Finding the Gradient of the Secant on a Curve: P and Q Coordinates Given

In summary, the conversation discusses finding the coordinates of two points on a given curve and using them to find the gradient of the secant. The solution involves evaluating the coordinates in terms of x and h and finding a formula for the gradient of the secant. It is important to note that the gradient of the secant is different from the gradient of the tangent at a specific point. The final step is to take the limit h->0 to find the gradient of the secant.
  • #1
zebra1707
107
0

Homework Statement



There are two parts to this problem.

On the curve 2x^2-5 lie two points P and Q. Let the abscissa of P be "x" and the abscissa of Q be "x+h". No numerical coordinates are given.

a) State the coordinates of P and Q.

b) Using these points find the gradient of the secant PQ


Homework Equations





The Attempt at a Solution



For the first part of the question I am assuming that the coordinates would be P (x, f(x)) and Q (x+h, f(x+h))

Now when it comes to finding the gradient of the secant - Can use a table to find a set of coordinates and then plug in the m = y2 - y1 / x2 - x1 ? Not should why the question would ask to use the above coordinates when the gradient of the secant is a different equation?

Could use some guidence on how I attack this problem. Many thanks.
 
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  • #2
Just evaluate the coordinates of P and Q in terms of x and h and using the explicit function f(x)=2x^2-5. Find a formula for the gradient of the secant. Nothing else to do.
 
  • #3
Dick said:
Just evaluate the coordinates of P and Q in terms of x and h and using the explicit function f(x)=2x^2-5. Find a formula for the gradient of the secant. Nothing else to do.

Okay - correct me here if necessary

f(x) = 2x^2 - 5
f(x+h) = 2(x+h)^2 - 5
= 2x^2+4xh+2h^2 - 5

f(x+h) - f(x) = 4xh+2h^2

f(x+h)-f(x) / h = 4x + 2h

f'(x) lim = 4x+2h
approach 0 = 4x gradient of the secant is 4x

Can you please take a moment to check my calculation to confirm my understanding.
 
  • #4
The calculation is fine. But what you have found is that the gradient of the secant is 4x+2h. Yes, if you take the limit h->0 you get 4x, but that's not the gradient of the secant. That's the gradient of the TANGENT at x. You are jumping ahead.
 
  • #5
Dick said:
The calculation is fine. But what you have found is that the gradient of the secant is 4x+2h. Yes, if you take the limit h->0 you get 4x, but that's not the gradient of the secant. That's the gradient of the TANGENT at x. You are jumping ahead.

Many thanks. So, 4x+2h is the gradient of the secant and I should leave the statement there?
 
  • #6
zebra1707 said:
Many thanks. So, 4x+2h is the gradient of the secant and I should leave the statement there?

Sure.
 
  • #7
Many thanks Dick - this is greatly appreciated.

Cheers
 

What is the gradient of the secant?

The gradient of the secant is defined as the slope of the line connecting two points on a curve. It is a measure of the rate of change of the curve between those two points.

How is the gradient of the secant calculated?

The gradient of the secant can be calculated by taking the difference in the y-coordinates of the two points and dividing it by the difference in the corresponding x-coordinates. This can be represented by the formula (y2-y1)/(x2-x1).

What is the significance of the gradient of the secant?

The gradient of the secant is important in calculus because it is used to approximate the slope of a curve at a specific point. It also helps in understanding the rate of change of a function over a given interval.

How does the gradient of the secant relate to the derivative?

The gradient of the secant is a precursor to the derivative. As the two points used to calculate the gradient get closer and closer together, the gradient of the secant approaches the value of the derivative at that point.

Can the gradient of the secant be negative?

Yes, the gradient of the secant can be negative. This indicates that the curve is decreasing in the interval between the two points. A positive gradient indicates an increasing curve, and a gradient of zero indicates a horizontal line.

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