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# Slope of a curve at a point

last edited by 15 years ago

The slope of a curve at a certain point is the same as the slope of the line that is tangent to the curve at that point (touches the curve at ONE point and does not cross over the curve). This is also the derivative (the instantaneous rate of change of a function). Therefore, to find the slope of a curve at point "P", you can take the derivative of the equation and plug in the X and Y values of point P on the curve f(x). For example:

P=(6,4)

f(x)=-x2+9x-14

f'(x)=-2x+9..... (you get this using the formula f'(x)=nx^(n-1))

...SO you then find the slope of the curve at (6,4) by finding the derivative at P:

f'(6)=-2(6)+9=-3

...This means that the slope of the curve f(x) at (6,4) is -3. Another way to discuss this concept is in terms of limits with the difference quotient (seen below). This is used to find the slope of the secant line, but as the "delta x" gets smaller, the secant line becomes increasingly similar to the tangent line. The slope of the tangent line is also known as the limit. When a curve has a tangent of zero, there is a maximum or minimum, or the graph is horizontal. There is no tangent of a curve (the function is not differentiable) if the function is not continuous (the limit of f(x) as x approaches a from the right does not equal the limit of f(x) as x approaches a from the left). There is a vertical tangent if and only if . Vertical tangents usually (but not always) occur at vertical asymptotes.

An applet demonstrating the slope of a curve at a point: calculusapplets.com/derivpoint.html

An applet about vertical tangents: http://www.analyzemath.com/calculus/derivative/vertical_tangent.html

Announcement: Janie Abernethy is forgoing college to give this woman a run for her money:

...And this is what Lee is going to school to be: http://www.roofrats.org/images/01.jpg

SOME OPPORTUNITIES TO ENRICH THE GARDEN (1.) The slope of the curve  (y^3)-x(y^2)=4 at the point where y=2 is

(A) -2

(B) 1/4

(C) -1/2

(D) 1/2

(E) 2

(2.) The slope of the curve (y^2)-xy-3x=1 at the point (0,-1) is

(A) -1

(B) -2

(C) 1

(D) 2

(E) -3

(3.) For what value of x does the function (you mean the graph of the equation) (5x^2)+2xy+(y^2)=16 have a vertical tangent line?

(A) x=2 only

(B) x=2 and x=0 only

(C) x=0, x=-1, and x=-2

(D) x=2 and x=-2 only

(E) never

(4.) Jasper is an extremely precocious feline and one day decides to sneak attack his owner and mortal enemy Janie "I Eat Babies in their Sleep" Abernethy. Just for kicks, he decides to map out his motion across the battlefield in path of the sinusoidal function y=sinx. He begins at the origin (conveniently located at the doorway to the monster's lair), and all is going smoothly until the Baby Eater lets out a massive snore in her sleep, when Jaspie is at x= on the graph. Jasper makes a bee-line for safety along the tangent line to the graph at this point. What is the slope of the line Jasper ran along to escape the fearsome wrath of this despicable being?

INSTRUCTIONS ON HOW TO SOW THE GARDEN

(1.) D... You must use implicit differentiation to find the derivative of the equation, then plug in the point (1,2) to find y'. **You know that the point is (1,2) because you plug y=2 into the original equation to get "x".**

(2.) A... Same technique as #1.

(3.) D... You begin the same way, with implicit differentiation, and you get: dy/dx = (-5x - y)/(x + y). The vertical tangent will appear at the vertical asymptote in this case, so when the denominator equals zero (y=-x). THEN plug this back into the original equation and solve for "x"... you get +2 and -2!

(4.) First you find y', which is cosx. Then you evaluate y'=cosx at x= , which is -1. Therefore, -1 is the slope of Jasper's linear escape route. #### Anonymous said

at 9:41 am on Feb 8, 2008

I didn't really understand any of your page (because you are a huge idiot) until I read the problem about Jasper and the baby eater. Thanks for finally making this concept clear to me.

Seriously: The recap of how to "sow the garden" was the most helpful. Maybe you should put that above the problems. I'm impressed by the Jasper problem. You really seem to understand the relationship of a slope of a curve at a point AND the relationship between me and my beloved feline. #### Anonymous said

at 9:54 am on Feb 8, 2008

Rachel, your page is very eloquent and manages to effectively blend humor with calculus. My only complaint is that your page seems a little busy. It might be easier to understand the concepts of slopes if you're page was a little easier on the eyes :) #### Anonymous said

at 9:56 am on Feb 8, 2008

Rachel, I think the dog video is poignantly autobiographical and I admire your self-effacing qualities. The enrichment of my garden as well as the recap was like cooling rain on the chia pet of my mind (it is now a rainforest). The all-caps and bolding was actually helpful in understanding, and now I understand Lee's true complex about meat (she is just bottling her passion). Thank you for your great explanations.
~Marissa #### Anonymous said

at 10:00 am on Feb 8, 2008 #### Anonymous said

at 10:00 am on Feb 8, 2008

Your explanation and example problems were really good, but I found the videos rather disturbing . . . I cannot believe a dog that ugly actually exists. #### Anonymous said

at 10:00 am on Feb 8, 2008

Thanks Pee Pants. You're a jerk.
-Rachelle #### Anonymous said

at 8:30 am on Feb 21, 2008

I actually disagree that the page is too busy. I like the changes in type and the overall look of the page. Which just shows you that typography and design is an art, too -- and people have different tastes. I love the problems, and I love the sense of humor. Excellent job! #### Anonymous said

at 8:31 am on Feb 21, 2008