differentaition word problem: tangents to y=4x^2

Becky4paws

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Feb 15, 2006
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Find all the points (x,y) on the graph of the function y=4x^2 with the property that the tangent to the graph at (x,y) passes thru the point (2,0).

I went about this by first finding the derivative 8x. Wouldn't that be slope?

Then I would apply point/slope formula y-0=8x(x-2)?

I think I missed something.
 
You have made what I believe to be the most common error on this type of problem. You have assumed the given point is ON the curve. It isn't.

You have:
1) A point
2) A general expression for slope, referencing the first derivative.

You do NOT have:
1) The point(s) where your line(s) are tangent to the curve.

Where does that leave you?
 
Re: differentaition word problem

Hello, Becky4paws!

From the way you went way off in your first step,
    \displaystyle \;\;I assume you could use a walk-through . . .

Find all the points (x,y)\displaystyle (x,y) on the graph of the function y=4x2\displaystyle y\,=\,4x^2
with the property that the tangent to the graph at (x,y)\displaystyle (x,y) passes thru the point (2,0).\displaystyle (2,0).
I don't suppose you made a sketch . . . [Naw, whatever for ??]
Code:
        *         |         */
                  |         /
                  |        o
           *      |      */
             *    |    * /
      -----------***----*------
                  |    /(2,0)
                  |   /
Here is one approach to this problem . . .

The line through (2,0)\displaystyle (2,0) has slope m.\displaystyle m.
Its equation is: y0=m(x2)        y=mx2m\displaystyle \,y\,-\,0\:=\:m(x\,-\,2)\;\;\Rightarrow\;\;y\:=\:mx\,-\,2m

Find the intersections of the parabola and this line:
\(\displaystyle \;\;4x^2\:=\:mx\,-\.2m\;\;\Rightarrow\;\;4x^2\,-\,mx\,+\,2m\:=\:0\)

Quadratic Formula: x=(m)±(m)24(4)(2m)2(4)  =  m±m232m8\displaystyle \,x\:=\:\frac{-(-m)\,\pm\,\sqrt{(-m)^2\,-\,4(4)(2m)}}{2(4)}\;= \;\frac{m\,\pm\,\sqrt{m^2\,-\,32m}}{8}\: [1]

We want the line to be tangent to the parabola,
    \displaystyle \;\;so there will be exactly one intersection.
This happens when the discriminant is zero: m232m=0\displaystyle \,m^2\,-\,32m\:=\:0

We have: \(\displaystyle \:m(m\,-\,32)\:=\0\;\;\Rightarrow\;\;m\:=\:0,\,32\)


Substitute m=0\displaystyle m\,=\,0 into [1]: x=0\displaystyle \,x\:=\:0
    \displaystyle \;\;Then: y=402=0\displaystyle \,y\:=\:4\cdot0^2\:=\:0
The line is tangent at (0,0).\displaystyle (0,0).


Substitute m=32\displaystyle m\,=\,32 into [1]: x=4\displaystyle \,x\:=\:4
    \displaystyle \;\;Then: y=442=64.\displaystyle \,y\:=\:4\cdot4^2\:=\:64.
The line is tangent at (4,64).\displaystyle (4,64).
 
Here's a slightly different approach than Soroban's:

You started out correctly:

y0=8x(x2)\displaystyle y-0=8x(x-2), but y=4x2\displaystyle y=4x^{2}

4x2=8x(x2)\displaystyle 4x^{2}=8x(x-2)

You have a quadratic:

4x216x=0\displaystyle 4x^{2}-16x=0

4x(x4)=0\displaystyle 4x(x-4)=0

As you can see, x=0 and 4 are the solutions.

Using y=mx+b and y=8x\displaystyle y=mx+b\ and\ y'=8x

If x=4, then 8(4)=32

0=32(2)+b\displaystyle 0=32(2)+b

b=-64

\(\displaystyle \H\\y=32x-64\)
 
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