Math Is Fun Forum

jigbobber

Member

Registered: 2008-08-13

Posts: 4

Help with some calculus

1.
a) Find the equation of the tangent to the curve

at the point where

b)the tangent in a) cuts the x-axis at

and the y-axis at

. Locate these points.
c)Find the area of the triangle

2.
Find the equation of the normal to the curve

at the point

. write this equation in general form.
3.
Let

a) find

b)find d^2y/dx^2, and show that (d^2y/dx^2)-b^2y=0
c)deduce an expression for d^ny/dx^n in terms of y

any help would be appreciated, thanks

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Help with some calculus

1.a)

x = 1

sub into y = mx+c

.b)

.c)
right angled triangle:
area = ½×base×height

I don't have time to go further.

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gnitsuk

Member

Registered: 2006-02-09

Posts: 121

Re: Help with some calculus

2)

Call this "Equation (1)".

Differentiation gives us the gradient of this curve at any point:

We know that two lines which are mutually perpendicular have gradients whose product is -1. Therefore the gradient of the line we require the equation of is:

This is the gradient of the perpendicular to the curve at any point on the curve with a given x ordinate.

So we know that the equation of the perpendicular to the curve at any given x ordinate will have the form:

Specifically, we are interested in the point where x = a, hence the equation at this point will be:

Call this "Equation (2)".

We just need to know the value for C. Well, we know one point on this line. The point (a,2a^3+7a-3) is on this line - this from equation (1).

So in Equation (2) we simply substitute in X = a and Y = 2a^3+7a-3 to give:

So putting this value of C into Equation (2) gives:

Call this Equation (3).

This is the equation we desire, it now has to be put into general form. At the moment it is in the form y = mx + c but general form would be ax + by + c = 0

So, working on Equation (3) yeilds:

This is our answer.

To see an instance of this result, let's set a=2. Then Equation (2) becomes:

A plot of this should yield a line which cuts the curve perpendicularly at x = 2.

Indeed it does

Last edited by gnitsuk (2008-09-10 23:57:49)

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Help with some calculus

3.
Let

a) find

b)find d^2y/dx^2, and show that (d^2y/dx^2)-b^2y=0
c)deduce an expression for d^ny/dx^n in terms of y

any help would be appreciated, thanks

a.

b.

c.

Last edited by luca-deltodesco (2008-09-10 23:52:16)

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The Beginning Of All Things To End.
The End Of All Things To Come.