Vertical tangent lines If a function f is con...

Question

Vertical tangent lines If a function f is continuous at a and lim x→a| f′(x)|=∞, then the curve y=f(x) has a vertical tangent line at a, and the equation of the tangent line is x=a. If a is an endpoint of a domain, then the appropriate one-sided derivative (Exercises 71–72) is used. Use this information to answer the following questions.
Graph the following curves and determine the location of any vertical tangent lines.
a. x²+y² = 9

Graph the following curves and determine the location of any vertical tangent lines.

a. x²+y² = 9

Accepted Answer

Below there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Sketch the graph of the function X2 + Y2 equals 25. And its vertical tangent lines. Awesome. So it appears for this particular problem we're asked to solve for two separate answers. Our first answer is we're trying to figure out what the sketch of this function is, and our second answer we're trying to solve for is we're trying to figure out what its vertical tangent lines are. Awesome. So now that we know what we're ultimately solving for, let's take a moment here to examine our function. And as you can see, X2 + Y2 equals 25. This given given function, as we should recall, will be the equation for a circle, and this circle will have a radius of R equals 5, because the square root of 25 is 5. So we'll have a radius of 5, and it will be centered at the origin. So creating a quick sketch. Of this particular graph. So if we have just a skeleton coordinate plane here, noting that we have a circle centered at the origin. Which the origin I will indicate in red. So the origin is at 0.0, that's our origin point, and it's going to be its radius will be 5, so it'll be at stretching from the origin point going out to 5 and then negative 5. And 5 and negative 5. So that's how far it will stretch out. So it's gonna be from the origin point going up 5, then going down 5 to the right to 5, and to the left 5 because of this radius of 5. Awesome. And once again, let's quickly note that our Y-axis is denoted by the vertical line and our X-axis is the horizontal line. Awesome. So with that said, now that we have a sketch of what our circle graph looks like, we now at this point need to find the vertical tangent lines, and in order to do that, we must recall that we have to solve for Y as a function of X. So with that said, we need to take Y2 is equal to 25 minus X2, which rearranging this equation to isolate and solve for Y all by itself will give us that Y is equal to plus or minus the square root of 25 minus X2. And now we can determine the vertical tangent lines we need to differentiate Y equals plus or minus the square root of 25 minus X2. So in order to in order to do that, we need to take Y, where Y is a fancy way of saying we're taking the first derivative of this function of Y. So Y is equal to D by DX where D by DX is a fancy way of saying we're taking the derivative with respect to X. So Y is equal to D by DX of plus or minus the square root of 25 minus X2. Fantastic. So, we can go ahead and write that Y is equal to Plus or minus 1 divided by 2 multiplied by the square root of 25 minus X squared. Multiplied by D by DX of negative X2. Fantastic. And we can take it a step further and we could write that Y is equal to + or minus 1 divided by 2 multiplied by the square root of 25 minus X2 multiplied by 2 X because 2 X is the derivative of negative X to the power of 2. Fantastic. So now, Our next step that we need to take Because we need to simplify a little bit more, and when we do, we will find that Y is equal to + or minus. X divided by the square root of 25 minus X2. Fantastic. And once again, we just simplified our derivative of this function of Y. Fantastic. So now our next step is we need to recall that the vertical tangent occurs where the derivative of the function is undefined or infinite, since the slope of a vertical line is undefined. So, in order to find where the derivative is undefined, we need to recall that we need to equate the denominator of the derivative to 0, meaning we need to take 25 minus X2. And we're taking the square root of 25 minus X to the power 2, and we need to set it equal to 0. So, When we start to solve, we will get 25 minus X squared is equal to 0. And once again, we're trying to rearrange this equation to isolate and solve for X all by itself, which I'm going to skip a few steps, but when you get X all by itself, you will find that X is equal to + or minus 5, and we need to recall and note that the function. X2 + Y2 equals 25, and it will be continuous at X equals plus or minus 5. And the derivative of the function will become undefined at X equals plus or minus 5, which confirms the presence of these vertical tangent lines. So, therefore, the function X2 + Y2 equals 25 will have vertical tangent lines at, and let's make a note of this, so we'll have vertical tangent lines at. X equals -5 and X equals 5. And that's it. We found our second answer. Hooray, we did it. So now we need to draw our graph that also has our vertical tangent lines in it, noting once again that our vertical tangent lines are at X equals -5 and X equals 5. So, when we go ahead and graph our expression, we will find, and I've gone ahead and Placed our graph into place the function that we are given, which is X2 + Y2 equals 25 into some graphing software to create this digital graph. Now as you can see, based off of our sketch we drew earlier, we have our circle, which is centered at the origin point. And as you can see, it goes to the to the right 5, and then to the left 5, it goes up 5 and down 5 since it has this radius of 5. And as you can see, we have our tangent lines at X equals -5 and X equals 5. Fantastic. And that's it. That is our final answer. So this is our final answer for our graph that includes the tangent lines. Hooray, we did it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Key Concepts

Vertical Tangent Lines

Implicit Differentiation

Endpoints and One-Sided Derivatives

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