Sample Problems, Lesson 1.7
Sample Problems for this Lesson Course Home Page To the Notes Menu   Assignment  

1.7, #12.     (a) Eliminate the parameter to find a Cartesian equation of the curve.
                   (b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases.
 

Here we use the meaning of the meaning of the natural logarithm to rewrite the relationship in exponential form.
Here we substitute the value for the parameter t into the second equation in place of t.
Taking a square root is equivalent to raising a quantity to the one-half power.
Simplifying using exponential laws.
Here is the plot of the equation: 
In Maple, this would be:

> plot(exp(x/2),x=1..5);
 

Here is the same plot, using the parametric values from a table of values in Excel.
Here is the same plot in parametric form:
Using Maple this is:

> plot([ln(x),sqrt(x),x=1..150]);
 


1.7, #25. Graph the curve: 
 
 

> implicitplot(x=y-3*y^3+y^5,x=-3..3,y=-3..3,numpoints=10000);

1.7, #30. If a projectile is fired with an initial velocity of  meters per second at an angle  above the horizontal and air resistance is assumed to be negligible, then its position after t secons is given by the parametric equations:   and  where g is the acceleration due to gravity 
        (a) If a gun is fired with , amd , when will the bullet hit the ground? How far from the gun will the bullet
             hit the ground? What is the maximum height reached by the bullet?

        (b) Use a graphing device to check your answers to part (a). Then graph the path of the projectile for several other values of the angle
to see where it hits the ground. Summarize your findings.

        (c) Show that the path is parabolic by eliminating the parameter.
 
 
 

Substitute in the  and the .
Simplify
Set equal to zero because we are trying to find when the bullet has zero height, and substitute the value for gravity, and simplify.
Factor the right side.
t = 0 could indicate before the bullet was fired. We are interested in the t = 51.02 sec.
Part (b) parametric plot. Using Maple this would be: 
> plot([250*sqrt(3)*t,250*t-4.9*t^2,t=0..52]);

(Exact values for the sine and cosine of 30 degrees were used)

Here you will see that an angle of 20 degrees was used. To plot correctly, a coversion is present to change to radians.

> plot([500*cos(20*3.14159/180)*t,500*
sin(20*3.14159/180)*t-4.9*t^2,t=0..52]);

 

Here you will see that an angle of 40 degrees was used. To plot correctly, a coversion is present to change to radians.

> plot([500*cos(40*3.14159/180)*t,500*
sin(40*3.14159/180)*t-4.9*t^2,t=0..52]);
 

If we can eliminate the parameter (c), then we can verify that this is paraboloic.
We substitute in the given values for  and , simplify, and solve for t.
Again, we substitute in the given values for  and . Now we substitute our value for t which we found in the last step. Then we simplify. The result is a paraboloa which opens downward.


Assignment
1-33, Odds
Lesson 1.7, Pages 81-83

 

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