4) * d/dt(-t/4))#

You will thus have

#s^'(t) = 40 - 0 + 160d/dt(e^(-t/4))#

#s^'(t) = 40 + 160 * [color(blue)(e^(-t/4) * d/dt(-t/4))]#

#s^'(t) = 40 + 160 * [e^(-t/4) * (-1/4)]#

which simplifies to

#s^'(t) = 40 - 40e^(-t/4)#

Therefore, the velocity of the skydiver, #v#, at a time #t# will be equal to

#color(darkgreen)(ul(color(black)(v = 40(1 - e^(-t/4)))))#