Pre-Calc Homework Solutions 176

Pre-Calc Homework Solutions 176 - 176 Section 4.6 dV dt dV...

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Unformatted text preview: 176 Section 4.6 dV dt dV dt dV dh dV , we have dh dt dt dV dr dV , we have dr dt dt d 2 (r h) dt dr h(2r) dt dr 2 rh dt 4. Use implicit differentiation. d (x sin y) dx d (1 dx dy x dx 3. (a) Since xy) y(1) sin y (b) Since (c) dV dt dV dt dV dt dP dt dP dt dP dt dP dt r2 dh . dt dr dt (x)(cos y) dy dx (sin y)(1) x cos y) dy dx dy dx dy dx 2 rh . (x y y sin y x x cos y y sin y x x cos y d 2 r h dt dh r2 dt dh r2 dt 5. Use implicit differentiation. d 2 x dx d tan y dx dy sec2 y dx 2x sec2 y 4. (a) 2x dy dx dy dx 2x cos2 y d (RI 2) dt dR d R I2 I2 dt dt dI dR R 2I I2 dt dt dI dR 2RI I2 dt dt dP dt (b) If P is constant, we have 2RI d (2x) dx dI dt 0, which means 2R dI I dt 2P dI . I 3 dt 6. Use implicit differentiation. d ln (x y) dx dy 1 1 dx x y dy 1 dx dy dx I2 dR dt 0, or dR dt 2 2(x 2x y) 2y 1 ds dt 1 2 x 2 5. ds dt d dt x2 y2 z2 7. Using A( 2, 1) we create the parametric equations x 2 at and y 1 bt, which determine a line passing through A at t 0. We determine a and b so that the line passes through B(4, 3) at t 1. Since 4 2 a, we have a 6, and since 3 1 b, we have b 4. Thus, one parametrization for the line segment is x 2 6t, y 1 4t, 0 t 1. (Other answers are possible.) 8. Using A(0, 4), we create the parametric equations x 0 at and y 4 bt, which determine a line passing through A at t 0. We now determine a and b so that the line passes through B(5, 0) at t 1. Since 5 0 a, we have a 5, and since 0 4 b, we have b 4. Thus, one parametrization for the line segment is x 5t, y 4 4t, 0 t 1. (Other answers are possible.) 9. One possible answer: 10. One possible answer: 2 3 2 d y 2 z dt dx dt dy dt dz dt 2 (x 2 y2 z 2) ds dt 1 2 x dx x dt 2 y dy y dt 2 z z 2 2x 2y 2z ds dt dz dt x2 y2 z2 6. dA dt dA dt dA dt d 1 ab sin dt 2 1 da db b sin a 2 dt dt 1 da db b sin a sin 2 dt dt sin ab cos ab d dt d sin dt t t 3 2 7. (a) Since V is increasing at the rate of 1 volt/sec, dV dt 1 volt/sec. 1 amp/sec, 3 2 (b) Since I is decreasing at the rate of dI dt 1 amp/sec. 3 Section 4.6 Exercises 1. Since 2. Since dA dt dS dt dA dr dA , we have dr dt dt dS dr dS , we have dr dt dt 2 r . 8 r . dr dt dr dt (c) Differentiating both sides of V dV dt IR, we have I dR dt R . dI dt ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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