MedVision ad

Calculus question! (2 Viewers)

yeezus

New Member
Joined
Jul 10, 2013
Messages
8
Gender
Male
HSC
N/A
Hey, I'm in year 10 right now, so forgive the lack of knowledge :p

Question 1: The curve y=root(x-3) has a tangent with gradient 1/2 at point N. Find co-ordinates for N
Question 2: The function f(x) =3rootx has f '(x) =3/4 evaluate x

Thanks!
 

AnimeX

Member
Joined
Aug 11, 2012
Messages
588
Gender
Male
HSC
N/A
Hey, I'm in year 10 right now, so forgive the lack of knowledge :p

Question 1: The curve y=root(x-3) has a tangent with gradient 1/2 at point N. Find co-ordinates for N
Question 2: The function f(x) =3rootx has f '(x) =3/4 evaluate x

Thanks!
Question 1: Differentiate y with respect to x, sub in dy/dx =1/2 and solve for x, then put that point back into y=sqrt[x-3] and you should have N.

Question 2: I'm not 100% sure what this question is asking, possibly it means find x. So, differentiate f(x) and make it equal 3/4, solve for x, should be the answer.

^.^
 

yeezus

New Member
Joined
Jul 10, 2013
Messages
8
Gender
Male
HSC
N/A
for question 1 I differentiate y=root(x-3) and get 1/2sqroot(x-3) and when i try sub in gradient = 1/2 I get a math error :L
 

bedpotato

Member
Joined
Jul 10, 2013
Messages
337
Gender
Undisclosed
HSC
2013
for question 1 I differentiate y=root(x-3) and get 1/2sqroot(x-3) and when i try sub in gradient = 1/2 I get a math error :L
What do you mean "sub in gradient". You let y' = 1/2, and solve for x. And then sub in the value of x back into y to find the y coordinate.
 

yeezus

New Member
Joined
Jul 10, 2013
Messages
8
Gender
Male
HSC
N/A
What do you mean "sub in gradient". You let y' = 1/2, and solve for x. And then sub in the value of x back into y to find the y coordinate.
Ah! OK it becomes 1/2= 1/2sqroot(x-3) and solve, thanks! didn't see it at first :(
 

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top