Mathematics - Stuck on a question
Hi Mature student here.
Find the gradient of the curve with the equation y = 3(sqrt(x)) at point where:
a: x = 4
WORKINGS SO FAR
let x = 4, therefore y = 3 * 2 = 6 (sqrt(4) = 2)
I define this P
I then choose another point to the right of the x = 4 location
let x = 16, therefore y = 3 * 4 = 12 (sqrt(16) =4)
I define this as Q
Q co ordinates are (16, 12) which are also (x2, y2)
P co ordinates are (4, 6) which are also x1, y1)
Use formula (y2 - y1) / (x2 - x1) to find m (gradient)
12 - 6 = 6 and 16 - 4 = 12, therefore 6 / 12 = 0.5
THE ANSWER PAGE IS SAYING 0.75. What am I doing wrong.
Find the gradient of the curve with the equation y = 3(sqrt(x)) at point where:
a: x = 4
WORKINGS SO FAR
let x = 4, therefore y = 3 * 2 = 6 (sqrt(4) = 2)
I define this P
I then choose another point to the right of the x = 4 location
let x = 16, therefore y = 3 * 4 = 12 (sqrt(16) =4)
I define this as Q
Q co ordinates are (16, 12) which are also (x2, y2)
P co ordinates are (4, 6) which are also x1, y1)
Use formula (y2 - y1) / (x2 - x1) to find m (gradient)
12 - 6 = 6 and 16 - 4 = 12, therefore 6 / 12 = 0.5
THE ANSWER PAGE IS SAYING 0.75. What am I doing wrong.
(edited 1 year ago)
Reply 1
3
Ok I think I got it...
I was having a brain malfunction...
So what I had to do was take the derivative of y = 3 sqrt(x) WHICH gives the >>>GRADIENT<<<
Which would have been f '(x) = 3/2(x)^(-(1/2)) or 3/2(x)^1 half
I then substitute x = 4 and hey presto we have 3/4
RaWR.
I was having a brain malfunction...
So what I had to do was take the derivative of y = 3 sqrt(x) WHICH gives the >>>GRADIENT<<<
Which would have been f '(x) = 3/2(x)^(-(1/2)) or 3/2(x)^1 half
I then substitute x = 4 and hey presto we have 3/4
RaWR.
Reply 2
Original post by kjhhowell
Ok I think I got it...
I was having a brain malfunction...
So what I had to do was take the derivative of y = 3 sqrt(x) WHICH gives the >>>GRADIENT<<<
Which would have been f '(x) = 3/2(x)^(-(1/2)) or 3/2(x)^1 half
I then substitute x = 4 and hey presto we have 3/4
RaWR.
I was having a brain malfunction...
So what I had to do was take the derivative of y = 3 sqrt(x) WHICH gives the >>>GRADIENT<<<
Which would have been f '(x) = 3/2(x)^(-(1/2)) or 3/2(x)^1 half
I then substitute x = 4 and hey presto we have 3/4
RaWR.
You can always use the power of 1/2 as a square root when doing differentiation and integration!
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