You are Here: Home >< Maths

# Differentiation Help! watch

1. Hey guys I am working on the following questions and I get stuck on them These are the questions I am stuck on:

1) The curve y=x^2 - 3x -4 crosses the x-axis at P and Q. The tangents to the curve at P and Q meet at R. The normals to the curve at P and Q meet at S. Find the distance RS.

2) At a particular point on the curve y= 5x^2 - 12x +1 the equation of the normal is x + 18y +c =0. Find the value of the constant c.

So for the first one, I know when y=0, the equation will factorise to (x+1)(x-4) = 0, hence x= -1,4. Then I do dy/dx to get 2x-3. When x = -1, gradient = -5 and when x = 4, gradient = 5. Now what do I do?
2. 1) Work out the two equations for the tangents and find where they intersect (R), similarly for the two normals (S), and then work out the distance between the two.

Rather a tediously long question, IMHO.

Edit: PS, by symmetry of the quadratic about its lower point, both the normals and tangents will intersect at x= 3/2
Hey guys I am working on the following questions and I get stuck on them These are the questions I am stuck on:

1) The curve y=x^2 - 3x -4 crosses the x-axis at P and Q. The tangents to the curve at P and Q meet at R. The normals to the curve at P and Q meet at S. Find the distance RS.

2) At a particular point on the curve y= 5x^2 - 12x +1 the equation of the normal is x + 18y +c =0. Find the value of the constant c.

So for the first one, I know when y=0, the equation will factorise to (x+1)(x-4) = 0, hence x= -1,4. Then I do dy/dx to get 2x-3. When x = -1, gradient = -5 and when x = 4, gradient = 5. Now what do I do?
1)write the equation of tangent lines
y-y0=m(x-x0) where (x0,y0) is coordinates of P or Q and m is the
THen solve the equations simultaneously you will get R
For normals use same equation but with -1/m gradient
an solve simultaneously ->S
THen calculate the length of vector RS
2) From the equation of normal the gradient is -1/8
Get the derivative of y and find that x value where the derivative
is -1/8. Calculate the value of y at x.
Substitue the coordinates of this point in the equation of normal
and you will get c.
4. (Original post by ghostwalker)
1) Work out the two equations for the tangents and find where they intersect (R), similarly for the two normals (S), and then work out the distance between the two.

Rather a tediously long question, IMHO.

Edit: PS, by symmetry of the quadratic about its lower point, both the normals and tangents will intersect at x= 3/2
Ok then, so Line 1 would be y = -5x + c, when x=-1, y=0, hence 5 + c =0, c + -5.
Line 2 would be y = 5x+c, when x=4, y=0, hence 20+c = 0, c = -20.
So -5x-5 = 5x-20, 10x = 15, x = 1.5. When x = 1.5, y = -12.5.

How would you do it for the normals?
Ok then, so Line 1 would be y = -5x + c, when x=-1, y=0, hence 5 + c =0, c + -5.
Line 2 would be y = 5x+c, when x=4, y=0, hence 20+c = 0, c = -20.
So -5x-5 = 5x-20, 10x = 15, x = 1.5. When x = 1.5, y = -12.5.

How would you do it for the normals?
Same as the tangents, except you get the gradient for the normal from the fact that "gradient of tangent times gradient of normal = - 1"
6. (Original post by ghostwalker)
Same as the tangents, except you get the gradient for the normal from the fact that "gradient of tangent times gradient of normal = - 1"
Ok then, so gradient of normal at P = 1/5 and gradient of normal at Q = -1/5.
y = 1/5x + c and when x= -1, y=0, hence -1/5 + c =0, c = 1/5.
And for the second line, when x = 4, y = 0, hence y = -4/5 + c = 0, c= 4/5.
And then 1/5x + 1/5 = -1/5x + 4/5,
2/5x = 3/5,
x = 1.5.
y = 1/5(1.5) + 1/5 = 1/2,
Distance = 1/2 + 12.5 = 13, correct?
Distance = 1/2 + 12.5 = 13, correct?
I'm not going to check all the arithmetic, but see attached - looks right.
Attached Images

8. (Original post by ghostwalker)
I'm not going to check all the arithmetic, but see attached - looks right.
Yup seems right thanks for the graph How would you approach question 2?
Yup seems right thanks for the graph How would you approach question 2?
In a nutshell:

Find the point on the curve where the normal has the same gradient as that straight line, then sub that point into that line to find c.

10. (Original post by ghostwalker)
In a nutshell:

Find the point on the curve where the normal has the same gradient as that straight line, then sub that point into that line to find c.

So dy/dx would be 10x - 12, but then what would you do?

And it's fine
So dy/dx would be 10x - 12, but then what would you do?

And it's fine
Does it's fine mean you've solved it now, or is that just replying to my "edit"?
12. (Original post by ghostwalker)
Does it's fine mean you've solved it now, or is that just replying to my "edit"?

Here's a synopsis of how to do it, as this is going to be my last post for a while.

So what's the gradient of your straight line which is the normal?

Hence what's the gradient of the tangent there?

Equate to dy/dx and solve for x to find the value of x that has that gradient for the tangent, and hence that other gradient for the normal.

Work out what the point on the curve is.

Sub into the equation for the normal since that point lies on the normal, and solve for c.
14. Sorry but once I get dy/dx, how can I find the gradient when I don't know the value of x?
Sorry but once I get dy/dx, how can I find the gradient when I don't know the value of x?
You seem to have ignored the first two steps in my synopsis.

Given the equation of a straight line, do you know how to find its gradient?
16. (Original post by ghostwalker)
You seem to have ignored the first two steps in my synopsis.

Given the equation of a straight line, do you know how to find its gradient?
Oh yes I do, so dy/dx of the tangent = 10x -12, but without a value of x to get the value of the gradient, how am I to get he gradient of the normal?
Oh yes I do, so dy/dx of the tangent = 10x -12, but without a value of x to get the value of the gradient, how am I to get he gradient of the normal?
You have the equation for the normal - it's given in the question - and that's where you get the value of the gradient for the normal from.

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: October 2, 2010
The home of Results and Clearing

### 2,341

people online now

### 1,567,000

students helped last year
Today on TSR

### University open days

1. Keele University
Sun, 19 Aug '18
2. University of Melbourne
Sun, 19 Aug '18
3. Sheffield Hallam University
Tue, 21 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams