Friday

October 9, 2015
Posted by **xxCeLoL** on Monday, November 26, 2007 at 7:00am.

Hope ya'll can help out.

1) for F(X) = 6x - 2x^2

Find the gradient of the chord joining the point where the X coorinates are 1 and (1+h) respectively.

b) hence find the gradient at x=1

2) Find the Coordinates of the point on the curve Y=1/2x^3 - 3/2x^2 +2x +1

at which the tangent:

a) is parallel to the X axis

b) makes an angle with the x axis whose tangent is 2

c) is parallel to y - 6x - 1 = 0

3) Find the equation of the tangent and normal to y= SQUARE-ROOT(x) at the point where X= 4

4) Find the equations of the tangents to

Y = 1/3x^3 - x^2 - x + 1

At the points where the tangent is parallel to Y - 7x = 5

5) Find dy/dx for y^3 - xy + 7 = 8

find the gradient when X = 0

I know this is quite a few questions.

I've missed out on so much from maths I don't understand A LOT of these questions. So any help in advance (even 1 question) will be MUCH appreciated!!!

- Calculus - Maths -
**Michael**, Monday, November 26, 2007 at 7:07am5) You'll have to implicitly differentiate y with respect to x. That means differentiate x normally and apply the chain rule when you differentiate y. Don't forget the product rule with the x*y. I don't know about the gradient part.

4) Find y' by differentiating. Find your slope from y - 7x = 5. (Parallel lines have equal slopes.)

3) Find the derivative. Evaluate the derivative at the given point, x=4. Solve for b (in y = mx+b).

2) Start by differentiating.

1) I don't know anything about gradients.

Work with those hints and see what you get. I can help you where you get stuck.

- Calculus - Maths -
**xxCeLoL**, Monday, November 26, 2007 at 7:14amI just looked up the quotient, product and product rule.

I'm screwed i cant remember all those for my test.!

- Calculus - Maths -
**Michael**, Monday, November 26, 2007 at 7:33amYes, you can...

The product rule is "first, derivative of the second + second, derivative of the first."

The quotient rule is "bottom, derivative of the top minus top, derivative of the bottom all over the bottom squared."

(The commas are multiplication.) Say those statements in your head every time you differentiate using the quotient or product rule.

- Calculus - Maths -
- Calculus - Maths -
**xxCeLoL**, Monday, November 26, 2007 at 7:50amCan i please get an elaboration of the questions.

- Calculus - Maths -
**Michael**, Monday, November 26, 2007 at 7:53amI can't do them for you. You can work on them and let us know where you're having difficulties understanding.

(I have to go now, so maybe someone else will take over.)

- Calculus - Maths -
- Calculus - Maths -
**bobpursley**, Monday, November 26, 2007 at 7:59amA gradient is the first derivative with respect to direcion, in this case, x.

Example: Temperature gradient.

dTemp/dx is the temperature gradient in the direction of x.

dTemp/dy is the termperature gradient in the direction of y.

Gradients are normally vectors.

- Calculus - Maths -
**xxCeLoL**, Monday, November 26, 2007 at 9:42pmthanks