Implicit Differentiation and Related Rates

Transcription

1 Math 3A Discussion Notes Week 5 October 7 and October 9, 05 Because of the mierm, we re a little behind lecture, but this week s topics will help prepare you for the quiz. Implicit Differentiation and Related Rates So far the derivatives we ve taken have been derivatives of functions, but we know that not all graphs are graphs of functions. For example, suppose we want to find the slope of the line tangent to the circle x + y at the point 3, ). How would we do this? Our answer will be implicit differentiation, which means differentiating an equation which defines y implicitly. We will treat y as if it were a function of x, and then take derivatives accordingly. In this example we have d x + y ) d ) x + y 0. Solving for y, we find that x. Just as before, the derivative continues to be interpreted y as the slope of a tangent line. In particular, the slope of the tangent line to x + y at 3, ) is ,) Example. Find an equation for the line tangent to the graph of xy + x y 6 at the point, ). Solution) We begin by differentiating the given equation. Notice our use of the product and chain rules. ) )y) + x + x)y ) + x ) y ) 0. So x + x y ) y xy In particular, the derivative at, ) is,). So an equation of the tangent line is given by y x ) +. y xy x + x y y + xy) x + xy) y x.

2 One of the most common uses for implicit differentiation is in related rates problems. These are problems where we have two quantities, related by some equation, and we want to know how quickly one of the quantities is changing, given the rate of change of the other quantity. There s a common strategy that will be helpful to you in most related rates problems:. Draw a picture of the situation.. Determine relevant variables. 3. Write down what you know as equations and formulae). In particular, relate your variables somehow.. Differentiate the equation relating your variables. 5. Solve for the unknown variables) and derivatives). We ll demonstrate most of) this strategy in the following, classical example. Example. A meter ladder leans against a wall. At time t 0, the bottom of the ladder is meter away from the wall, and the bottom of the ladder begins moving away from the wall at a constant rate of 0.5m/s. Find the velocity of the top of the ladder at time t. Solution) The part of our strategy that we ll skip is awing a picture. The picture you aw should be a triangle, with the hypotenuse representing the length of the ladder. Determine relevant variables. At time t, let s call the height of the ladder yt) and the distance of the base of the ladder from the wall xt). As in the statement of the problem, t will represent the time elapsed since the ladder began falling, measured in seconds. Write down what you know. Using the pythagorean identity, we can relate x and y: xt)) + yt)). ) We also know the position of the base of the ladder at time t 0: x0), and we know the constant rate at which the base is moving: 0.5m/s. Differentiate the equation relating your variables. We can differentiate ) to obtain xt) + yt) 0. Solve for the unkown. The quantity we re interested in is the velocity of the top of the ladder, which is represented by. We can solve for this by rearranging the above equation. xt) yt) xt) yt) 0.5m/s.

3 In particular, we want to know the velocity at time t, so this will require us to compute x) and y). Since the base of the ladder is moving at a constant rate of 0.5m/s, Then, using ), So x) x0) + 0.5m/s s m + m m. y) 6 x)) 6 3m m/s 3 m/s. So after seconds the ladder is falling at a rate of / meters per second. Example. Gas is escaping from a spherical balloon at the rate of 000 cubic inches per minute. At the instant when the radius is 0 inches, at what rate is the radius decreasing? At what rate is the surface area decreasing? Solution) We ll call the radius of the balloon rt) and the volume of the balloon V t). These are related by the formula V 3 πr3, and we know that dv 000in 3 /min at all times. We may differentiate the above equation to find that dv πr dv πr. Substituting in our radius, we find that when the radius of the balloon is 0 inches, 00π 000) 5 π in/min. So the radius is decreasing at a rate of 5/π) inches per minute. For the surface area we have the formula A πr. Differentiating this we have Then, when r 0in, da 8πr. da 8π0) 5 ) 00in /min. π So when the radius is 0 inches, the surface area is decreasing at a rate of 00 square inches per minute. 3

4 Example. The thin lens equation in physics is s + S f, ) where s represents the distance of an object from a lens, S represents the distance from the lens of the image of this object, and f is the focal length of the lens. Suppose that a certain lens has a focal length of 6cm and that an object is moving toward the lens at a rate of cm/s. How fast is the image distance changing at the instant when the object is 0cm from the lens? Is the image moving away from the lens or toward the lens? Solution) Since the image distance is represented by S, we re looking for ds, so we differentiate ): ds s + ds S 0. The focal length of the lens is constant.) Then When s 0cm, we have 0 + S 6 ds S /s ds/. S 5cm, so ds 5/00.5cm/s. Since the derivative of the distance is positive, the image is moving away from the lens at a rate of.5 centimeters per second. Example. Coffee is aining from a conical filter into a cylinical coffeepot at a rate of 0 in 3 /min. The filter is a circular cone with a heigh of 6 inches and a 6 inch diameter. The cylinical coffeepot has a diameter of 6 inches. a) How fast is the level in the pot rising when the coffee in the cone is 5 inches deep? b) How fast is the level in the cone falling at this point in time? Solution) a) Denote the volume in the pot at time t by V p t), and the height by h p t). Then V p and h p are related by V p π3in) h p V p 9in )πh p.

5 Differentiating this equation with respect to time, we obtain dv p 9in )π dh p dh p dv p 9in )π. Since the coffee is aining into the pot at a rate of 0 in 3 /min, dv p / 0in 3 /min, so dh p dv p 9in )π 0 9π in/min. Note that this rate is constant, and thus does not depend on the height of the coffee in the cone. b) As with the cylinder, we have a formula relating the volume of the liquid in the filter at time t to the height of the liquid in the filter at time t: V f t) πr f t)) h f t), where r f t) is the radius of the surface of the liquid in the cone. We also immediately have dv f 0in 3 /min. The radius depends linearly on the height of the liquid, and can be written as r f t) 3in h ft) 6in h ft). So the radius is 0 when the height is 0 and the radius is 3 inches when the height is 6 inches, in agreement with our original assumptions.) The formula for the volume now becomes ) hf t) V f t) π h f t) π h ft)) 3. Differentiating this with respect to t gives dv f 3π h ft)) dh f dh f 3πh f t)) dv f. So, when h f 5 in, dh f hf 5 3π 5in 0 in3 /min) 8 5π in/min. 5