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What is this question asking?

  • #1

Homework Statement



Find the equation of the curve for which y'' = 8 if the curve is tangent to the line y = 11x at (2, 22).

Homework Equations



?

The Attempt at a Solution


[/B]
y' = 8x + c1

y = 4x2 + c1x + c2

What exactly is the question asking me to do, especially with the y = 11x bit? ∫ydx?
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



Find the equation of the curve for which y'' = 8 if the curve is tangent to the line y = 11x at (2, 22).

Homework Equations



?

The Attempt at a Solution


[/B]
y' = 8x + c1

y = 4x2 + c1x + c2

What exactly is the question asking me to do, especially with the y = 11x bit? ∫ydx?
It wants your curve to be tangent at (2,22). That means it must pass through that point and have the same slope there as the line.
 
  • #3
It wants your curve to be tangent at (2,22). That means it must pass through that point and have the same slope there as the line.
But doesn't tangent mean derivative in this case? My curve should have a slope of 11 at (2,22), fine; how does that apply to y'' and y' and y as I said? My curve would be y, right, because y'' = 8 is the double derivative of that? How to solve it with two different constants, then?
 
  • #4
LCKurtz
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You have two constants with which you can make the curve agree with point and slope.
 
  • #5
You have two constants with which you can make the curve agree with point and slope.
So y = 4x2 + c1x + c2 = 11? Without knowing the slope for y', how can we solve for this?

I mean, so far, I've got -4x2 - c1x + 11 = c2, but now what?
 
  • #6
LCKurtz
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You have ##y = 4x^2 + c_1x + c_2## with two unknown constants. What do the constants have to be so that ##y(2) = 22## and ##y'(2) = 11##?
 
  • #7
You have ##y = 4x^2 + c_1x + c_2## with two unknown constants. What do the constants have to be so that ##y(2) = 22## and ##y'(2) = 11##?
Oh, y' = 11 as well? Because y = 11x is a line?
 
  • #8
LCKurtz
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There's really nothing more to tell you without working it for you. Just do it.
 
  • #9
There's really nothing more to tell you without working it for you. Just do it.
Got it. WileyPlus accepted my answer.

So basically, being tangent to a line means y' = 11 at that x-value, so it's solvable. If someone else asks a similar question, answer like that, and it might just make a lot more sense to them.
 

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