is the intuition: the derivative tells us the change in output y (from f(b)) in response to a change of input x at x = b.

Transcription

1 Uses of differentials to estimate errors. Recall the derivative notation df d is the intuition: the derivative tells us the change in output y (from f(b)) in response to a change of input at = b. Eamples. y = f( b + y ) f(b). = f (b) (approimately) The radius of a sphere is measured to be r = 84 cm with a possible error of r = ±.5 cm. What is the surface area of the sphere? Recall S = 4πr, so ds dr = 8πr. We have S = 4 π 84 = 88, 668. cm The uncertainty of r = ±.5 cm in the measurement of the radius r, means there uncertainty in the area is S = ( 8 π r ) r = 8 π 84 (±.5) cm =, 55.5 cm What is the volume of the sphere? Recall V = 4 πr, so dv dr = 4πr. We have V = 4 π 84 =, 48, 7.6 cm The uncertainty of r = ±.5 cm in the measurement of the radius r, means the uncertainty in the volume is V = ( 4 π r ) r = 4 π 84 (±.5) cm = 44.4 cm

2 Uses of the tangent line. We give two important uses of the tangent line to a graph: Accurate estimates of the function. A method (Newton s method) to determine where the graph of a function crosses the -ais. Accurate estimates of the function. If P = (b, f(b)) is a graph point of a function f, and m = f (b) is the tangent slope at P, then the tangent line: y = m ( b) + f(b) gives very accurate estimates of the function near the input = b. Eample. Consider the function the implicitly defined function y = y() which is defined by the equation: cos ( y ) + sin( ) = We looked at this in week 5. The point P = ( π 4, π ) lies on the graph, and the tangent slope at P is:

3 .5 cos(y-) + sin() = We use implicit differentiation to find the tangent slope at P : d ( ) d cos ( y ) + sin( ) = d d ( ) = sin ( y ) ( dy d ) + cos () = ( dy d ) + cos () sin ( y ) = So, dy d = + cos () sin ( y ), and the tangent line at P = ( π 4, π ) is dy d ( π 4,π ) = + cos ( π 4 ) sin ( π π 4 ) = y = T () = ( π 4 ) + π The net table gives values of y() and T () for near π 4. actual = f( π 4 + ) est. = T ( π 4 + ) error relative error actual est. ( actual est. ) This eample shows that the tangent line (at input = b) can be used to estimate the values of a function for inputs near = b. Not only will the error: error = ( true value at input (b + ) ) ( tangent line value at (b + ) ) go to zero as, but the relative error error goes to zero too.

4 Use the tangent line of the function y = to give an estimate value for 4.. We know 4 =, so P = (4, ) lies on the graph of the function. We have The tangent line at P is: dy d =, The tangent line estimate for 4. is thus: dy d =4 = 4 = 4 T () = 4 ( 4 ) + T (4.) = 4 ( 4. 4 ) + =.5. The actual value of 4. is Newton s method. This method was discovered by Sir Isaac Newton in the late 6 s to numerically solve for roots of equations. Illustration of Newton s method. The function f() = + has irrational roots. One of the roots is betwwen.5 and. ^ - + old new Newton s method is to take a guess for the root we take old =. If the guess is not a root, then follow the tangent line at P = ( old, f( old )) to where it crosses the -ais and call that point new.

5 So, Since f () =, the tangent slope at P = ( old, f( old )) is m P = old. Then, m p = f( old ) old new. new = old f( old) m P = old old old + old new = old old For our guess =, the new guess is then = 5 = 9 = Newton s method is to then take as our guess, and compute a new guess in the same fashion, so = = old new There is a root at ^ - + old new

6 The equation + = has three roots. We can use Newton s method to determine the approimate have of the roots. We take initial guesses of and.5 and.5. We get: The three roots of + = are approimately.588,.4796, and Newton s method: Suppose a function f is a differentiable on an interval [a,b], and the graph crosses the -ais at some point in the interior of the interval. Suppose is an initial guess of a root f( root ) = in the interval. Then if f is suitably nice, and the inital guess is close enough to the root, the sequence,,... of roots of tangent lines given by new = old f( old) f ( old ) will converge to (have limit) root. We call the function I() = f() f () the iteration function, and to the inital guess, we have: = I( ), = I( ), = I( ), 4 = I( ),...

7 Eample: Consider the set of points which satisfy + y 6y = The line y = intersects the graph in three points. Find numerical estimates of the three points. When y =, the equation + y 6y = becomes = + 8 ; so we need to find the roots of f() = + 8 =. We have f () =. If we use old as a guess for a root, then Newton s method says the net guess should be net = old f( old) f ( old ) = old old old + 8 old = old 8 old = old 4 old The three roots of + 8 = are approimately.6447,.69459, and

8 Eample where Newton s method fails. We take g to be the odd continuous function g() = < which is differentiable for and < g () = < If we take initial guess old = a >, the new guess is new = a g(a) a g (a) = a = a a = a ( < ) Similarly if we take initial guess old = a <, the new guess is new = a g( a) g ( a) = a a = a + a = a ( > ) a a It follows that if we choose any b, the sequence of guesses will just endlessly switch back and forth b, b, b, b,