What is the instantaneous velocity of an object moving in accordance to # f(t)= (2t-1,e^t-7)# at # t=1 #?

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Answer 1 · 2018-04-07T00:03:24

#e/2#

We have:

#x=2t-1# and #y=e^t-7#

Let's turn these equations around a bit.

#=>x+1=2t#

#=>(x+1)/2=t#

#=>y+7=e^t#

#=>ln(y+7)=t#

#=>(x+1)/2=ln(y+7)#

#=>e^((x+1)/2)=y+7#

#=>e^((x+1)/2)-7=y#

#=>d/dx(e^((x+1)/2)-7)=d/dx(y)#

Some rules:

#d/dx(e^x)=e^x#

#d/dx(C)=0# where #C# is a constant.

#d/dx(x^n)=nx^(n-1)# where #n# is a constant.

#d/dx(g(h(x))=g'(h(x))*h'(x)#

#=>e^((x+1)/2)*d/dx((x+1)/2)=dy/dx#

#=>e^((x+1)/2)*1/2*d/dx((x+1))=dy/dx#

#=>e^((x+1)/2)*1/2*1*x^(1-1)=dy/dx#

#=>e^((x+1)/2)*1/2*x^(0)=dy/dx#

#=>(e^((x+1)/2))/2=dy/dx#

Now...

If #t=1#:

#x=2*1-1#

#x=1#

#=>(e^((1+1)/2))/2=dy/dx#

#=>(e)/2=dy/dx#