# Equation of line & Gradient qus.

• Apr 16th 2006, 12:19 PM
Equation of line & Gradient qus.
Sorry for so many questions. Im just stuck on these and they are meant to be simple!

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1) Find the equation of the line through the origin which is parallel to the line 4x - 27 - 5 = 0

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2) The equation of a curve is y = (x-3)(x+4). Find the gradient of the curve:
a) at the point the curve crosses the y-axis
b) at each of the points where the curve crosses the x-axis

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3) Find the coordinates of the point(s) on the curve y=3x^3 - x + 8 at which the gradient is
a) 8
b) 0

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4) The equation of a curve s=4t^2 +5t. Find the gradient of the normal at each of the points where the curve crosses the t-axis.

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5) If the gradient of a curve is y=2x^2 -3x -2. Find
a) the gradient at the point where x=0
b) the co-ordinates of the point where y=0
c) the gradient at each of the points found in b)
• Apr 16th 2006, 12:27 PM
ThePerfectHacker
Quote:

Originally Posted by dadon
Sorry for so many questions. Im just stuck on these and they are meant to be simple!

------------------------------------------------------

1) Find the equation of the line through the origin which is parallel to the line 4x - 27 - 5 = 0

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This is a basic line $x=k$ thus, any parallel line also has the same equation, wince it passes through the origin it must be $x=0$
• Apr 16th 2006, 12:32 PM
ThePerfectHacker
Quote:

Originally Posted by dadon

2) The equation of a curve is y = (x-3)(x+4). Find the gradient of the curve:
a) at the point the curve crosses the y-axis
b) at each of the points where the curve crosses the x-axis

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Thus, we have,
$y=(x-3)(x+4)=x^2+x-12$
If the gradient means derivative then,
$y'=2x+1$

It passes the y-axis when $x=0$ thus
$y=0^2+0-12=-12$

It passes the x-axis when $y=0$ thus,
$0=x^2+x-12=(x-3)(x+4)$
Thus,
$x=\left\{ \begin{array}{c}-4\\3\end{array}\right$
• Apr 16th 2006, 12:36 PM
ThePerfectHacker
Quote:

Originally Posted by dadon

3) Find the coordinates of the point(s) on the curve y=3x^3 - x + 8 at which the gradient is
a) 8
b) 0

If,
$y=3x^3-x+8$
Then,
$y'=9x^2-1$
Yuu need it to be zero thus,
$0=9x^2-1$
Thus,
$9x^2=1$
Thus,
$x^2=1/3$
Thus,
$x=\pm \sqrt{1/3}=\pm \frac{\sqrt{3}}{3}$
• Apr 16th 2006, 01:24 PM
CaptainBlack
Quote:

Originally Posted by dadon
1) Find the equation of the line through the origin which is parallel to the line 4x - 27 - 5 = 0

You probably have a typo here, given the relative location of the y and 7
keys I will guess this should have been:

" Find the equation of the line through the origin which is parallel to the line 4x - 2y - 5 = 0"

This may be rewritten:

$y=2x-5/2$,

and hence has slope $2$, and so the equation through the origin
parallel to this also has slope $2$ and so has equation:

$
y=2x
$
.

RonL
• Apr 16th 2006, 01:30 PM
CaptainBlack
Quote:

Originally Posted by dadon
2) The equation of a curve is y = (x-3)(x+4). Find the gradient of the curve:
a) at the point the curve crosses the y-axis
b) at each of the points where the curve crosses the x-axis

Expanding the terms on the RHS and then differentiation gives:

$
y=x^2+x-12
$
,

so:

$
\frac{dy}{dx}=2x+1
$

a)The curve crosses the y axis when $x=0$, hence the gradient
of the curve is obtained by plugging this value of $x$ into the expression
for the derivative and so the required gradient is $1$

b)The curve crosses the x axis at $x=3$ and $x=-4$.
Plugging these into the expression for the derivative give gradients of $7$ and $-7$ respectively.

RonL
• Apr 16th 2006, 01:42 PM
CaptainBlack
Quote:

Originally Posted by dadon
3) Find the coordinates of the point(s) on the curve y=3x^3 - x + 8 at which the gradient is
a) 8
b) 0

The derivative of $y=3x^3-x+8$ is:

$
\frac{dy}{dx}=9x^2-1
$

a) Now this is $8$ when:

$
9x^2-1=8
$
,

or:

$
x^2=1
$

or $x=\pm 1$, which may be plugged into the equation for y to
give $y=10$ when $x=1$ and $y=6$ when $x=-1$.
So the required points are $(1,10)$ and $(-1,6)$.

b)Now this is $0$ when:

$
9x^2-1=0
$
,

or:

$
x^2=1/9
$

or $x=\pm 1/3$.

Plugging these values back into the equation for $y$ gives the other
coordinate, and so we find the required points are: $(1/3,8)$ and $(-1/3,8)$.

RonL
• Apr 16th 2006, 01:49 PM
CaptainBlack
Quote:

Originally Posted by dadon
4) The equation of a curve s=4t^2 +5t. Find the gradient of the normal at each of the points where the curve crosses the t-axis.

The points at which the curve crosses the t-axis correspond to the roots of:

$
4t^2+5t=0
$

which are $t=0$ and $t=-5/4$.

Now:

$
\frac{ds}{dt}=8t+5
$
,

so the slopes at the two points are $5$ and $-5$ respectively.

Now the gradient of the normal is -1 over the gradient of the tangent, and so
the gradients of the two normals are $-1/5$ and $1/5$ respectively.

RonL
• Apr 16th 2006, 02:00 PM
CaptainBlack
Quote:

Originally Posted by dadon
5) If the gradient of a curve is y=2x^2 -3x -2. Find
a) the gradient at the point where x=0
b) the co-ordinates of the point where y=0
c) the gradient at each of the points found in b)

This has a confusing use of y for the gradient of the curve, and assuming
this is right there is sufficient information to find only the x-coords for the
answer to part b), and c) is redundant as it is given as 0.

RonL
• Apr 17th 2006, 02:32 AM