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Salary: <｜begin▁of▁sentence｜># 1. (a) (i) Use the linear approximation formula y f(x) f(x + x) f(x) + f (x) x Size: px Start display at page: Download "1. (a) (i) Use the linear approximation formula y f(x) f(x + x) f(x) + f (x) x" Error: - Melvyn Randall - 5 years ago - Views: Transcription 1 1. (a) (i) Use the linear approximation formula y f(x) f(x + x) f(x) + f (x) x with a suitable choice of f(x) to show that e θ 1 + θ for small values of θ. (ii) Use the result obtained in part (a) to approximate e 0.1. (iii) Approximate e 0.1 using differentials. (b) If A = 2πr 2 + 2πrh and r and h are each subject to a maximum percentage error of 1%, find the approximate maximum percentage error in A. (a) (i) Let f(θ) = e θ. Then f (θ) = e θ. For small θ, f(θ) f(0) + f (0)θ = 1 + θ. (ii) e (iii) Let y = e x. Then dy = e x dx. When x = 0, dx = 0.1, dy = 0.1. So e (b) A = 2πr 2 + 2πrh. da = (4πr + 2πh)dr + 2πrdh. da A = (4πr + 2πh)dr + 2πrdh 2πr 2 + 2πrh = (2r + h)dr + rdh r(r + h) = 2r + h r + h dr r + r r + h dh h. So da A 2r + h r + h 0.01 r + r r + h 0.01 h = 0.01( 2r + h r + h + r r + h ) = 0.01( 2r + h + r r + h ) = 0.01( 3r + h r + h ). 1 2 2. (a) Find the critical points of the function f(x, y) = x 3 + 3xy y 3. (b) Classify each critical point as a local maximum, local minimum or a saddle point. (a) f x = 3x 2 + 3y = 0, f y = 3x 3y 2 = 0. From f x = 0, y = x 2. Substitute into f y = 0: 3x 3x 4 = 0, 3x(1 x 3 ) = 0, so x = 0 or x = 1. Then y = 0 or y = 1. So critical points: (0, 0) and (1, 1). (b) f xx = 6x, f yy = 6y, f xy = 3. At (0, 0): f xx = 0, f yy = 0, f xy = 3, so D = = 9 < 0, saddle point. At (1, 1): f xx = 6, f yy = 6, f xy = 3, so D = 36 9 = 27 > 0, and f xx > 0, so local minimum. 2 3 3. (a) Evaluate the double integral 1 0 x 0 ex/y dy dx by reversing the order of integration. (b) Evaluate the double integral 1 0 x 0 ex/y dy dx by using the substitution u = x/y. (a) Region: 0 x 1, 0 y x. So also: y from 0 to 1, and for fixed y, x from y to 1. So I = 1 0 [ 1 y e x/y dx] dy. For fixed y, let u = x/y, then x = yu, dx = y du, and when x = y, u = 1; when x = 1, u = 1/y. So inner integral: 1/y 1 e u y du = y(e 1/y e). So I = 1 0 y(e1/y e)dy. This is not simpler. Alternatively, reverse order: x from 0 to 1, y from 0 to x, so region also: y from 0 to 1, x from y to 1. So I = 1 0 [ 1 y e x/y dx]dy. For fixed y, in inner integral, let u = x/y, then x = yu, dx = y du, limits: x = y u = 1, x = 1 u = 1/y. So inner integral becomes 1/y 1 e u y du = y(e 1/y e). So I = 1 0 y(e1/y e)dy. This is not simpler. Maybe the intended order reversal is different? Actually, the original order is dy dx, with y from 0 to x. So region: 0 x 1, 0 y x. That is triangle. If we reverse: x from y to 1, y from 0 to 1. So I = 1 0 [ 1 y e x/y dx]dy. Now, for fixed y, integrate with respect to x: treat y as constant. Then e x/y dx. Let u = x/y, then dx = y du, limits: x = y gives u = 1, x = 1 gives u = 1/y. So inner integral = 1/y 1 e u y du = y(e 1/y e). So I = 1 0 y(e1/y e)dy = 1 0 ye1/y dy e 1 0 y dy. Now, 1 0 ye1/y dy: let t = 1/y, then y = 1/t, dy = 1/t 2 dt, when y 0, t ; y = 1, t = 1. So 1 0 ye1/y dy = 1 (1/t)e t ( 1/t 2 )dt = 1 e t /t 3 dt. That diverges? Actually, as y 0, e 1/y, so the integral is improper. Check: near y=0, e 1/y is huge. So maybe the reversal is not simpler. Alternatively, maybe the intended is to reverse to dx dy? But then the inner integral becomes e x/y dx, which is similar. So part (b) suggests using substitution u = x/y. So for the original