# Differentiation: Tangents and Normals help please

• Oct 29th 2008, 10:11 AM
Gregory
Differentiation: Tangents and Normals help please
I have looked through loads of various notes and stuff and I'm sure I'm doing nothing wrong but my textbook keeps telling me I'm wrong.

Can someone help me with these two questions, very clearly showing ALL working please?

1) Find the equation of the tangent to the curve y=x[squared]+(-7/x[squared]) at the point (1, -6)
--- I keep finding the answer as "y=2x-8" but the book says "y=16x-22"

2) Find the equation of the normal to the curve y=x[squared]+(-8/[square root]x) at the point (1, -6)
--- I keep finding the answer as "x+8y-100=0" but the book says "17y+2x-212=0"

Thanks
• Oct 29th 2008, 11:27 AM
Ivar_sb
Question 1
The line required is of the form y=ax+b where a,b are constants.

Let the original function y be represented by f(x) = x^2 - 7/(x^2)

The line required will have gradient a = f '(1). That is, the gradient is equal to the derivative of f(x) taken at the given point (1,-6).

Now for the derivative

f '(x) = 2x + 14/(x^3)

(NOTE: if you have trouble differentiating the second section, you can write it first as -7x^(-2). Multiply -7 by the power of x (-2 x -7 = 14), the power of x goes down one and becomes -3. Put it back in the denominator and there is your x^3).

Hence the gradient is given by

a = f '(1) = 2*1+14/(1^3) = 16

We now have y=16x+b.

To find b, substitute x,y for the given point (1,-6):

-6=16*1+b

=> b=-22

and we have found the tangent to f(x) at the point (1,-6) to be

y=16x-22

I hope this helps :)
• Oct 29th 2008, 12:20 PM
Opalg
Quote:

Originally Posted by Gregory
2) Find the equation of the normal to the curve y=x[squared]+(-8/[square root]x) at the point (1, -6)
--- I keep finding the answer as "x+8y-100=0" but the book says "17y+2x-212=0"

There's something wrong here, because the point (1,-6) does not lie on the curve $y = x^2 -\frac8{\sqrt x}$. Please check the wording of the question!
• Oct 30th 2008, 01:25 AM
Gregory
Quote:

Originally Posted by Ivar_sb
The line required is of the form y=ax+b where a,b are constants.

Let the original function y be represented by f(x) = x^2 - 7/(x^2)

The line required will have gradient a = f '(1). That is, the gradient is equal to the derivative of f(x) taken at the given point (1,-6).

Now for the derivative

f '(x) = 2x + 14/(x^3)

(NOTE: if you have trouble differentiating the second section, you can write it first as -7x^(-2). Multiply -7 by the power of x (-2 x -7 = 14), the power of x goes down one and becomes -3. Put it back in the denominator and there is your x^3).

Hence the gradient is given by

a = f '(1) = 2*1+14/(1^3) = 16

We now have y=16x+b.

To find b, substitute x,y for the given point (1,-6):

-6=16*1+b

=> b=-22

and we have found the tangent to f(x) at the point (1,-6) to be

y=16x-22

I hope this helps :)

Thanks a lot :)

Quote:

Originally Posted by Opalg
There's something wrong here, because the point (1,-6) does not lie on the curve "y=x[squared]+(-8/[square root]x)." Please check the wording of the question!

You're right, okay I'll double check ...
sorry it was (4,12) - I looked at the point for the question above. sorry lol, can you help anyway?