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?