order: I = 1 0 [ x 0 ex/y dy] dx. For fixed x, let u = x/y, then y = x/u, dy = x/u 2 du. When y=0, u ; when y=x, u=1. So inner integral becomes: 1 e u ( x/u 2 ) du = x 1 e u /u 2 du. So I = 1 0 x [ 1 e u /u 2 du] dx = ( 1 0 x dx)( 1 e u /u 2 du) = (1/2) 1 e u /u 2 du. Now, 1 e u /u 2 du. Integrate by parts: let v = e u, dw = du/u 2, then dv = e u du, w = 1/u. So = [ e u /u] 1 e u /u du = (e 1 /1 e / ) 1 e u /u du. But e / =? Actually, limit as u, e u /u = 0. So = e e u /u du. And 1 e u /u du is the exponential integral, which is not elementary. So that gives I = 1/2 (e 1 e u /u du). That is not a nice number. Possibly the limits are different? Alternatively, maybe the intended is to reverse the order first? Let's try: I = 1 0 x 0 ex/y dy dx. Region: 0 x 1, 0 y x. So also: y from 0 to 1, x from y to 1. So I = 1 0 [ 1 y e x/y dx] dy. For fixed y, let u = x/y, then x = yu, dx = y du, limits: x=y gives u=1, x=1 gives u=1/y. So inner = 1/y 1 e u y du = y(e 1/y e). So I = 1 0 y(e1/y e)dy = 1 0 ye1/y dy e 1 0 y dy = 1 0 ye1/y dy e/2. Now, 1 0 ye1/y dy. Let t = 1/y, then y=1/t, dy = 1/t 2 dt, when y=0, t ; y=1, t=1. So = 1 (1/t)e t ( 1/t 2 )dt = 1 e t /t 3 dt. That diverges as t because e t /t 3 0 actually, as t, e t decays, so it's convergent at infinity. But at t=1, it's fine. So it's an improper integral from 1 to. That integral is not elementary. So that approach gives I = ( 1 e t /t 3 dt) e/2. That is not simpler. Given part (b) says: "Evaluate the double integral by using the substitution u = x/y." So likely the intended is to use that substitution directly in the original order. But then we got I = 1/2 1 e u /u 2 du, which is not elementary. Wait, maybe the limits are different? Possibly the integral is 1 0 x 0 ex/y dy dx. Could it be that the region is: 0 y 1, y x 1? Then it would be 1 0 [ 1 y ex/y dx]dy. That is what we had. That doesn't simplify. Alternatively, maybe the intended substitution is u = x/y, but then also change order? I'll re-read: "3. (a) Evaluate the double integral 1 0 x 0 ex/y dy dx by reversing the order of integration. (b) Evaluate the double integral 1 0 x 0 ex/y dy dx by using the substitution u = x/y." So both parts are for the same integral. Perhaps the integral is actually 1 0 x 0 ex/y dy dx. That is what I used. Maybe there is a mistake: e x/y is problematic when y=0. So the integral is improper. Perhaps the intended is to evaluate it as an iterated integral in the order given. For fixed x, inner integral: 0 x e x/y dy. Let u = x/y, then y = x/u, dy = -x/u^2 du, when y=0, u -> infinity; when y=x, u=1. So becomes: 1 e u (x/u^2) du = x 1 e u /u^2 du. So then I = 1 0 x ( x 1 e u /u^2 du) dx = ( 1 0 x^2 dx)( 1 e u /u^2 du) = (1/3) 1 e u /u^2 du. Then integration by parts: 1 e u /u^2 du = [ -e u /u] 1 + 1 e u /u du = ( -e/1 + lim_{u-> } e u /u) + 1 e u /u du = -e + 1 e u /u du. And 1 e u /u du is the exponential integral Ei(1) but from 1 to infinity? Actually, 1 e u /u du is divergent because at infinity, e u /u ~? Wait, check: as u -> infinity, e u /u -> infinity? Actually, e u grows faster than any power, so 1 e u /u du diverges. So that suggests that the integral is divergent. But maybe it's not: because when y is near 0, e x/y is huge. So the inner integral 0 x e x/y dy, for fixed x>0, as y->0, x/y -> infinity, so e x/y -> infinity. So the inner integral is improper and likely diverges. For example, take x=1: 0 1 e 1/y dy. Let t=1/y, then y=1/t, dy=-1/t^2 dt, when y->0, t->infty; y=1, t=1. So becomes 1 e t (1/t^2) dt, and as t->infty, e t /t^2 -> infty, so