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Unformatted text preview: Homework 1 – A Brief Calculus I and II review Due Thursday, January 21st 1. Chain Rule (a) Let h ( t ) := sin(cos(tan t )). Find the derivative with respect to t Solution : d dt ( h ( t )) = d dt (sin(cos(tan t ))) = cos(cos(tan t )) d dt (cos(tan t )) = cos(cos(tan t ))( sin(tan t )) d dt (tan t ) = cos(cos(tan t ))( sin(tan t ))sec 2 t (b) Let s ( x ) := 4 √ x where x ( t ) := ln( f ( t )) and f ( t ) is a differentiable function. Find ds dt . Solution : Using Leibniz notation, ds dt = ds dx dx dt . So, ds dt = 1 4 x 3 / 4 · f ( t ) f ( t ) . But we need make sure ds dt is a single variable function of f . So, ds dt = 1 4[ln( f ( t ))] 3 / 4 · f ( t ) f ( t ) . 2. Use implicit differentiation to find dy dx when sin( x + y ) = y 2 cos x. Solution : Recall that we treat y as a function of x . I.e. y ( x ). sin( x + y ) = y 2 cos x d dx (sin( x + y ) = y 2 cos x ) d dx (sin( x + y ( x ))) = d dx ( y 2 ( x )cos x ) cos( x + y ) d dx ( x + y ( x )) = 2 y dy dx cos x + y 2 ( sin x ) cos( x + y ) 1 + dy dx ¶ = 2 y cos x dy dx y 2 sin x ) cos( x + y ) + cos( x + y ) dy dx = 2 y cos x dy dx y 2 sin x cos( x + y ) dy dx 2 y cos x dy dx = cos( x + y ) y 2 sin x dy dx = cos( x + y ) y 2 sin x ) cos( x + y ) 2 y cos x 3. Parametrized curves (a) Describe the curve given parametrically by ( x := 5sin(3 t ) y := 3cos(3 t ) , when 0 ≤ t < 2 π 3 . What happens if we allow t to vary between 0 and 2 π ? Solution : Note that ‡ x 5 · 2 + ‡ y 3 · 2 = sin 2 (3 t ) + cos 2 (3 t ) = 1. Hence, this parametrizes (at least part of) the ellipse ‡ x 5 · 2 + ‡ y 3 · 2 = 1. By examining differing values of t in 0 ≤ t ≤ 2 π 3 , we see that the parametrization above travels the ellipse in a clockwise fashion exactly once....
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This note was uploaded on 02/21/2011 for the course MATH 241 taught by Professor Kim during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Kim
 Math, Chain Rule, Derivative

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