MHB What are the essential precalculus concepts for success in calculus 1?

[mathdad]

Find values of t such that the points (-2, 3) and (t, 1) are 6 units apart.

The set up is:

6 = sqrt{(t + 1)^2 + {(1 - 3)^2}

I now square both sides to solve for t.

Correct?

 

[MarkFL]

Find values of t such that the points (-2, 3) and (t, 1) are 6 units apart.

The set up is:

6 = sqrt{(t + 1)^2 + {(1 - 3)^2}

I now square both sides to solve for t.

Correct?

You have the right idea to use the distance formula, however you haven't applied it correctly. What you want is:

$$6=\sqrt{(t-(-2))^2+(1-3)^2}$$

You see, the distance formula states that the distance $d$ between the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by:

$$d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$$

 

[mathdad]

You have the right idea to use the distance formula, however you haven't applied it correctly. What you want is:

$$6=\sqrt{(t-(-2))^2+(1-3)^2}$$

You see, the distance formula states that the distance $d$ between the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by:

$$d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$$

Are you saying that 6 should be squared?

 

[MarkFL]

Are you saying that 6 should be squared?

You will want to square both sides to solve for $t$, but do you see how the equation you stated is incorrect?

 

[mathdad]

I see my error in the radicand. I usually post my questions after work between 1:30am and 3am. I am tired after midnight. I have a split days off schedule. On such a schedule, I must take advantage of every minute to get things done. I am off on Tuesday and Friday. It's a horrible schedule but what can I do?

 

[MarkFL]

The problem is equivalent to saying:

Consider the circle centered at (-2,3) with radius 6 and the line y = 1. What are the x-coordinates of the point(s) of intersection between the circle and the line?

Let's look at a graph:

View attachment 6650

We have the following equation in $t$:

$$(t+2)^2+2^2=6^2$$

Arrange in standard quadratic form:

$$t^2+4t-28=0$$

Applying the quadratic formula, we obtain:

$$t=-2\pm4\sqrt{2}$$

 

[mathdad]

The problem is equivalent to saying:

Consider the circle centered at (-2,3) with radius 6 and the line y = 1. What are the x-coordinates of the point(s) of intersection between the circle and the line?

Let's look at a graph:
We have the following equation in $t$:

$$(t+2)^2+2^2=6^2$$

Arrange in standard quadratic form:

$$t^2+4t-28=0$$

Applying the quadratic formula, we obtain:

$$t=-2\pm4\sqrt{2}$$

Sorry but I cannot see the line y = 1.

 

[MarkFL]

Sorry but I cannot see the line y = 1.

If your browser isn't rendering images well, you can sketch the circle and line pretty easily. Plot the point (-2,3) then one at a time move up, right, down, left 6 units from the center and plot points there and then connect the 4 points with a circle. Then draw the horizontal line y = 1 to get an idea where the values of t will be, as they will be the x-coordinates of the points of intersection between the circle and line.

 

[mathdad]

The problem is equivalent to saying:

Consider the circle centered at (-2,3) with radius 6 and the line y = 1. What are the x-coordinates of the point(s) of intersection between the circle and the line?

Let's look at a graph:
We have the following equation in $t$:

$$(t+2)^2+2^2=6^2$$

Arrange in standard quadratic form:

$$t^2+4t-28=0$$

Applying the quadratic formula, we obtain:

$$t=-2\pm4\sqrt{2}$$

Ok. I can now see the entire circle and the line y = 1.

- - - Updated - - -

If your browser isn't rendering images well, you can sketch the circle and line pretty easily. Plot the point (-2,3) then one at a time move up, right, down, left 6 units from the center and plot points there and then connect the 4 points with a circle. Then draw the horizontal line y = 1 to get an idea where the values of t will be, as they will be the x-coordinates of the points of intersection between the circle and line.

Mark,

Thank you for your help. I've decided not to post every question from the David Cohen precalculus textbook. This task will take forever to accomplish. I will instead focus on the questions every student should know how to solve before going to calculus 1. I love my precalculus textbook but it is simply too lengthy.