diverges. So the integral is divergent. So perhaps the intended is actually 1 0 x 0 e y/x dy dx? That would be different. Or maybe it's 1 0 x 0 e x/y dy dx is actually convergent? Check: For fixed x, near y=0, e x/y is enormous. But then dy, so we need to see the behavior: y -> 0, let u = x/y, then y = x/u, dy = -x/u^2 du, so the integrand becomes e u * (x/u^2) du. As u -> infinity, e u /u^2 -> infinity, so the integral from some large U to infinity is like U e u du which diverges. So indeed, the inner integral diverges for each x>0. So the double integral is divergent. Given that the problem asks to evaluate it by two methods, it's likely that the integral is actually something like 1 0 x 0 e y/x dy dx. Because that one is common: 1 0 x 0 e y/x dy dx = 1 0 x(e-1) dx = (e-1)/2. And then reversing order: y from 0 to 1, x from y to 1, gives 1 0 (1-y)e y/x? Not sure. Alternatively, it could be 1 0 x 0 e x/y dy dx is not standard. I will check: If it were 1 0 x 0 e y/x dy dx, then inner: 0 x e y/x dy = x(e-1). Then outer: 1 0 x(e-1)dx = (e-1)/2. And reversing order: y from 0 to 1, x from y to 1, then inner: y 1 e y/x dx. That is not as straightforward. And using substitution u = y/x gives: for fixed y, let u = y/x, then x = y/u, dx = -y/u^2 du, when x=y, u=1; when x=1, u=y. Then inner becomes: 1 y e u (y/u^2) du = y y 1 e u /u^2 du. Then double integral becomes 1 0 y [ y 1 e u /u^2 du] dy. That doesn't simplify to (e-1)/2 easily. So that is not likely. Another possibility: 1 0 x 0 e x/y dy dx. I'll check online memory: There is an integral: ∫_0^1 ∫_0^x e^(x/y) dy dx. Actually, it might be that the intended is to evaluate by changing order and then using substitution. Given that part (b) says "using the substitution u = x/y", that suggests that in the inner integral, we set u = x/y. For the original order, that gave I = 1 0 x [ 1 e u /u^2 du] dx = (1/2) 1 e u /u^2 du, which is not elementary. If we reverse order first, we got I = 1 0 y(e1/y e)dy, and then use substitution u=1/y, we get I = e u /u^3 du - e/2, also not elementary. So both methods lead to an integral that is related to the exponential integral. So maybe the integral is actually convergent? Let's check the inner integral for a fixed x>0: I(x) = 0 x e x/y dy. As y -> 0+, let z = x/y, then y = x/z, dy = -x/z^2 dz, and when y -> 0, z -> infinity. So I(x) = infinity e^z * (x/z^2) dz = x infinity e^z/z^2 dz, which diverges because for large z, e^z/z^2 -> infinity. Wait, careful: The substitution: y from 0 to x. When y is very small, say y = ε, then x/y is large. So we set u = x/y, then y = x/u, dy = -x/u^2 du. When y = ε, u = x/ε which is large. When y = x, u = 1. So I(x) = u=1 to u=x/ε e^u * (x/u^2) du. As ε -> 0, x/ε -> infinity, so I(x) = x 1 infinity e^u/u^2 du. And 1 infinity e^u/u^2 du diverges because e^u grows. So indeed, for each x>0, the inner integral diverges. So the double integral is divergent. Therefore, I suspect that the intended integral is actually 1 0 x 0 e y/x dy dx. But then part (b) says "using the substitution u = x/y". If it were e^(y/x), then the substitution u = x/y would give u = x/y, so y = x/u, dy = -x/u^2 du, and then e^(y/x) = e^(1/u). Then inner becomes: 0 x e^(1/u) dy = infinity^? Let's do: For fixed x, inner: 0 x e^(y/x) dy. If we set u = x/y, then when y=0, u -> infinity; when y=x, u=1. Then y = x/u, dy = -x/u^2 du, so the integral becomes: u=infinity to 1 e^(1/u) * (x/u^2) du = x 1 infinity e^(1/u)/u^2 du. That is convergent but not simpler. So

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