Grade 11 Math Help Quadratic functions/ physics

[Zacko282]

1. Determine the equation that represents the relationship between the power and the current when the electric potential difference is 24v and the resistance is 1.5 Ω. 2. Draw a graph of the parabola that corresponds to the equation found in (a). 3. Determine the current needed in order for the device to use maximum power.

2. P = IV - I^2 R. P = IV3. P = 16 x 24 - 256 x 1.5 = 384 - 384 = 0
P (I) = 256I + 16I - 1.5

Ive been stuck on the questions for a month now and not even really sure how to turn a physics equation into a quadratic function. Any help is appreciated.

 

[Mark44]

1. Determine the equation that represents the relationship between the power and the current when the electric potential difference is 24v and the resistance is 1.5 Ω. 2. Draw a graph of the parabola that corresponds to the equation found in (a). 3. Determine the current needed in order for the device to use maximum power.

2. P = IV - I^2 R. P = IV3. P = 16 x 24 - 256 x 1.5 = 384 - 384 = 0
P (I) = 256I + 16I - 1.5

Ive been stuck on the questions for a month now and not even really sure how to turn a physics equation into a quadratic function. Any help is appreciated.

Substitute the values for V and R in the first equation in item 2 above. Then you will have a quadratic equation in P and I. I don't know why you substituted 16 for I in the equation you wrote.

After you have the equation with P as a function of I, sketch a graph, which I on the horizontal axis and P on the vertical axis. From the graph you should have a good idea about what current is associated with the maximum power value.

 

[Zacko282]

Substitute the values for V and R in the first equation in item 2 above. Then you will have a quadratic equation in P and I. I don't know why you substituted 16 for I in the equation you wrote.

After you have the equation with P as a function of I, sketch a graph, which I on the horizontal axis and P on the vertical axis. From the graph you should have a good idea about what current is associated with the maximum power value.

so the end equation would look something like this? P = I24 - I^2 1.5 ? or would it be P = I (24) - (1.5)^2 ?
and for example if i put it into an equation of y = a (x - h)^2 + k would it be y = 24 (x - 0) + 1.5 ? Thanks for the help btw.

 

[Mark44]

so the end equation would look something like this? P = I24 - I^2 1.5 ?

This one, but I would write it as $P = 24 I - 1.5 I^2$ with the coefficients in front of the variables.

or would it be P = I (24) - (1.5)^2 ?

No. You lost the $I^2$ term. In that term I should be squared, not 1.5.

and for example if i put it into an equation of y = a (x - h)^2 + k would it be y = 24 (x - 0) + 1.5 ? Thanks for the help btw.

No. For one thing, don't switch from I and P to x and y. For another thing your coefficients are in the wrong places.

Just factor the correct equation, and that will give you the current (I) values for which P is zero. The I value for maximum current will be exactly in the middle of the two I-intercepts.

 

[Zacko282]

This one, but I would write it as $P = 24 I - 1.5 I^2$ with the coefficients in front of the variables.
No. You lost the $I^2$ term. In that term I should be squared, not 1.5.

No. For one thing, don't switch from I and P to x and y. For another thing your coefficients are in the wrong places.

Just factor the correct equation, and that will give you the current (I) values for which P is zero. The I value for maximum current will be exactly in the middle of the two I-intercepts.

Ok, so The answer i get should be P = 22.5I then?

 

[Mark44]

Ok, so The answer i get should be P = 22.5I then?

No, you're way off. Are you thinking that $24 I - 1.5 I^2 = 22.5 I$? If so, you need to review your algebra.

 

[Zacko282]

No, you're way off. Are you thinking that $24 I - 1.5 I^2 = 22.5 I$? If so, you need to review your algebra.

Do i simplify both sides of the equation and then isolate the variable ?

 

[Mark44]

Do i simplify both sides of the equation and then isolate the variable ?

No.
Your equation is $P = 24 I - 1.5 I^2$, which is already a quadratic equation in the variable I.
Do you know to graph a quadratic equation? The graph will be a parabola.
Do you know how to find the vertex of a parabola? That point is where P has its maximum value.

To be honest, I'm starting to see why you have been working on this problem for a month without success.

 

[Zacko282]

No.
Your equation is $P = 24 I - 1.5 I^2$, which is already a quadratic equation in the variable I.
Do you know to graph a quadratic equation? The graph will be a parabola.
Do you know how to find the vertex of a parabola? That point is where P has its maximum value.

To be honest, I'm starting to see why you have been working on this problem for a month without success.

Yes, i believe so, I find the x- intercepts of the equation and then use that to determine the a of the equation.

