HOMEWORK 7 Math75 solutions ----Warren D. Smith----------- #65 p.127: Circle has radius (in cm) r(t) = 10 t (t in seconds). So the area enclosed is 2 2 Area(t) = Pi r(t) = 100 Pi t (square cm) and the rate of increase in that area is Area'(t) = 200 Pi t (in square cm per second). When r = 20 cm, t = 2 sec, and Area'(t) = 400 Pi SQUARE CM / SEC. #23 p.136: If F(w) = w arcsin(w) then w F'(w) = arcsin(w) + ----------- 2 1/2 (1 - w ) . Why? The first step is to use the product rule F'(w) = 1 arcsin(w) + w arcsin'(w) . 2 -1/2 The next step is to use the formula that arcsin'(S) = ( 1-S ) which is from the book page 135. This formula also can be derived (if you did not have the book handy or did not memorize this) by implicitly differentiating both sides of arcsin(sin x) = x using chain rule to get arcsin'(sin(x)) cos(x) = 1 2 1/2 then write S = sin(x) and cos(x) = (1-S ) (from Pythagoras's 2 2 sin(x) + cos(x) = 1 ) to get the formula. #49 p.137: The question was: If R(t) is the radius of a sphere and R'(t) = 2 cm/sec is the rate of increase of that radius with time t, then what is the rate of change of volume ("rate of inflow of air into balloon") when R(t) = 10 cm? Answer: 3 Volume = 4 Pi R(t) / 3 [book, frontspiece] in cubic cm (cubic cm are called cc for short) 2 so Volume'(t) = 4 Pi R(t) R'(t) in cc per second. Why: because I used 3 the chain rule to differentiate R(t) = outer(inner(t)) where 3 2 outer(z) = z inner(t) = R(t) outer'(z) = 3 z inner'(t) = R'(t) 2 3 so outer'(inner(t)) inner'(t) = 3 R(t) R'(t) = derivative of R(t) . Now: If R(t) = 10 cm, then Volume'(t) = 4 Pi 100 2 = 800 Pi cc / second = 2513 cc/second approximately (which was answer in back of book). #11 p.140: If x and y are related by 3 x ln(y) + y = ln(x) (defining a function y(x) implicitly) then what is y'(x)? Answer: do implicit differentiation. The 3 terms are: x ln(y) is handled by product rule to get 1 ln(y) + (x/y) y' where note ln(y(x))=outer(inner(x)) is done by chain rule: outer(z)=ln(z), inner(x)=y(x), outer'(z)=1/z, inner'(x)=y'(x) so its derivative is (d/dx) (ln(y(x)) = outer'(inner(x)) inner'(x) = (1/y(x)) y'(x) 3 y(x) = outer(inner(x)) is handled by chain rule, 3 2 outer(z)=z inner(x)=y(x) outer'(z)=3z inner'(x)=y'(x) 3 2 2 so (d/dx) y(x) = outer'(inner(x)) inner'(x) = 3 y(x) y'(x) = 3 y y'. ln'(x) = 1/x. So... 2 ln(y) + (x/y) y' + 3 y y' = 1/x Rewrite as 2 [(x/y) + 3 y ] y' = 1/x - ln(y) So finally: 1/x - ln(y) y'(x) = ------------ 2 (x/y) + 3 y Final problem: suppose x and y are related by 3 x tan(y) + ln(x) y = 7 (thus implicitly defining a function y(x)) . Determine y'(x) in terms of y and x (Hint: use implicit differentiation). 3 NOTE: unfortunately I wrote x^3 instead of x in the problem statement since I was in COW mode! Sorry. So if you read it as 3x instead, I will give credit for that problem instead!) 3 x tan(y(x)) + ln(x) y(x) = 7 differentiate - use product & chain rule to handle 1st term, product rule to handle second term 2 3 2 y(x) 3 x tan(y(x)) + x sec(y(x)) y'(x) + ln(x) y'(x) + ---- = 0 x which is the same thing as 2 3 2 y 3 x tan(y) + x sec(y) y' + ln(x) y' + --- = 0 x which is 3 2 -y 2 [x sec(y) + ln(x)] y' = --- - 3 x tan(y) x so that (solving for y') we have 2 -y/x - 3 x tan(y) y' = ------------------- 3 2 x sec(y) + ln(x) ALTERNATE SOLUTION (TO ALTERNATE PROBLEM): 3 x tan(y(x)) + ln(x) y(x) - 7 differentiate implicitly - use product & chain rule to handle 1st term, product rule to handle second term 2 y(x) 3 tan(y(x)) + 3 x sec(y(x)) y'(x) + ln(x) y'(x) + ---- = 0 x which is the same thing as 2 y 3 tan(y) + 3 x sec(y) y' + ln(x) y' + --- = 0 x which is 2 -y [3 x sec(y) + ln(x)] y' = --- - 3 tan(y) x so that (solving for y') we have -y/x - 3 tan(y) y' = -------------------- 2 3 x sec(y) + ln(x) . /////////////////////////////////////