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 July 19th, 2013, 04:16 AM #1 Newbie   Joined: Mar 2013 Posts: 14 Thanks: 0 Calculus makes me want to pull my hair out ! Please help :) Hi guys and gals could someone please explain to me how I would go about solving these calculus problems, Im really confused 1. If the gradient of the curve y= x^3 + ax^2 +9x-4 is -3 at x= 2, calculate a. 2.Find the co-ordinates if a point on the curve y=2x^2-6x+10, where the tangent to the curve is parallel to y=2x-1 Thanks so much in advance you brilliant soul ! P.S how awkward would it be if no one answered... haha
 July 19th, 2013, 06:28 AM #2 Newbie   Joined: Jul 2013 Posts: 12 Thanks: 0 Re: Calculus makes me want to pull my hair out ! Please help For your second problem, are you asking what are the coordinates of the point whose tangent is parallel to the line $y=2x-1$? If so... We're finding a tangent line parallel to $y=2x-1$. The slope of this line is 2, so the tangent's slope must also be 2. The derivative tells us the slope of the tangent line. So we differentiate the first function. $\frac{d}{dx}(2x^2-6x+10)=4x-6$ Now set the derivative to 2, and solve to find the x-coordinate of the intersection of y and the tangent you're finding. $4x-6=2 x=2$ So now you know the tangent parallel to $y=2x-1$ intersects y at x=2, so just put the x-coordinate into the original function to find the y-coordinate. $f(2)=2(2)^2-6(2)+10=6$ And now you've found the point at which the tangent intersects y, (2,6). If you are trying to find the equation of the tangent, just put your points into $y=mx+b$. $6=2(2)+b b=2$ And substituting for b, $y=2x+2$ I hope this is what you were asking for, and I hope it helped (and I also hope I did this right lol).
 July 19th, 2013, 06:39 AM #3 Newbie   Joined: Jul 2013 Posts: 12 Thanks: 0 Re: Calculus makes me want to pull my hair out ! Please help And for the first one, as far as I know the gradient is the slope of the tangent line at a point on a curve. So you have a function, $y=x^3+ax^2+9x-4$ treat $a$ as any other constant (e.g. 1, 2, 3, $\pi$), and differentiate to find the equation for finding the slope of the tangent, which is the gradient. Once you've done that, you can substitute, and solve for $a$. Hopefully my terminology is as I remember it.
July 19th, 2013, 06:47 AM   #4
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 Originally Posted by AzraaBux Hi guys and gals could someone please explain to me how I would go about solving these calculus problems, Im really confused 1. If the gradient of the curve y= x^3 + ax^2 +9x-4 is -3 at x= 2, calculate a.
Some of the first things you should have learned in Calculus:
a) The gradient of a sum is the sum of the gradients- to find the gradient of x^3+ ax^2+ 9x- 4, take the gradient of each term separately
b) The gradient of ax^n is an x^(n-1).

So, what is the gradient of x^3 (n=3)? What is the gradient of ax^2? What is the gradient of 9x (n=1)? What is the gradient of -4 (n= 0)?
Add those together to find the gradient of the function and set x= 2 (or set x= 2 in each then add- its the same thing). The result will, of course, depend on a. Set that equal to -3 and solve for a.

Quote:
 2) Find the co-ordinates if a point on the curve y=2x^2-6x+10, where the tangent to the curve is parallel to y=2x-1
Another thing you should have learned early:
c) The gradient of a function (called "derivative" in the U.S.A.), evaluated at a specific value of x, is the slope of the tangent line to the graph of the functon at that value of x.

I presume you know that the slope of y= 2x- 1 is 2. So this problem is asking you to find the gradient of 2x^2- 6x+ 10, set it equal to 2 and solve for x.

Quote:
 Thanks so much in advance you brilliant soul ! P.S how awkward would it be if no one answered... haha

 July 19th, 2013, 06:59 AM #5 Newbie   Joined: Jul 2013 Posts: 12 Thanks: 0 Re: Calculus makes me want to pull my hair out ! Please help I recommend https://mooculus.osu.edu/ for practicing more problems like these, and there are things in "Week 3" that help with comprehension too. I used this last year and thought it to be a very good source.

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