limits and slope of tangents

[lilyz]

this question uses stuff from the beginning chaper of calculus. We've only learned how to rationalize radical expressions, calculate the slope of a tangent, the rate of change (average and instantaneous), how to evaluate the limit of a function, properties, and how to see if a graph or function is continuous (or if a certain point on that graph is continuous). This said, we got this question:

A ski boat travels in a parabolic curve. Let the vertex of the parabola be the origin and let the parabola open upward. The boat is currently at a point 100 m west and 100 m north of the origin, travelling toward the origin. The dock is situated 100m east and 50 m north of the origin. at what point should the skier release the tow rope to head straight for the dock?

(so far all I've figured out is that the equation of the parabola would be: y=(1/100) x^2
Hellp!! plz^^

 

[galactus]

The vertex is at (0,0) and the parabola passes through (-100,100).

Thus, using \(\displaystyle y=a(x-h)^{2}+k\), the parabola has equation \(\displaystyle y=\frac{x^{2}}{100}\)

The slope at any point on the parabola is \(\displaystyle y'=m=\frac{x}{50}\)

Now, use the given information to solve \(\displaystyle y-y_{1}=m(x-x_{1})\) for x.

Where \(\displaystyle (x_{1},y_{1})\) is the point the tangent line passes through. In other words, the location of the dock.

You have \(\displaystyle x_{1}, \;\ y_{1}, \;\ y, \;\ m\). Sub them in and solve for x. y follows.

 

[lilyz]

thank you^^

<<Now, use the given information to solve

for x.
Where

is the point the tangent line passes through. In other words, the location of the dock.
You have

. Sub them in and solve for x. y follows. >>

how is it that the tangent line passes through the dock? and with what information am i using to solve

for x?
and also, what is the question asking for? is the point the skier lets go of the rope a point on the parabola? if so, how do i find it?

thank you^^

 

[lilyz]

still in trouble...

"Now, use the given information to solve

for x.
Where

is the point the tangent line passes through. In other words, the location of the dock.
You have

. Sub them in and solve for x. y follows. "

thanks but:
how is it known the tangent line passes through the dock?
also, with what info am i using to solve

for x?
what are

?

finally, the question asks AT WHAT POINT THE SKIER SHOULD LET GO OF ROPE TO HEAD FOR DOCK. this point is on the parabola right? how do i find it?

thanks

 

[lilyz]

The vertex is at (0,0) and the parabola passes through (-100,100).

Thus, using \(\displaystyle y=a(x-h)^{2}+k\), the parabola has equation \(\displaystyle y=\frac{x^{2}}{100}\)

The slope at any point on the parabola is \(\displaystyle y'=m=\frac{x}{50}\)

Now, use the given information to solve \(\displaystyle y-y_{1}=m(x-x_{1})\) for x.

Where \(\displaystyle (x_{1},y_{1})\) is the point the tangent line passes through. In other words, the location of the dock.

You have \(\displaystyle x_{1}, \;\ y_{1}, \;\ y, \;\ m\). Sub them in and solve for x. y follows.

thanks
but, i still have a bit of trouble. what is the info given that can solve for

for x?
and since when did i have

? Sub them in (what??) and solve for x? y follows?
mostly, what i don't get is, it's asking for a point on the parabola right? what does it mean straight to the dock though? and how would i find that point? the skier can let go at any point and still head in a straight line for the dock no???!?!?

 

[galactus]

The skier is traveling along the parabola. At some point he is going to let go of the rope and head for the dock. This is the tangent line. The point on the parabola where he leaves go is the point where the line is tangent to the parabola, and the point the line passes through is the dock coordinates.
When they leave go of the rope, they keep heading straight for a time. This is the tangent line.

\(\displaystyle y-y_{1}=m(x-x_{1})\)

\(\displaystyle y=\frac{x^{2}}{100}, \;\ y_{1}=50, \;\ m=\frac{x}{50}, \;\ x_{1}=100\)

Sub these in. It is entirely in terms of x (the x coordinate of the 'leave go" point). Solve for x. This is the point where they leave go of the rope and head toward the dock at (100,50). The y coordinate can then be easily found by subbing your newly found x coordinate into the parabola equation.

You can even go a step further by finding the equation of this tangent line. You now have two points it passes through.

 

[lilyz]

srry, one last time :S
i've managed this so far, not sure if it's right
when using

, this is formula for slope right?
so how is it known that y = x^2/ 100??
also, when solving for x, there were two outcomes, both positive. I just chose the one closer to the question, but could both values/points work to answer the question, or is it just one of them?

also thank you soo much
you're a big help ^^

 

[pappus]

srry, one last time :S
i've managed this so far, not sure if it's right
when using

, this is formula for slope right? <--- this is the equation of a straight line in point-slope-form
so how is it known that y = x^2/ 100?? <--- you are going to calculate the coordinates of the "release-point" which is situated on the parabola. Therefore it's coordinate must satisfy the equation of the parabola.
also, when solving for x, there were two outcomes, both positive. I just chose the one closer to the question, but could both values/points work to answer the question, or is it just one of them?

also thank you soo much
you're a big help ^^

.... also, when solving for x, there were two outcomes, both positive ...

Your results are OK. Since you get two tangents to a parabola you get of course two tangent-points. The tangents pass through the point (100,50). And certainly only one tangent is plausible with your problem.

 

[HallsofIvy]

View attachment 1717
srry, one last time :S
i've managed this so far, not sure if it's right
when using

, this is formula for slope right?

That is the equation of a line, through \(\displaystyle (x_1, x_2)\) with slope m

so how is it known that y = x^2/ 100??

There are two different paths involved here- the parabola the skier follows as long as he is being towed by the boat and the straight line he will follow once he releases the tow rope. You were told "A ski boat travels in a parabolic curve. Let the vertex of the parabola be the origin and let the parabola open upward. The boat is currently at a point 100 m west and 100 m north of the origin, travelling toward the origin." A parabola with vertex at the origin has form \(\displaystyle y= ax^2\) for some value of a. Knowing that when x=100, y= 100 tells you that \(\displaystyle 100= a(100)^2\) so a= 1/100. The linear equation above is for the skier after he releases the tow rope.

also, when solving for x, there were two outcomes, both positive. I just chose the one closer to the question, but could both values/points work to answer the question, or is it just one of them?

also thank you soo much
you're a big help ^^

In a sense both are solutions. (I assume you meant "closer to the origin (dock)"). However, one of them has x positive, the other x negative. Since he is "currently at x= 100", waiting for the other point he would have to go through the dock!

 

[lilyz]

thank you all soo much
it was a big help ^^