Differentiation Help!

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

(edited 13 years ago)

Reply 2

ztibor

10

adil12

Hey guys I am working on the following questions and I get stuck on them :s-smilie:

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? :s-smilie:

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
relating gradient.
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.

Reply 3

OP

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?

Reply 4

adil12

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"

Reply 5

OP

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? :smile:

Reply 6

adil12

Distance = 1/2 + 12.5 = 13, correct? :smile:

I'm not going to check all the arithmetic, but see attached - looks right.

OP

ghostwalker

I'm not going to check all the arithmetic, but see attached - looks right.

Yup seems right thanks for the graph :biggrin:

How would you approach question 2?

Reply 8

adil12

Yup seems right thanks for the graph :biggrin:

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.

Edit: Misread. :frown:

Corrected.

(edited 13 years ago)

Reply 9

OP

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.

Edit: Misread. :frown:

Corrected.

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

And it's fine :biggrin:

Reply 10

adil12

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

And it's fine :biggrin:

Does it's fine mean you've solved it now, or is that just replying to my "edit"?

Reply 11

OP

ghostwalker

Does it's fine mean you've solved it now, or is that just replying to my "edit"?

Oh sorry I was just replying to your edit :rolleyes:

Reply 12

adil12

Oh sorry I was just replying to your edit :rolleyes:

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.

Reply 13

OP

Sorry but once I get dy/dx, how can I find the gradient when I don't know the value of x? :s-smilie:

Reply 14

adil12

Sorry but once I get dy/dx, how can I find the gradient when I don't know the value of x? :s-smilie:

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?

(edited 13 years ago)

Reply 15

OP

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? :s-smilie:

Reply 16