Exam 2: Limits and Derivatives

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Find the derivative of the function. f(x)=−x2+x+2f ( x ) = - x ^ { 2 } + x + 2

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Find the linearization L(x)L ( x ) of the function at aa . f(x)=x2/3;a=64f ( x ) = x ^ { 2 / 3 } ; \quad a = 64

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Find the point(s) on the graph of ff where the tangent line is horizontal. f(x)=x2e−xf ( x ) = x ^ { 2 } e ^ { - x }

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Two chemicals react to form another chemical.Suppose that the amount of chemical formed in time t (in hours) is given by x(t)=11[1−(23)3t]1−14(23)3tx ( t ) = \frac { 11 \left[ 1 - \left( \frac { 2 } { 3 } \right) ^ { 3 t } \right] } { 1 - \frac { 1 } { 4 } \left( \frac { 2 } { 3 } \right) ^ { 3 t } } where x(t)x ( t ) is measured in pounds. a. Find the rate at which the chemical is formed when t=4t = 4 . Round to two decimal places. b. How many pounds of the chemical are formed eventually?

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The level of nitrogen dioxide present on a certain June day in downtown Megapolis is approximated by A(t)=0.03t3(t−7)4+64.80≤t≤7A ( t ) = 0.03 t ^ { 3 } ( t - 7 ) ^ { 4 } + 64.8 \quad 0 \leq t \leq 7 where A(t)A ( t ) is measured in pollutant standard index and tt is measured in hours with t=0t = 0 corresponding to 7 a.m. What is the average level of nitrogen dioxide in the atmosphere from 1 a.m. to 2 p.m. on that day? Round to three decimal places.

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The top of a ladder slides down a vertical wall at a rate of 0.15 m/s0.15 \mathrm {~m} / \mathrm { s } . At the moment when the bottom of the ladder is 1.5 m1.5 \mathrm {~m} from the wall, it slides away from the wall at a rate of 0.3 m/s0.3 \mathrm {~m} / \mathrm { s } . How long is the ladder?

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s(t)s ( t ) is the position of a body moving along a coordinate line; s(t)s ( t ) is measured in feet and tt in seconds, where t≥0t \geq 0 . Find the position, velocity, and speed of the body at the indicated time. s(t)=4tt2+1;t=3s ( t ) = \frac { 4 t } { t ^ { 2 } + 1 } ; \quad t = 3

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Two curves are said to be orthogonal if their tangent lines are perpendicular at each point of intersection of the curves.Show that the curves of the given equations are orthogonal. y−74x=π2,x=74cos⁡yy - \frac { 7 } { 4 } x = \frac { \pi } { 2 } , \quad x = \frac { 7 } { 4 } \cos y  Two curves are said to be orthogonal if their tangent lines are perpendicular at each point of intersection of the curves.Show that the curves of the given equations are orthogonal.  y - \frac { 7 } { 4 } x = \frac { \pi } { 2 } , \quad x = \frac { 7 } { 4 } \cos y

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Find the derivative of the function. f(x)=0.2x−13f ( x ) = 0.2 x ^ { - 13 }

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The volume of a right circular cone of radius rr and height hh is V=π3r2hV = \frac { \pi } { 3 } r ^ { 2 } h . Suppose that the radius and height of the cone are changing with respect to time tt . a. Find a relationship between dVdt,drdt\frac { d V } { d t } , \frac { d r } { d t } , and dhdt\frac { d h } { d t } . b. At a certain instant of time, the radius and height of the cone are 12in12 \mathrm { in } . and 13 in. and are increasing at the rate of 0.2in./sec0.2 \mathrm { in } . / \mathrm { sec } and 0.5in./sec0.5 \mathrm { in } . / \mathrm { sec } , respectively. How fast is the volume of the cone increasing?

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The curve with the equation x2/3+y2/3=25x ^ { 2 / 3 } + y ^ { 2 / 3 } = 25 is called an asteroid. Find an equation of the tangent to the curve at the point (486,1)( 48 \sqrt { 6 } , 1 ) .  The curve with the equation  x ^ { 2 / 3 } + y ^ { 2 / 3 } = 25  is called an asteroid. Find an equation of the tangent to the curve at the point  ( 48 \sqrt { 6 } , 1 ) .

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If ff is the focal length of a convex lens and an object is placed at a distance vv from the lens, then its image will be at a distance uu from the lens, where f,vf , v , and uu are related by the lens equation 1f=1v+1u.\frac { 1 } { f } = \frac { 1 } { v } + \frac { 1 } { u } . Find the rate of change of vv with respect to uu .

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Find f′f ^ { \prime } in terms of g′g ^ { \prime } . f(x)=x7g(x)f ( x ) = x ^ { 7 } g ( x )

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If ff is a differentiable function, find an expression for the derivative of y=x3f(x)y = x ^ { 3 } f ( x ) .

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Use differentials to estimate the amount of paint needed to apply a coat of paint 0.0017 cm0.0017 \mathrm {~cm} thick to a hemispherical dome with diameter 70 m70 \mathrm {~m} .

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The graph shows the percentage of households in a certain city watching television during a 24-hr period on a weekday ( t=0t = 0 corresponds to 6 a.m. ). By computing the slope of the respective tangent line, estimate the rate of change of the percentage of households watching television at a-12 p.m. Note that dy=0.03d y = 0.03  The graph shows the percentage of households in a certain city watching television during a 24-hr period on a weekday (  t = 0  corresponds to 6 a.m. ). By computing the slope of the respective tangent line, estimate the rate of change of the percentage of households watching television at a-12 p.m. Note that  d y = 0.03

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Suppose that ff and gg are functions that are differentiable at x=1x = 1 and that f(1)=1,f′(1)=−3,gf ( 1 ) = 1 , f ^ { \prime } ( 1 ) = - 3 , g (1)=2( 1 ) = 2 , and g′(1)=5g ^ { \prime } ( 1 ) = 5 . Find h′(1)h ^ { \prime } ( 1 ) . h(x)=xf(x)x+g(x)h ( x ) = \frac { x f ( x ) } { x + g ( x ) }

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Two sides of a triangle are 2 m2 \mathrm {~m} and 3 m3 \mathrm {~m} in length and the angle between them is increasing at a rate of 0.06rad/s0.06 \mathrm { rad } / \mathrm { s } . Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is π3\frac { \pi } { 3 } .

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Find equations of the tangent lines to the curve y=x−8x+8y = \frac { x - 8 } { x + 8 } that are parallel to the line x−y=8.x - y = 8 . A) x−y=−18.5x - y = - 18.5 B) x−y=−4.5x - y = - 4.5 C) x−y=−1.5x - y = - 1.5 D) x−y=−12.5x - y = - 12.5 E) x−y=−15x - y = - 15

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Suppose the total cost in maunufacturing xx units of a certain product is C(x)C ( x ) dollars. a. What does C′(x)C ^ { \prime } ( x ) measure? Give units. b. What can you say about the sign of C′C ^ { \prime } ? c. Given that C′(3000)=11C ^ { \prime } ( 3000 ) = 11 , estimate the additional cost in producing the 3001 st unit of the product.

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