 

[Mark44]

Yes, i believe so, I find the x- intercepts of the equation and then use that to determine the a of the equation.

Isn't it obvious that a = -1.5? Also, you want the I-intercepts. The axis of symmetry of the parabola is halfway between the two I-intercepts. You can use the halfway value to find the high point of the parabola.

 

[Zacko282]

Isn't it obvious that a = -1.5? Also, you want the I-intercepts. The axis of symmetry of the parabola is halfway between the two I-intercepts.

No, the work document sent to me by the teacher doesn't even teach me how to solve this kind of question =/. and then does that mean for the equation in vertex form it would be y = 1.5 (x - 24) + 0 or (x - 0) + 24? I'm sorry I'm kind of slow =( This is the only problem all unit I've had trouble with.

 

[Mark44]

Please stop using x and y. You'll just confuse yourself, as the equation you're working with involves I and P.

No, the work document sent to me by the teacher doesn't even teach me how to solve this kind of question =/. and then does that mean for the equation in vertex form it would be y = 1.5 (x - 24) + 0 or (x - 0) + 24?

The equation is $P = 24 I - 1.5 I^2$, or $P = -1.5 I^2 + 24 I$.

If I translate what you have above, I get $P = 1.5(I - 24)$ or $(I - 0) + 24$
I don't know what you are doing here, since an equation doesn't have "or" in it, and neither of these can be transformed to get back to $P = -1.5 I^2 + 24 I$.

Have you worked with quadratic functions before? Do you know about completing the square to find the vertex of a parabola that is represented by a quadratic function? If you know this technique, it's a lot harder than what I've suggested, about finding the axis of symmetry.

 

[Zacko282]

Please stop using x and y. You'll just confuse yourself, as the equation you're working with involves I and P.The equation is $P = 24 I - 1.5 I^2$, or $P = -1.5 I^2 + 24 I$.

If I translate what you have above, I get $P = 1.5(I - 24)$ or $(I - 0) + 24$
I don't know what you are doing here, since an equation doesn't have "or" in it, and neither of these can be transformed to get back to $P = -1.5 I^2 + 24 I$.

Have you worked with quadratic functions before? Do you know about completing the square to find the vertex of a parabola that is represented by a quadratic function? If you know this technique, it's a lot harder than what I've suggested, about finding the axis of symmetry.

i have, idk why i find this one so hard to understand =/ and ok so P = 24I - 1.5I^2 is the equation. How would i find the I intercepts to be able to find the axis of symmetry? (again thanks for helping me) (edit, would the vertex be (8,96))

 

[Mark44]

i have, idk why i find this one so hard to understand =/ and ok so P = 24I - 1.5I^2 is the equation. How would i find the I intercepts to be able to find the axis of symmetry? (again thanks for helping me) (edit, would the vertex be (8,96))

Yes, that's the vertex.
Using the technique I suggested:
$P = 24 I - 1.5 I^2 = I (24 - 1.5 I)$
The I-intercepts (where P = 0) are obtained by setting P = 0.
$I (24 - 1.5 I) = 0 \Rightarrow I = 0 \text{ or } I = 16$
The axis of symmetry is the vertical line I = 8, which is halfway between 0 and 16. Substitute this value for I in the equation to get $P(8) = 24 * 8 - 1.5 * 64 = 96$

 

[Zacko282]

Yes, that's the vertex.
Using the technique I suggested:
$P = 24 I - 1.5 I^2 = I (24 - 1.5 I)$
The I-intercepts (where P = 0) are obtained by setting P = 0.
$I (24 - 1.5 I) = 0 \Rightarrow I = 0 \text{ or } I = 16$
The axis of symmetry is the vertical line I = 8, which is halfway between 0 and 16. Substitute this value for I in the equation to get $P(8) = 24 * 8 - 1.5 * 64 = 96$

THANK GOD! Thanks a lot for the help, and how would i graph this parabola? Nvm found out how to graph it! now how do i find the max power?

 

[Mark44]

THANK GOD! Thanks a lot for the help, and how would i graph this parabola? Nvm found out how to graph it! now how do i find the max power?

Graph the parabola on axes labeled I (horizontal) and P (vertical). Which way does it open -- up or down? Is there a high point or a low point?

 

[Zacko282]

Graph the parabola on axes labeled I (horizontal) and P (vertical). Which way does it open -- up or down? Is there a high point or a low point?

It opens down and the maximum value is 96

 

[Mark44]

It opens down and the maximum value is 96

Right

 

[Zacko282]

THANKYOU