# slope of a tangent to a curve

• Apr 7th 2008, 02:59 PM
OzzMan
slope of a tangent to a curve
I could use some help on this one. Calculate the slope of a line tangent to the curve of each of the functions $\displaystyle y = f(x)$ for the given point $\displaystyle P$.

$\displaystyle Y = x^2$ ; $\displaystyle P$ is $\displaystyle (2,4)$
• Apr 7th 2008, 03:05 PM
Mathstud28
Quote:

Originally Posted by OzzMan
I could use some help on this one. Calculate the slope of a line tangent to the curve of each of the functions $\displaystyle y = f(x)$ for the given point $\displaystyle P$.

$\displaystyle Y = x^2$ ; $\displaystyle P$ is $\displaystyle (2,4)$

First find the derivative $\displaystyle f'(x)=2x$...then you want to find the slope of the tangent line at the point $\displaystyle (2,4)$ and the derivative evaluated at a point x has the same slope as the tangent line at that point so the slope of the tangent is $\displaystyle f'(2)=2\cdot{2}=4$
• Apr 7th 2008, 03:17 PM
OzzMan
Uh I'm new to calculus so if you could kind of walk me through it a little bit more that be great.
• Apr 7th 2008, 03:28 PM
Mathstud28
ok
Quote:

Originally Posted by OzzMan
Uh I'm new to calculus so if you could kind of walk me through it a little bit more that be great.

If you haven't learned differntiation rules then we will go with the quotient difference...ok the slope of a curve is given by $\displaystyle f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$..so to get the devative of $\displaystyle f(x)=x^2$ before you know differentiation rules is to go through this thus the derivative of $\displaystyle x^2$ is $\displaystyle \lim_{h \to 0}\frac{(x+h)^2-x^2}{h}=\lim_{h \to 0}\frac{x^2+2xh+h^2-x^2}{h}=\lim_{h \to 0}\frac{2xh+h^2}{h}$$\displaystyle =\lim_{h \to 0}\frac{2xh}{h}+\lim_{h \to 0}\frac{h^2}{h}=2x+\lim_{h \to 0}h=2x$...so the slope at the point x is $\displaystyle 2\cdot{x}$...so the slope of the curve of $\displaystyle x^2$ at x=2 is $\displaystyle 2\cdot{2}=4$...so the slope of the tangent line at the point x=2 has the same slope as the curve so its slope is 4
• Apr 7th 2008, 03:31 PM
OzzMan
wow lol. and im only in chapter 2. i got 7 more chapters to go tonight and tomorrow.
• Apr 7th 2008, 03:36 PM
Mathstud28
here let me make it a little easier
Quote:

Originally Posted by OzzMan
wow lol. and im only in chapter 2. i got 7 more chapters to go tonight and tomorrow.

all that work just to get $\displaystyle x^2$...but I'll tell you a little trick if $\displaystyle f(x)=x^{n}$...then the derivative of f(x) or f'(x) is $\displaystyle f'(x)=nx^{n-1}$...so to get the derivative of x²...just multiply the coefficient by the exponent and reduce the exponent by one..so it would be $\displaystyle f'(x)=2*x^{2-1}=2x^1=2x$
• Apr 7th 2008, 03:39 PM
OzzMan
I'm a little confused though where you got those equations from. In my book they seem to be a little different. Damn I'm lost.
• Apr 7th 2008, 03:49 PM
Mathstud28
?
Quote:

Originally Posted by OzzMan
I'm a little confused though where you got those equations from. In my book they seem to be a little different. Damn I'm lost.

WHich equations?
• Apr 7th 2008, 04:03 PM
OzzMan
The 4th post in this thread
• Apr 7th 2008, 04:04 PM
Mathstud28
If you mean
Quote:

Originally Posted by OzzMan
The 4th post in this thread

the limit stuff...hmm...thats something you just have to learn...here I think this will really help you its a great site..good luck! if you have any more questions just ask

Visual Calculus
• Apr 7th 2008, 04:15 PM
OzzMan
Well what I mean is that it just appeared a little differently in my book. It's all good though.
• Apr 7th 2008, 04:16 PM
Mathstud28
Ok then
Quote:

Originally Posted by OzzMan
Well what I mean is that it just appeared a little differently in my book. It's all good though.

Well good luck!