What is the Formula for Finding the Slope of a Tangent Line on a Curve?

[Lucretius]

Well, seeing as I will be working on calculus for a long time, and I figure at the rate I've been going I'll get stuck a lot and need help, I'll just make one thread for all my calculus troubles.

I just read the chapter on Tangent Lines and Slopes. I did a problem, double-checked the math, then turned to the back of the book and found out the answer was completely incorrect. I retraced again, and found no problem, but the book is a bit confusing so I might have missed something.

The problem reads: Find a formula that gives the slope at any point P (x,y) on the given curve. y=4-x^2 P:(1,-1)

My work is as follows:

1)I begin by making a secant line from point P to a point Q, which I arbitrarily place. The coordinates are Q:(-1+\Delta x, 4-(-1+\Delta x)^2

2)To find my slope: \frac{\Delta y}{\Delta x} = \frac{4-(-1+\Delta x)^2}{\Delta x}

3)This becomes: \frac{4-1+2\Delta x-\Delta x^2}{\Delta x}

4)Finally, I get 5-\Delta x=m Getting rid of the \Delta x my slope is m=5

The back of the book says the answer is -2x.

What happened?

 

[amcavoy]

I'm new here, but I might be able to help.

Since you read the section on tangent lines, did you read about:

f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}?

Try using this formula on your original equation f(x)=4-x^2 and that will give you the tangent line at any point.

 

[Lucretius]

The section on limits and derivatives comes right after this one.

 

[VietDao29]

Hmmm, I don't understand what you are doing here...
P(1, -1) is not on the curve.
Let P(\varepsilon, 4 - \varepsilon ^ 2)be a point on a curve.
Let Q(\varepsilon + \Delta x, 4 - (\varepsilon + \Delta x)^ 2)be another point on a curve.
You have as \Delta x \rightarrow 0 => Q \rightarrow P
Can you find \frac{Q_y - P_y}{Q_x - P_x} as \Delta x \rightarrow 0?
Er, I think limits should be learn before tangent line.
Viet Dao,

 

[Lucretius]

Should I skip past this chapter for now, learn limits, and then return to it?

You mentioned P(1,-1) is not on the curve — and you're right. It's (-1,3). The math I did uses (-1,3) though, I just mistyped it here.

Can you make that itexed slope formulation tex-ed? It's too small, can't read it.

Should I substitute 0 for \Delta x?

 

[RadiationX]

Well, seeing as I will be working on calculus for a long time, and I figure at the rate I've been going I'll get stuck a lot and need help, I'll just make one thread for all my calculus troubles.

I just read the chapter on Tangent Lines and Slopes. I did a problem, double-checked the math, then turned to the back of the book and found out the answer was completely incorrect. I retraced again, and found no problem, but the book is a bit confusing so I might have missed something.

The problem reads: Find a formula that gives the slope at any point P (x,y) on the given curve. y=4-x^2 P:(1,-1)

My work is as follows:

1)I begin by making a secant line from point P to a point Q, which I arbitrarily place. The coordinates are Q:(-1+\Delta x, 4-(-1+\Delta x)^2

2)To find my slope: \frac{\Delta y}{\Delta x} = \frac{4-(-1+\Delta x)^2}{\Delta x}

3)This becomes: \frac{4-1+2\Delta x-\Delta x^2}{\Delta x}

4)Finally, I get 5-\Delta x=m Getting rid of the \Delta x my slope is m=5

The back of the book says the answer is -2x.

What happened?

i don't know how far along you are, but all you need to do is find the derivative of the given function:

y=4-x^2 the derivative of this is -2x.

the derivative of a function is the slope of the tangent line at a certain point.
for the point (1,1) the equation of the tangent line is :

y-1=-2(x-1)

 

[Curious3141]

I think the OP is just beginning with calculus and is not allowed to "just differentiate".

So, going back to first principles, the slope of a tangent to a curve at a point is defined by \frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}

We have y = 4 - x^2. ---eqn(1)

Then

y + \Delta y = 4 - (x + \Delta x)^2 = 4 - x^2 - 2x\Delta x - (\Delta x)^2 ---eqn(2)

using the binomial expansion.

Take eqn(2) - eqn(1),

we have \Delta y = -2x\Delta x - (\Delta x)^2

Divide that by \Delta x,

\frac{\Delta y}{\Delta x} = -2x - \Delta x

And taking the limit as \Delta x goes to zero, we have :

\frac{dy}{dx} = -2x

as we would expect. This is the equation that gives the slope of the tangent to the curve at any point (x,y)

If you want the equation of a line that has that slope and that passes through the point (1, -1), the equation of that line is simply,

y - (-1) = -2X(x - 1)

or y = -(2X)x + 2X - 1

Note that that line will NOT actually be a tangent to any point on the curve, it will only be parallel to the tangent to the curve at the chosen point (X,Y).

If you want the equation of the actual tangent line to the curve at a point (X,Y), then it would be given by

y - Y = -2X(x - X)

y - (4 - X^2) = -2X(x - X)

which rearranges to

y = -(2X)x + X^2 + 4

 

[Lucretius]

It seems as if you are all suggesting that I use derivatives and/or limits. I believe I will skip this chapter, learn how to do limits and derivatives, and then return to it at a later time. Wish me luck with derivatives and limits. I will need it.

 

[geosonel]

choose an arbitrary x
then the point P = (x, 4 - x^2) is on the curve
now move to another x a short distance Δx away: x + Δx
then the point Q = (x + Δx, 4 - (x + Δx)^2 ) is also on the curve
the slope of the line connecting P and Q is:

\mbox{slope \overline{PQ} } \ = \ \frac {\mbox{ (y value of Q) } \ - \ \mbox{ (y value of P) } } {\mbox{ (x value of Q) } \ - \ \mbox{ (x value of P) } } \ = \ \displaystyle \frac { ( 4 - (x + \Delta x)^2) \ - \ ( 4 - x^2 ) } { (x + \Delta x) \ - \ (x)} \ = \ \displaystyle \frac { - 2x\Delta x \ - \ (\Delta x)^2 } { \Delta x} \ = \ -2x \, - \, \Delta x

now what happens to Δx when point P approaches point Q?
if you answered that Δx → 0, then what is the slope of the tangent to the curve?