Question 6 complete not graded suppose a particles

Suppose a particle's position is given by \( f(t) = \sqrt{t+3} \), where \( t \) is measured in seconds and \( f(t) \) is given in
centimeters. 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 above
The 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 expression
or 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: \( \frac{1}{\sqrt{5}\,x^2} \)
Question
Complete
Select one:
6
Not graded
Find the derivative of \( f(x) = \frac{-1}{\sqrt{5}\,x} \).
Select one:

Question
Complete
Find the equation of the line tangent to the curve \( y=\frac{1}{2x} \) when \( x = 1 \).
Select one:
7
Not graded
Question
Complete
8
Not graded
The 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 \( (1, \frac{1}{2}) \).
\( y- \frac{1}{2} = -\frac {1}{2} (x-1) \qquad \qquad Point-slope\,\,form\\ y- \frac{1}{2} = -\frac {1}{2}x + \frac {1}{2} \qquad \qquad
Simplify.\\ y = -\frac {1}{2} x + 1 \qquad \qquad\qquad Add \,\,\frac{1}{2}\,\, to\,\, both\,\, sides. \)