Integration help (1 Viewer)

ALOZZZ

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Hey guys i am having trouble with the following questions:
1) given d2y/dx2=8 and dy/dx=0 and y=3 when x=1, find the equation of y in terms of x.
2) The tangent to a curve with d2y/dx2= 2x-4 makes an angle of inclination of 135 with the x-axis at the point (2,-4). Find the equation
3) A function has a tangent parallel to the line 4x-y-2=0 at the point (0,-2), and f''(x)=12x^2-6x+4. Find the equation of the function
4) A curve has d2y/dx2=6 and a tangent at (-1,3) is perpendicular to the line 2x+4y-3=0. Find the equation of the curve.
 

A1La5

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1.



Integrating, we get:



Using the information from the question, when




Further integrating the function:

Using the information from the question, when .



Hence, the equation of the curve is: .

Edit: Renamed constants.
 
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A1La5

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4.

Integrating, we get:


In order to find the value of this constant , we need to understand that the gradient of the tangent at (-1,3) is perpendicular to the gradient of the line .
Rearranging:

So, the gradient of the tangent is , as the product of perpendicular gradients is .

Substituting this into :




Integrating again, we get:

The point (-1,3) is on this curve. Using this information in the equation:




Hence, the equation of the curve is:

Edit: Renamed constants.
 
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A1La5

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Considering no one's tried to help you since I last posted I'll attempt 2 as it helps both of us.



Integrating:



We can source the value of the gradient function at through knowing the angle of inclination.

Capture.JPG

in this case is 135 degrees. Using the relationship above, the gradient function = -1.

Plugging that into the above equation and solving for C:




Integrating again:

Knowing that the point is on the curve, it just now boils down to simple substitution.




The equation of the curve is:
 
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A1La5

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Just a pedantic correction, it may be better to name your constants and rather than both as this implies they are both the same.
Trust me, I would've if I knew how to. I'm still pretty new to this whole LaTeX business. Thanks though.
 

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