- Grade new problem set by next Thursday
- Example: position (s), velocity (v), and acceleration (a)
- v = ds/dt
- a = dv/dt
- Antiderivatives
- s = 1/2 at2 + Ct + K
- Meteor
- Height = 60 miles
- Speed = 30000 mi/hr
- g (=a) = 32 ft/sec2
- s = -16 t2 - 43400 t + 316800
- (Units are feet and seconds)
- = 316800 - 43400 t - 16 t2
- What if you don’t have someone to tell you that C is initial velocity and K initial height, all you know is height = K - Ct - 16t2?
- s = -16 t2 - 43400 t + 316800
- Solve for constants using measured numbers
- Notation and rules for antiderivatives
- Normals (perpendiculars) to curves
- Lighting effects in computer graphics, e.g., highlights from
partially reflected light source, darkening of surface where
light is less direct
- Calculating brightness is these cases depends on direction
light is coming from relative to “normal”, aka
direction perpendicular to surface
- Suppose profile of flared tube above is described by equation y = √x - 1
- What is slope (or equation) of normal at any x value?
- Basic idea:
- Differentiate equation for curve
- Take negative reciprocal (get slope of perpendicular)
- Line w/ slope m, perpendicular has slope -1/m
- Plug in x values to get slope of normal at any point (and to calculate point)
- d( √x - 1 )/dx = 1/2√x
- Slope of normal = -2√x
- What’s equation for normal line at x = 1?
- m = -2, point on line = (1,0)
- y = -2x + 2
- Line w/ slope m through point (x0,y0) has equation (y-y0) = m(x-x0)
- Basic idea:
- Lighting effects in computer graphics, e.g., highlights from
partially reflected light source, darkening of surface where
light is less direct
- Product and quotient rules
- Higher-order derivatives
- “Products and Quotients” part of section 3.3
- Top of page 122 - bottom of 124