The electric potential at a point (x;y)
on the line segment extending

from (0;3) to (2;0) is given by $\displaystyle P = 3x^2+2y^2$

At what point on this segment is the potential a minimum?

(Hint: the equation of the line segment is $\displaystyle y=-\frac 3 2x + 3$)

2. Originally Posted by joey1
The electric potential at a point (x;y)
on the line segment extending

from (0;3) to (2;0) is given by $\displaystyle P = 3x^2+2y^2$

At what point on this segment is the potential a minimum?

(Hint: the equation of the line segment is $\displaystyle y=-\frac 3 2x + 3$)
Hi joey,

try using the hint to replace $\displaystyle y$ with $\displaystyle -\frac{3}{2}x+3$

Then you will have a quadratic in x.

If you differentiate the quadratic with respect to x,
you can equate that to zero to discover which x causes the tangent
to the curve to be parallel to the x-axis.

Solve for the derivative=0

When you have found that x, use it to find the vertical co-ordinate corresponding to it on the line segment..

4. $\displaystyle P=3x^2+2y^2=3x^2+2\left(-\frac{3}{2}x+3\right)\left(-\frac{3}{2}x+3\right)=3x^2+2\left(\frac{9}{4}x^2-\frac{9}{2}x-\frac{9}{2}x+9\right)$

If you simplify that and write the quadratic equation, you can use calculus
to find the minimum point on the quadratic.

If you haven't done calculus, you will need to find the axis of symmetry of the curve.

Respond if you are still stuck with that.

5. okay this is what i have so far
$\displaystyle P=\frac {15}{2}x^2- 18x+18$
$\displaystyle p'=15x-18$
equate p'=0 therfore $\displaystyle x=\frac{18}{15}$
hope i am right so far
thereafter i went on to think that the co ordinates given that the values of x=0 and x=2
when x=2
p'=$\displaystyle 15(2)-18=12$
since the answer is >0 it is the minimum
to find the min value at x=2
$\displaystyle P=\frac{15}{2}(4)-18(2)+18=24$
the min values are (2:24),please tell me that is right,or i need serious help

6. Originally Posted by joey1
okay this is what i have so far
$\displaystyle P=\frac {15}{2}x^2- 18x+18$
$\displaystyle p'=15x-18$
equate p'=0 therfore $\displaystyle x=\frac{18}{15}$
hope i am right so far yes

thereafter i went on to think that the co ordinates given that the values of x=0 and x=2 this is where the original line segment crosses the x and y axes. You don't need to refer to these points at all.

when x=2
p'=$\displaystyle 15(2)-18=12$
since the answer is >0 it is the minimum the logic is incorrect
to find the min value at x=2
$\displaystyle P=\frac{15}{2}(4)-18(2)+18=24$
the min values are (2:24),please tell me that is right,or i need serious help
You are examining the quadratic equation to find the value of x at the minimum of the quadratic equation.

The quadratic equation generates the curve for P.

You found the minimum x co-ordinate on that.

$\displaystyle x=\frac{18}{15}=\frac{6}{5}=1.2$

The original question is asking you to find the point on the line segment given
at which the voltage is a minimum.

Place the x you discovered, corresponding to minimum potential back into the line equation to discover the corresponding value of y.

7. Originally Posted by Archie Meade
You are examining the quadratic equation to find the value of x at the minimum of the quadratic equation.

The quadratic equation generates the curve for P.

You found the minimum x co-ordinate on that.

$\displaystyle x=\frac{18}{15}=\frac{6}{5}=1.2$

The original question is asking you to find the point on the line segment given
at which the voltage is a minimum.

Place the x you discovered, corresponding to minimum potential back into the line equation to discover the corresponding value of y.
please forgive my ignorance but do i put the x value of 1.2 into this equation
$\displaystyle y=-\frac{3}{2}x +3$
if i do that then the y value is 1.2
therefore the points would be (1.2;1.2)
sorry to be a pest

8. is that right

9. Hi Joey,

1. "Do you understand why we substitute y into the equation for P ?"

2. "Do you understand why we differentiate the resulting quadratic ?"

3. "Do you understand why we substitute the minimum x back into the line equation ?"

The reason I'd rather ask those questions is because when you understand
why you are taking those steps in that order,
you are then mastering the technique and you will confidently answer
such questions in the future.

You will get the right answer (barring blunders) if you understand the technique.

Yes, the answer you got is correct.
If you try working it through again from start to finish, it will be worthwhile.

You could also find the minimum x without calculus, since quadratic curves are symmetrical about the turning point.

$\displaystyle \frac{15}{2}x^2-18x+18=18\ when\ x=0$

To find the other value of x making P=18

$\displaystyle \frac{15}{2}x^2-18x=18-18=0$

$\displaystyle x\left(\frac{15}{2}x-18\right)=0$

$\displaystyle \frac{15}{2}x-18=0$

$\displaystyle 15x=36$

$\displaystyle 5x=12$

$\displaystyle x=\frac{12}{5}$

The axis of symmetry which goes vertically through the minimum
lies halfway between 0 and 2.4, which is x=1.2

10. i finally understand after looking at the graph
thanks alot Archie Meade,i really appreciate it
you have been a great help
you are 1 of the greatest assets to this forum