# 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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