Introduction to Derivatives
When we have a formula for a function, we can determine the slope of the tangent line at a point by calculating the slope of the secant line through the points and , , and then taking the limit of as approaches (Fig. 8) :
Example 2: Find the slope of the line tangent to the graph of at the point . (Fig. 9).
Solution: In this example , so and
The slope of the tangent line at is
The tangent line to the graph of at the point has slope 4.
We can use the point–slope formula for a line to find the equation of the tangent line:
so and .Practice 4: Use the method of Example 2 to show that the slope of the line tangent to the graph of at the point is . Also find the values of at and .
It is possible to find the slopes of the tangent lines one point at a time, but that is not very efficient. You should have noticed in the Practice 4 that the algebra for each point was very similar, so let's do all the work once for an arbitrary point and then use the general result for our particular problems. The slope of the line tangent to the graph of at the arbitrary point is
The slope of the line tangent to the graph of at the point is . We can use this general result at any value of without going through all of the calculations again. The slope of the line tangent to at the point (4,16) is and the slope at is . The value of determines where we are on the curve ( at ) as well as the slope of the tangent line, , at that point. The slope is a function of and is called the derivative of .
Simply knowing that the slope of the line tangent to the graph of is at a point can help us quickly find the equation of the line tangent to the graph of at any point and answer a number of difficult sounding questions.
Example 3: Find the equations of the lines tangent to at (3,9) and .
Solution: At , the slope of the tangent line is , and the equation of the line is so and .
At , the slope of the tangent line is , and the equation of the line is so and .
Example 4: A rocket has been programmed to follow the path in space (from left to right along the curve), but an emergency has arisen and the crew must return to their base which is located at coordinates . At what point on the path should the captain turn off the engines so the ship will coast along the tangent to the curve to return to the base? (Fig. 10)
Solution: You might spend a few minutes trying to solve this problem without using the relation , but the problem is much easier if we do use that result.
Lets assume that the captain turns off the engine at the point on the curve , and then try to determine what values and q must have so that the resulting tangent line to the curve will go through the point . The point is on the curve , so , and the equation of the tangent line, found in Example 3, is .
To find the value of so that the tangent line will go through the point , we can substitute the values and into the equation of the tangent line and solve for :
The only solutions of are and , so the only possible points are and . You can verify that the tangent lines to at and go through the base at the point (Fig. 11). Since the ship is moving from left to right along the curve, the captain should turn off the engines at the point . Why not at ?
Practice 5: Verify that if the rocket engines in Example 4 are shut off at , then the rocket will go through the point .