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

Question

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

Luke Alvarez · Accepted Answer

#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#

Charles Anctil · Answer

undefined To find the instantaneous velocity of an object moving according to the function $ f(t) = (2t - 1, e^t - 7) $ at $ t = 1 $, we need to calculate the derivative of the position function $ f(t) $ with respect to time $ t $, and then evaluate it at $ t = 1 $.

The position function $ f(t) $ is given by two components: $ f_1(t) = 2t - 1 $ and $ f_2(t) = e^t - 7 $.

To find the derivative of $ f_1(t) $, we differentiate $ 2t - 1 $ with respect to $ t $, yielding $ f_1'(t) = 2 $.

To find the derivative of $ f_2(t) $, we differentiate $ e^t - 7 $ with respect to $ t $, yielding $ f_2'(t) = e^t $.

Thus, the velocity function $ v(t) $ is $ v(t) = (2, e^t) $.

Now, to find the instantaneous velocity at $ t = 1 $, we evaluate $ v(t) $ at $ t = 1 $, which gives $ v(1) = (2, e^1) = (2, e) $.

So, the instantaneous velocity of the object at $ t = 1 $ is $ (2, e) $.