Tangent to the Curve at a Point

[Jason76]

I'm thinking the following problem is simply the derivative:

Problem:

Find the equation of the tangent line to the curve

$\displaystyle y= x^{2}$

at the point

$\displaystyle (2, 4)$

My reasoning on the problem:

The tangent line is the slope of the curve. Well, actually the curve has no actual slope, but it's tangent line does. The tangent line slope is considered the average slope of the curve -- the average of the slopes of many secant lines drawn thru the curve.

 

[wjm11]

I'm thinking the following problem is simply the derivative:

Problem:

Find the equation of the tangent line to the curve

$\displaystyle y= x^{2}$

at the point

$\displaystyle (2, 4)$

My reasoning on the problem:

The tangent line is the slope of the curve. Well, actually the curve has no actual slope, but it's tangent line does. The tangent line slope is considered the average slope of the curve -- the average of the slopes of many secant lines drawn thru the curve.

Yes, the derivative will give you the slope at (2,4) once you plug in the 2. After finding the slope, you must write the equation of the line that contains (2,4).

 

[mmm4444bot]

The tangent line is the slope of the curve.

This statement is not correct. A tangent line is not a slope. A tangent line is a line; it is a line that touches a curve locally at a single point only. It seems that what you're trying to say is, "The slope of a tangent line is the same as the slope of the curve at the tangent point."

Well, actually the curve has no actual slope, but it's tangent line does.

This is not correct, either. A curve does have slope; the slope generally changes at every point, as the curve's behavior changes at every point. (There are some exceptions to the notion that a curve has slope at every point -- namely, the exceptions occur where the derivative fails to exist -- and these exceptions are things like discontinuities, cusps, & "vertical" inflection points.)

The tangent line slope is considered the average slope of the curve -- the average of the slopes of many secant lines drawn thru the curve.

The slope of a tangent line is not considered an average slope; it is the slope of the curve at that point. The demonstration you saw, where a sequence of secant lines increasingly moves toward tangency at some specific point, does not represent an average. It represents a limit procedure.

I hope that these clarifications help you to solidify your understanding of these concepts. Please do ask continuing questions, if any uncertainty remains. Cheers:cool:

 

[Jason76]

$\displaystyle x = 2$

Main Formula:

The following equation divided by h:

$\displaystyle f(x + h) - f(x)$

Apply the 2.


$\displaystyle 2(x + h)^{2} - (2x^{2})$

Foil the square root.

$\displaystyle 2[(x + h)(x+ h)] - (2x^{2})$

$\displaystyle 2(x^{2} + xh + xh + h^{2}) - (2x^{2})$

Distributive property.

$\displaystyle 2(x^{2} + 2xh + h^{2}) - (2x^{2})$

$\displaystyle 2x^{2} + 4xh + 2h^{2} - 2x^{2}$

$\displaystyle 4xh+ 2h^{2}$

Dividing each term by the denominator of h (not listed but part of derivative formula)

$\displaystyle 4x + 2h$

Apply the limit of 0 to h to rid the problem of 2h.

Derivative = $\displaystyle 4x$



I think I did something wrong here.

 

[mmm4444bot]

Hi Jason. Do not let x = 2 at the start. That substitution comes later, after you've found the derivative.

If you don't feel like typing the ratios with h in the denominator, then you may factor out that denominator. (Don't skip typing the h altogether, in your steps.)

[f(x+h) - f(x)]/h

1/h [f(x+h) - f(x)]

1/h [(x+h)^2 - x^2]

1/h [x^2 + 2xh + h^2 - x^2]

1/h [2xh + h^2]

Now expand.

(1/h)(2xh) - (1/h)(h^2)

2x - h

The derivative is the limit as h goes to zero:

2x - 0

2x

Now it is time to substitute x=2, to calculate the slope (derivative value) of the curve of y at the point (2,4).

Questions?

 

[Jason76]

$\displaystyle x = 2$

Main Formula:

The following equation divided by h:

$\displaystyle f(x + h) - f(x)$


$\displaystyle (x + h)^{2} - (x^{2})$

Foil the square root.

$\displaystyle [(x + h)(x+ h)] - (x^{2})$

$\displaystyle (x^{2} + xh + xh + h^{2}) - (x^{2})$

$\displaystyle (x^{2} + 2xh + h^{2}) - (x^{2})$

$\displaystyle x^{2} + 2xh + h^{2} - x^{2}$

$\displaystyle 2xh + h^{2}$

Now divide whole thing by h in the denominator

$\displaystyle 2x + h$

$\displaystyle 2x$ = derivative

So the slope at the point of $\displaystyle 2, 4$ is 4 (from 2 being plugged into x)

 

[mmm4444bot]

$\displaystyle x = 2$ You don't need to write this line

It's already clear that we're at the point (2,4)

Main Formula:

The following equation divided by h: That should be "expression", not "equation"

$\displaystyle f(x + h) - f(x)$


$\displaystyle (x + h)^{2} - (x^{2})$

Foil the square root That should be "squared binomial", not "square root"

(I sure hope that you know what a square root is!)

$\displaystyle [(x + h)(x+ h)] - (x^{2})$

$\displaystyle (x^{2} + xh + xh + h^{2}) - (x^{2})$

$\displaystyle (x^{2} + 2xh + h^{2}) - (x^{2})$

$\displaystyle x^{2} + 2xh + h^{2} - x^{2}$ You don't really need to write this line, but ok

$\displaystyle 2xh + h^{2}$

Now divide whole thing by h in the denominator

$\displaystyle 2x + h$

$\displaystyle 2x$ = derivative You forgot to say, "Take limit as h goes to zero"

So the slope at the point of $\displaystyle 2, 4$ is 4 (from 2 being plugged into x)

Yes -- and it's standard to write (2,4) versus writing "of 2, 4"

Your terminology is sloppy, but your reasoning looks good.

You found y${\displaystyle You\subset stitutedx=2\int oy}$ to get 4

Now you know that the slope of the line tangent to curve y at point (2,4) is m = 4.

Finish the exercise by completing what wjm11 instructed: write the tangent line equation.

 

[Jason76]

The Point Slope form of the line is

(y2 - y1) = m(x2 - x1)

So

$\displaystyle (y - 4) = 4(x - 2)$


in slope-intercept form:

$\displaystyle 4 = 2(2) + b$

 

[mmm4444bot]

Your equation in Point-Slope form (below) is a correct answer to this exercise.

If you would like to report this answer in Slope-Intercept Form, instead, then solve your equation for y:

y - 4 = 4(x - 2)

Expand right-hand side, then add 4 to both sides and simplify. :cool: