| 
  • If you are citizen of an European Union member nation, you may not use this service unless you are at least 16 years old.

  • You already know Dokkio is an AI-powered assistant to organize & manage your digital files & messages. Very soon, Dokkio will support Outlook as well as One Drive. Check it out today!

View
 

Slope of a curve at a point

This version was saved 16 years, 2 months ago View current version     Page history
Saved by PBworks
on January 31, 2008 at 12:51:44 am
 

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:

http://www.youtube.com/watch?v=VnOVAXJ3Xlw&NR=1

..........and her.... http://www.youtube.com/watch?v=DCfTyDrZzPs

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

 

 

 

SOME OPPORTUNITIES TO ENRICH THE GARDEN THAT IS YOUR MIND...

 

(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. 

Comments (0)

You don't have permission to comment on this page.