Graph sketching (differentiation)

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March 6th 2013, 05:19 AM
alexander9408

Graph sketching (differentiation)

(1) Given the curve x=6t^2 , y=t^3 -2t where t is not = 0, show that dy/dx = (t^2 -1)/4t. Hence, find the point in the curve at which the tangents to the curve and parallel to the x-axis.
(2) y= (2x-5)/(x^2 -4) . Prove that y>=1 or y<= for all real values of x.
(3) Sketch the curve of y^2 = x^2 (1-x).

Please show me the steps about how can i solve these. Especially about the graph sketching part. Thanks.
March 6th 2013, 06:19 AM
MINOANMAN

Re: Graph sketching (differentiation)

There are many mistakes in your post

anyway to find the dy/dx differentiate separate first and find dy/dt then dx/dt and then find dy/dx=(dy/dt)/(dx/dt) . verify your post because you may have typed wrongly the parametric equations. anyway if you do corect the calculations you will find the dy/dx.

For the second question something is missing...the curve y=(2x-5)/(x^2-4) has as domain R-{-2,+2} and has 3 branches. The part of the curve which is defined in the domain (-2,+2) has range y biger or equal to 1.... with the lines x=-2 and x=+2 to be the vertical asymptotes .
I don't understand what you want...

For the last question try to find a graph calculator to see the shape of the graph and then try to graph it.
March 6th 2013, 05:02 PM
hollywood

Re: Graph sketching (differentiation)

For (1), you need to use $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ . And of course, the curve is parallel to the x-axis when $\frac{dy}{dx}=0$ .

For (2), you kind of need to sketch the graph to figure out what's going on. The numerator is positive for $x>\frac{5}{2}$ and negative for $x<\frac{5}{2}$ . You can do the same thing for the denominator and then figure out what the function's doing.

For (3), you might start by describing what $x^2(1-x)$ does in the ranges $-\infty<x<0$ , $0<x<1$ , and $1<x<\infty$ . If, for example, it is negative, then there will be no solutions with those values of x.

- Hollywood