Geometrical interpretation of derivatives Name 1. Interpret the function π΄(π₯) = π₯ 2 as the area of a square with dimensions π₯ by π₯. a) Label the square with the dimensions. b) Now assume each dimension is increased by a tiny amount ππ₯. Label each component of the additional area. c) Using the previous labeled drawing, write down an expression for the increase in area, ππ΄. d) If ππ₯ were truly a tiny increase, which components of your expression for ππ΄ are dominant? Which can ignored? e) Solve your expression for ππ΄⁄ππ₯ . 2. Repeat the steps in problem 1 for π΄(π₯) = (π₯ + 1)(π₯ + 3). 3. Repeat the steps in problem 1 for the general function π΄(π₯) = π(π₯)π(π₯) to determine ππ΄⁄ππ₯ . Begin by assuming the dimensions are increased by ππ and ππ. Once you have an expression for ππ΄, assume ππ ππ that a tiny increase ππis equal to ππ₯ ππ₯ and that a tiny increase ππ is equal to ππ₯ ππ₯. . 4. Repeat the steps in problem 1 for the π(π₯) = π₯ 3 to determine ππ⁄ππ₯ . Interpret π(π₯) as the volume of a cube with dimensions of π₯ by π₯ by π₯. 5. a) Draw a unit circle on an x-y coordinate plane. Draw an angle π in the first quadrant in standard position. If an angle is represented in radians, the angle is equal to length of the arc in the unit circle intercepted by the angle. Label this arc π. b) Construct a right triangle (Triangle 1) using the angle π and the terminal side as a hypotenuse. Label the point on the terminal side of the angle that crosses the unit circle using cos π and sin π. c) Imagine zooming in on the piece of the arc right next to the terminal side of the angle. Increase the arc, π, by amount ππ. Assume that ππ is small enough that in this region the arc can be assumed to be a straight line. Also think about increasing the arc as moving the point on the terminal side by small amount in the x direction, π cos π, and y direction, π sin π. Construct a right triangle (Triangle 2) using ππ, π sin π, and π cos π. d) How does Triangle 2 compare to Triangle 1? e) Use Triangle 2 to compute an expression for π sin π⁄ππ. 6. a) Graph a function π(π₯) = π₯ 2 . Shade the area from 0 to π₯. Call this area π΄(π₯). b) Increase the shaded area by a tiny amount ππ₯. Write down an expression for ππ΄. c). Solve your expression for ππ΄ for ππ΄⁄ππ₯ . d) What does your expression for ππ΄⁄ππ₯ tell you about the function π΄(π₯)?