Asked by xxCeLoL
I got a few questions.
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!!!
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!!!
Answers
Answered by
Michael
5) 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.
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.
Answered by
xxCeLoL
I just looked up the quotient, product and product rule.
I'm screwed i cant remember all those for my test.!
I'm screwed i cant remember all those for my test.!
Answered by
Michael
Yes, 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.
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.
Answered by
xxCeLoL
Can i please get an elaboration of the questions.
Answered by
Michael
I 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.)
(I have to go now, so maybe someone else will take over.)
Answered by
bobpursley
A 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.
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.
Answered by
xxCeLoL
thanks