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Question 6 complete not graded suppose a particles

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Suppose a particle's position is given by \( f(t) = \sqrt{t+3} \), where \( t \) is measured in seconds and \( f(t) \) is given incentimeters. What is the velocity of the particle when \( t=1 \)?Select one:a. \( \frac{1}{2}\, cm/sec \)b. \( \frac{1}{4}\, cm/sec \)c. \( 2\, cm/sec \)d. \( -\frac{1}{2}\, cm/sec \)e. None of the aboveThe velocity at t = 1 is f'(1). This can be found either by ±rst computing f'(t) and then substituting 1 for t in the resulting expressionor by computing f'(1).\( f'(1) = lim_{\Delta t \to 0} \frac{f(1+\Delta t)-f(1) }{\Delta t} \quad Use\,\, the\,\, de±nition\,\, of \,\, the\,\, derivative\,\,\\ \qquad\qquad \qquad \qquad \qquad \qquad\quad to\,\, compute\,\,f'(1) \\ = lim_{\Delta t \to 0} \frac{\sqrt{1+\Delta t +3} -\sqrt{1+3}}{\Delta t} \qquad Substitute.\\ = lim_{\Delta t \to 0} \frac{\sqrt {4+\Delta t }-2}{\Delta t } \qquad\qquad Simplify.\\ = lim_{\Delta t\to 0} \frac{4+\Delta t - 4}{\Delta t(\sqrt{4+\Delta t}+2)} \qquad Multiply\,\, the\,\,numerator\,\, and \,\, denominator\,\,by\\ \qquad\qquad \qquad \qquad \qquad\quad \sqrt{4+\Delta t}+ 2.\\ = lim_{\Delta t \to 0} \frac{1}{\sqrt{4+\Delta t} +2}\qquad Simplify.\\ =\frac{1}{2+2}\qquad\qquad \qquad \quad Evaluate\,\,the\,\,limit.\\ =\frac{1}{4} \)The velocity at t =1 is \( \frac{1}{4} \) cm/sec.The correct answer is: \( \frac{1}{4}\, cm/sec \)
\( -\frac{1}{\sqrt{5}\,x} \) is the same as \( - \frac{1}{\sqrt{5}}. \frac{1}{x} \).The derivative of \( \frac{1}{x} \) is \(- \frac{1}{x^2} \)Thus, the derivative of\( -\frac{1}{\sqrt{5}\,x} \) is \( \frac{1}{\sqrt{5}\,x^2} \)The correct answer is: \( \frac{1}{\sqrt{5}\,x^2} \)
QuestionCompleteSelect one:6Not gradedFind the derivative of \( f(x) = \frac{-1}{\sqrt{5}\,x} \).Select one:
QuestionCompleteFind the equation of the line tangent to the curve \( y=\frac{1}{2x} \) when \( x = 1 \).Select one:7Not graded
QuestionComplete8Not gradedThe derivative of the function\( f(x) = \frac{1}{2x} \) gives the slope of the tangent lint to the curve\( y= \frac{1}{2x} \).\( f'(x) = lim_{\Delta x \to 0} \frac {\frac{1}{2x+2 \Delta x } - \frac {1}{2x}}{\Delta x} \qquad Find\,\, the\,\, derivative \\ \qquad =lim_{\Delta x \to 0} \frac {x-x-\Delta x \Delta x}{2x(x+\Delta x) \Delta x} \qquad Simplify.\\ \qquad = lim_{\Delta x \to 0} \frac {-1}{2x^2 + 2x \Delta x} \qquad \quad Simplify.\\ \qquad = -\frac{1}{2x^2} \qquad \qquad \qquad Evaluate\,\, the\,\, limit.\\ f'(1) = -\frac{1}{2(1)^2} = -\frac{1}{2} \qquad \qquad Evaluate\,\, the\,\, the\,\, derivative\,\, at\,\, x =1.\\ \)When x =1, \( y= \frac{1}{2} \), so the tangent line passes through the point \( (1, \frac{1}{2}) \).\( y- \frac{1}{2} = -\frac {1}{2} (x-1) \qquad \qquad Point-slope\,\,form\\ y- \frac{1}{2} = -\frac {1}{2}x + \frac {1}{2} \qquad \qquadSimplify.\\ y = -\frac {1}{2} x + 1 \qquad \qquad\qquad Add \,\,\frac{1}{2}\,\, to\,\, both\,\, sides. \)

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Term
Fall
Professor
idk
Tags
Derivative, Continuous function, lim g

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