Solving for Coordinates at 35 Degree Angle: Derivative & Next Steps

[Arienne]

We are currently interpreting derivatives using the delta process in my analytical geometry class. The problem is: Calculate the coordinates (x,y) of the point at which the graph y= 0.027x^2 + 8.3x -7.5 is at a 35 degree angle from the horizontal. I have figured the derivative to be equal to 0.054x + 0.027 delta x +8.3
Question #1: Is the derivative correct?
Question #2: What do I do now?
Thanks

 

[HallsofIvy]

No, that is not the derivative. What you have calculated is the "difference quotient":
(f(x+ deltax)- f(x))/detax. To find the derivative itself, take the limit as deltax goes to 0. In this particular case, that's very easy: it is exactly the same as setting deltax equal to 0: 0.054x +8.3.

You should know that the derivative is the same as the "slope of the tangent line" and that the slope is equal to tan(θ) where θ is the angle the line makes with the x-axis (horizontal). In this problem you are told that that is 35 degrees.

Find the tangent of 35 degrees, set 0.054x +8.3 equal to that, and solve for x.

 

[arildno]

I'd like to add to HallsofIvy's comment a link to concepts you are probably familiar with from analytical geometry:

When your teacher introduced you to "the derivative", I would think that he/she first talked about secants/secant lines and then about tangents/tangent lines. (?)

In particular, you may have learned that the slope of the tangent line at some point of a curve may be found as the limit of the slopes of secant lines associated with that point when the distance between the two points (on the curve) defining a secant line goes to zero.

To make maths out of this:
1.
Let the two points on the curve be:
$$P_{1}=(x,y(x)), P_{2}=(x+\bigtriangleup{x},y(x+\bigtriangleup{x}))$$
2.
Since you've got two points, $$P_{1},P_{2}$$, you can evidently draw a straight line between them!
This straight line is called the secant line S with respect to the points $$P_{1},P_{2}$$.

3.
Now, I would think that you know that a straight line L in the plane usually can be represented as a function Y(X)=Ax+B, where Y(X) is the vertical coordinate Y at a point at L, while X is the horizontal coordinate of the same point at L.
A and B are constants for L (equal values for all choices of X!); B is called the Y-intercept (lies on the Y-axis, X=0), while A is called the slope of L.
4.
Going back to our secant line S, how can I find its slope?
We know 2 things about S:
a) If X=x, then Y(X)=y(x) (i.e, we're at the point $$P_{1}$$)
b) If $$X=x+\bigtriangleup{x}$$, then $$Y(X)=y(x+\bigtriangleup{x})$$ (i.e, we're at the point $$P_{2}$$)

Going back to the general equation for a line L, we must therefore have for S:
$$A*x+B=y(x), A*(x+\bigtriangleup{x})+B=y(x+\bigtriangleup{x})$$
Solving these equations for A and B, we find:
$$A=\frac{y(x+\bigtriangleup{x})-y(x)}{\bigtriangleup{x}}$$
$$B=y(x)-A*x$$

Hence, we may represent the Y-coordinate of a point on S, $$S_{Y}$$, as a function of the horizontal coordinate, X, like this:
$$S_{Y}(X)= \frac{y(x+\bigtriangleup{x})-y(x)}{\bigtriangleup{x}}X+y(x)-A*x$$

This is the way in which the secant line S can be represented in the usual manner of a line L.

5.
We are interested in the slope of S, $$\frac{y(x+\bigtriangleup{x})-y(x)}{\bigtriangleup{x}}$$
This is called the quotient of differences, as HallsofIvy says.
6.
In order to find the slope of the tangent line at point $$P_{1}$$, we evaluate the slope expression from S as we let the difference between the values of horizontal coordinates of $$P_{1},P_{2}$$ shrink to zero.
(That difference is $$\bigtriangleup{x}$$).

Geometrically, this limiting process has the interpration that we evaluate the slopes of different secant lines which have $$P_{1}$$ in common, but where each secant line's $$P_{2}$$ is chosen to be progressively closer to $$P_{1}$$.
The tangent line's slope is found when $$P_{2}$$ becomes $$P_{1}$$.

The derivative of y at x, $$\frac{dy}{dx}$$, is the name of the slope of the tangent line at $$P_{1}$$.