'O' level integration: equation of curve

donkhead333 SG

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 SG

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 SG

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

how did you arrive at this equation?

wee_ws SG

Question as below:

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 SG

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 SG

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 SG

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 SG

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