Grade 11 Math Help Quadratic functions/ physics

  • #1
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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 = IV


3. 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.
 
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  • #2
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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 = IV


3. 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.
 
  • #3
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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.
 
  • #4
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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.
Zacko282 said:
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.
Zacko282 said:
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.
 
  • #5
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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?
 
  • #6
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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.
 
  • #7
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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 ?
 
  • #8
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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.
 
  • #9
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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.
 
  • #10
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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.
 
  • #11
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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 doesnt 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.
 
  • #12
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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 doesnt 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.
 
  • #13
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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))
 
  • #14
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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##
 
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  • #15
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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?
 
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  • #16
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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?
 
  • #17
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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
 
  • #19
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THANKYOU
 
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