Using derivatives to find equation for the graph

[donjt81]

Hi I have this problem where I have to find the equation of the graph using derivatives or anti-derivatives... I'm not sure... I really need some help on this...

Find the equation for the graph that passes through the point (-2,3) with the slope 1 given that d^2y/dx^2 = 6x/5

can someone point me in the right direction on how to start the solution.

Thanks in advance

 

[radou]

Hi I have this problem where I have to find the equation of the graph using derivatives or anti-derivatives... I'm not sure... I really need some help on this...

Find the equation for the graph that passes through the point (-2,3) with the slope 1 given that d^2y/dx^2 = 6x/5

can someone point me in the right direction on how to start the solution.

Thanks in advance

This is a second order differential equation which can be solved easily by direct integration. After integrating, you'll have two unknown constants, which you will calculate with the help of the given slope and point.

 

[HallsofIvy]

Integrate
$$\frac{d^2y}{dx^2}= \frac{6}{5} x$$
to find dy/dx (plus a constant).
Integrate that to to find y (plus another constant).
Use the information given to determine what the two constants must be.

 

[donjt81]

ok so y' = 3x^2/5 + C
y = x^3/5 + Cx + D

and also we have the point (-2,3) and slope 1
if we substitute the values for x and y here we get

3 = -8/5 -2C + D

?? how do we get the values of C & D... I'm stuck...

 

[HallsofIvy]

ok so y' = 3x^2/5 + C
y = x^3/5 + Cx + D

and also we have the point (-2,3) and slope 1
if we substitute the values for x and y here we get

3 = -8/5 -2C + D

?? how do we get the values of C & D... I'm stuck...

"also we have the point (-2,3) and slope 1" so you know that
y(-2)= 3 and y'(-2)= 1.

You have correctly put those values into the equation for y(x) to get one equation but you haven't used the fact that the slope there is 1. Now put x= -2, y'(-2)= 1 into the equation for y'(x) to get a second equation. It's easy to solve two linear equations for the two unknown values C and D.

 

[donjt81]

ohhh ok so...

y' = 3x^2/5 + C
1 = 3(-2)^2/5 + C
C = -1.4

y = x^3/5 + Cx + D
3 = -2^3/5 - 1.4 * -2 + D
D = 1.8

So the eqn of the graph is
y = x^3/5 - 1.4x + 1.8

is that correct?

 

[HallsofIvy]

Check it yourself. If x= 1 what is y(x)? If x= 1, what is y'(1)? Does the second derivative satisfy the equation you were given?