Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter?

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Question 4:

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Answer 1 · 2018-05-16T15:12:23

#y=24x-40#

Given #x=f(t)# and #y=g(t)#, we can generalise the tangent equation as
#y=(g'(t))/(f'(t))x+(g(t)-f(t)((g'(t))/(f'(t))))#

#dy/dx=dy/dt*dt/dx=(2t-2)*(2sqrtt)=4(t-1)sqrtt#

#t=4# gives us:
#dy/dx=4(4-1)sqrt4=24#

#f(4)=sqrt4=2#
#g(4)=4^2-2(4)=8#

#8=2(24)+c#

#c=8-48=-40#

#y=24x-40#