'O' level integration: equation of curve

• donkhead333

  31 Aug 09, 18:57

  Question as below:

  A curve is such that dy/dx= kx^2 - 1, where k is a constant. Given that the tangent to the curve at the point (-1,0) passes through the point (-1.5, 2.5), find the value of k and the equation of the curve.

  Ans provided as k = 6, y = 2x^3 - x +1.

  don't really understand...could not find k, kept substituting wrong equations

  ty for any help

• TenSaru

  31 Aug 09, 22:03

  sub x=-1 into the given equation => gradient of tangent is k-1

  (2.5-0)/(-1.5-(-1)) = k-1.... solve for k

  then integrate everything and sub the given coordinates (-1,0) into the the equation to find C

• donkhead333

  1 Sep 09, 11:34

  Originally posted by TenSaru:

   

  (2.5-0)/(-1.5-(-1)) = k-1.... solve for k

   

  how did you arrive at this equation?

• wee_ws

  1 Sep 09, 12:30

  Originally posted by donkhead333:

  Question as below:

  A curve is such that dy/dx= kx^2 - 1, where k is a constant. Given that the tangent to the curve at the point (-1,0) passes through the point (-1.5, 2.5), find the value of k and the equation of the curve.

  Ans provided as k = 6, y = 2x^3 - x +1.

  don't really understand...could not find k, kept substituting wrong equations

  ty for any help

  Hi,

    At (-1, 0), dy/dx = k - 1 and this is the gradient of the tangent at this point.

    Equation of tangent at (-1, 0) is thus given by

    y - 0 = (k - 1)(x + 1).

    Since this tangent passes through (-3/2, 5/2), we obtain

    5/2 = (k - 1)(-3/2 + 1),

    from which k can be found.

    Now, with k = 6, dy/dx = 6x^2 - 1.

    The equation of y can be found via integration of the expression 6x^2 - 1. The constant of integration can be found by the fact that (-1, 0) lies on the curve.

    Hope it's clearer. Thanks!

  Cheers,
  Wen Shih

• donkhead333

  1 Sep 09, 15:34

  hi wen shih!

  thanks for the help, and happy teacher's day to everyone who has served in educating

  just another random question for kinematics in differentiation/integration:

  if the question asks for a calculation of the minimum/maximum velocity, how am i supposed to tackle it?

  for example, if they ask for instantaneous rest, i think of v = 0.

  for maximum velocity, a = 0.

  what about minimum? do i have to do a long winded derivative test for it?

  thanks!

• donkhead333

  1 Sep 09, 16:03

  Originally posted by wee_ws:

  Hi,

    At (-1, 0), dy/dx = k - 1 and this is the gradient of the tangent at this point.

    Equation of tangent at (-1, 0) is thus given by

    y - 0 = (k - 1)(x + 1).

    Since this tangent passes through (-3/2, 5/2), we obtain

    5/2 = (k - 1)(-3/2 + 1),

    from which k can be found.

    Now, with k = 6, dy/dx = 6x^2 - 1.

    The equation of y can be found via integration of the expression 6x^2 - 1. The constant of integration can be found by the fact that (-1, 0) lies on the curve.

    Hope it's clearer. Thanks!

  Cheers,
  Wen Shih

  er, is it possible to explain using the basic formula, y=mx +c ? personally i dont use much of the other formula

• wee_ws

  1 Sep 09, 17:44

  Hi,

    The basic formula y = mx + c was obtained by

    y - y1 = m (x - x1),

    where m is the gradient of the line and (x1, y1) is a point on the line.

    In differentiation, this result is most applicable.

    Thanks!

  Cheers,
  Wen Shih

• wee_ws

  1 Sep 09, 17:47

  Originally posted by donkhead333:

  hi wen shih!

  thanks for the help, and happy teacher's day to everyone who has served in educating

  just another random question for kinematics in differentiation/integration:

  if the question asks for a calculation of the minimum/maximum velocity, how am i supposed to tackle it?

  for example, if they ask for instantaneous rest, i think of v = 0.

  for maximum velocity, a = 0.

  what about minimum? do i have to do a long winded derivative test for it?

  thanks!

  Hi,

    For max/min velocity, dv/dt (or a) = 0. In general, we always set the derivative to zero when finding stationary points.

    To determine max/min nature, you may use either the first derivative test or the second derivative test.

    Thanks!

  Cheers,
  Wen Shih