OK, after getting a nice answer for my last calculus problem, I found out that the equation I wrote down was wrong, and using the new equation I end up getting stuck finding the slope. I needed to find a tangent line to two points, the first is y=x^2, the second (which I had written incorrectly), is y= -x^2+6x-5. I got all the coordinates of the two points in terms of C by setting the derivatives equal to each other (since the slope at each point is the same) and getting [-c+3 , (-c+3)^2] for the point on the curve y=x^2; (c, -c^2+ 6c-5) for the point on the second curve. Then I solved for the slope with rise over run. In the original equation, I got the slope coming to an even 4/3, but for the new equation, I get the slope as being (-2c^2 +12c -14)/(2c-3)
How do I continue now that I have a variable in the slope? Am I doing something incorrectly, because in the original problem I had all the variables cancelling each other out…
Because the teacher told me that to solve the problem I had to get all the points in terms of one variable, so I just used c. Setting the derivatives equal got me 2a= -2c+6 —> a=-c+3
Those aren’t points, they’re functions. The slope of the tangent at any point along a function is the derivative of that function. For the first function it’s slope=2x. So at x=4, the slope of the tangent line is 8. I’m not sure if you learned how to take derivatives of functions, and I’ll be damned if I remember the formula. I do remember there is a shortcut, which is to take each term, multiply by the current power of x (in this case 2), then lower the power of x by 1. So x^2 became 2x. x^3 would become 3x^2. A number, say 5 is equivalent to 5x^0, so it becomes 0. Thus your second equation; y = -x^2+6x-5 becomes -2x + 6 + 0, or just 2x + 6. So that will give you the slope of the tangent to that function at any point along the function. Using the shortcut is a good way to check your answer from the long way.
You probably need to do it the long way for class. This may be what your teacher was talking about where you need to get everything in terms of one variable.
I’m not sure why you’d need to set the derivatives equal to each other, though. Unless maybe you’re looking for a point along the two functions where the tangent has the same slope.
On preview, I see that this is probably what you want, so you wouldn’t make up new variables, you would just set 2x = -2x + 6 and solve for x. You should be able to do that without help.
I don’t understand the question. You say that you need to find the tangent lines to two points, but you give the points as y=x^2 and y=-x^2+6x-5 – those are equations of curves, not points.
Maybe you mean that you need the tangent line at the point where the two curves touch, but the curves don’t necessarily touch, and even if they do, the tangents are not necessarily equal at the intersection point. These are things you need to check and understand before you proceed with the problem.
Yes, I know that those are equations for curves. What I need to find is the equation of a line that’s tangent to both of the curves (theres supposedly two that exist, I only need one). The curves don’t intersect, just picture a U and an upside down U somewhat close to each other with the tangent line going from the bottom right of the U to the top left of the upside down U.
… and since it’s a tangent line shared by two curves, I can’t just find the derivative for one curve like normal. But since the point on each curve that has the tangent line through it share the slope (a line obviously can’t change slope), I set the derivatives (or slopes) equal. That helped me get EACH point in terms of C. Now that I had the 4 coordinates of the two points on the graph, I just did a simple rise over run equation to find the slope of the line. But this time, unlike with the last equation, had the c variable remaining after calculating the rise/run. Last time the c’s conveniently cancelled out.
It sounds like you want what I said at the end of my post. You’re trying to find the point(s) where the tangent line on both functions has the same slope.
You just set the derivative functions equal to each other. There’s no need to make up a c variable, just use x.
2x = -2x + 6
4x = 6
x = 6/4 = 3/2
So at x = 3/2, both functions have the same tangent slope.
x=3/2 is the x-coordinate where both curves have the same slope. But the tangent line to both curves is going to intersect y=x[sup]2[/sup] at some(x[sub]1,y[sub]1), and will intersect the other function at some (x[sub]2,y[sub]2), where x[sub]2 and x[sub]1 aren’t equal.
The y intercept of the tangent line. I have to write the equation of the tangent line (y=mx+b) to the two points that the line goes through.
RT, I don’t get how you would find what the points are in order to get the slope using what you gave me. If I plugged in 3/2 into each equation, I would have (3/2, #) and (3/2, #), meaning that the tangent line is straight up and down, obviously not the case when looking at the graph.
I see. Right now, you can get the slope of the tangent line at the point x=3/2, at which both functions’ tangents’ have the same slope.
The problem is that they both don’t have the same y intercept, since the point at which the tangent slopes are equal is not the intersection of the functions (or is it?). So you’re going to end up with 2 different tangent lines in the mx+b form. However, the m will be the same for both.
Let’s use function 1 for example:
at x=3/2, y = x^2 evaluates to: y = 9/4. This is the coordinate along the first function at which we are finding the tangent function. (3/2, 9/4).
Now, the derivative function is slope of tangent = 2x. The slope of the tangent line is thus 3/2 * 2, or 3. So we have the tangent line function, y = 3x + b. To find b, we need an x and y value along that function. Conveniently, we have the answer to #1.
Thus we can say that 9/4 = 3(3/2) + b. Solve for b to get the b value for the mx+b function that describes the tangent to function 1 at x = 3/2.
Repeat the same for function 2.
I don’t think the two functions you gave have a point where they have the exact same tangent line, but if they do, then that point is x = 3/2, and you’ll see you get the same tangent line equation for both.
Man, I wish I had my TI-82 so I could graph these and see.
I forgot I have Matlab here. I graphed the two functions, and in fact, they DO NOT intersect. They come really close, at x = 3/2, but there’s definitely space in between them. You can see from the picture that they have the same tangent slope at that point though, so basically everything so far is correct.
I must admit I’d be interested in a verbatim statement of the problem, from your text, handout, or whatever.
I wasn’t supplying a method to find the points in question - I was simply explaining to c_goat why x=3/2 couldn’t be a solution to the problem as described.
I guess if the question is “find the tangent line that is tangent to both of these curves” then everything I’ve done so far is wrong. But you can’t blame me, I was just following orders
Maybe I should just let the pros handle it next time. I got a little excited trying to remember my calculus.
So how would you find the line that is tangent to both curves, not necessarily at the same x position though? I can’t seem to come up with anything.
