Archive for July, 2015

Problem 899

Use the compound interest formula to find the future value A, for the following values.

P = 3000

i = 0.09

n = 35

 

Solution:-

 

A = P(1+i)n

Using the compound interest formula, replace the unknowns with corresponding values.

A = 3000(1+0.09)35

Calculate the future value.

A = 61241.9

 

Problem 898

The given  point is on the curve. Find the lines that are

a. Tangent

b. Normal to the curve at the given point.

x2  +xy – y2 = -4 ,(2,4)

 

Solution:-

 

a. Give the equation of the that is tangent to the curve at the given point.

y =  \frac{4}{2}x+\frac{4}{3}

 

b. Given the equation of the line that is normal to the curve at  the given point.

y = \frac{-3}{4}x +\frac{11}{2}

 

Problem 897

Find the derivative of the function s = \frac{1}{3\pi } sin(3t)-\frac{1}{7\pi } cos(7t).

 

Solution:-

 

Use the rules of differentiation to find \frac{ds}{dt}.

s = \frac{1}{3\pi } sin(3t)-\frac{1}{7\pi } cos(7t).

\frac{ds}{dt}=\frac{1}{3\pi }*\frac{d}{dt}(sin(3t))-\frac{1}{7\pi }*\frac{d}{dt}(cos(7t))

To find \frac{d}{dt} sin(3t) and \frac{d}{dt}cos(7t), use the Chain Rule.

\frac{d}{dt}(sin(3t)) = 3 cos(3t)

To find \frac{d}{dt}(cos(7t)), use the Chain Rule.

\frac{d}{dt}(cos(7t)) = -7sin(7t)

Substitute the expressions for \frac{d}{dt} (sin(3t)) and and \frac{d}{dt}(cos(7t)).

\frac{ds}{dt}=\frac{1}{3\pi }*\frac{d}{dt}(sin(3t))-\frac{1}{7\pi }*\frac{d}{dt}(cos(7t))

=\frac{1}{3\pi }* (3cos(3t))-\frac{1}{7\pi }* (-7sin(7t))

=\frac{1}{\pi }(cos(3t))+sin(7t)

Thus, \frac{ds}{dt}=\frac{1}{\pi }(cos(3t)+sin(7t))

 

Problem 896

Find \frac{dy}{dx}

y= 6(cos x+sin x) (cos x – sin x )

 

Solution:-

 

\frac{dy}{dx} = -24 sin x cos x

 

Problem 895

The value V = \frac{4}{3} πr3 of a spherical balloon change with the radius.

a. At with rate (ft3/ft) dose the volume change with respect to the radius when r = 3 ft?

b. Using the rate from part a, by approximately how much dose the volume increase when the radius change from 3 to 3.3 ft?

 

Solution:-

 

a. At what rate (ft3/ft) does the volume change with respect to the when r = 3 ft?

36π ft3

 

b. Using the rate from part a, by approximately how much does the volume increase when the radius change from 3 to 3.3 ft?

33.93 ft3

 

Problem 894

A body moves on a coordinate line such that it has a position s= f(t) = t2-5t+4 on the interval 0 ≤t≤8, with in meters and t in seconds.

a. Find the body’s displacement and average velocity for the given time interval.

b. Find the body’s speed and acceleration at the endpoint of the interval.

c. When , if ever, during the interval dose the body change direction?

 

Solution:-

 

The body’s displacement for the given time interval is 24 m.

The body’s average velocity for the given time interval is 3 m/sec.

The body’s speed at the left and right endpoint of the interval are 5 m/sec and 11 m/sec, respectively.

 

The body’s acceleration at the left and right endpoint of the interval are 2 m/sec2 and 2 m/sec2, respectively.

When, if ever, during the interval does the body change direction? Selection correct choice below and fill in any answer boxes within your choice.

The body change direction at t = 5/2 sec.

Problem 893

An object is dropped from the top of a cliff 650 meters high. Its height above the ground t seconds after is dropped is 650 – 4.9t2. Determine its speed 4 seconds after it is dropped.

 

Solution:-

 

The speed of the object 4 second after it is dropped is \frac{196}{5} m/sec.

 

 

 

Problem 892

Find an equation for the line tangent to y = -3 – 3x2 at (-2,-15).

 

Solution:-

 

The equation for the line tangent to y = -3 – 3x2 at (-2,-15) is y = 12x + 9.

 

 

Problem 891

Estimate the slope (in y-units per x-unit) of the tangent line to the curve.

tangent

 

Solution:-

 

First look at the tangent line on the curve.

When moving from left to right the line decreases.

Therefore the slope of the line will be negative.

The slope of a line L is defined as

sloe = \frac{rise}{run}= \frac{y-y_{o}}{x-x_{o}}

where (x,y) and (x_{0},y_{0}) are points on the line.

To estimate the slope, start at any point (x_{0},y_{0}) on the line and move one unit to the right on the x-axis. Estimate the new value yo of the line at x = xo + 1. The slope will be the difference y – y0.

If we start at xo = 7 we can estimate that y0 \approx 6.

Now we move 1 unit to the right on the x-axis to x = 8.

At x = 8 we estimate that y\approx 0.

Therefore,

slope \approx  \frac{y-y_{o}}{x-x_{o}} \approx  \frac{0-6}{8-7} \approx  -6

 

 

 

Problem 890

Define g(7) for the given function so that it is continuous at x = 7.

g(x) = \frac{4x^{2}-196}{4x-28}

 

Solution:-

 

The function g(x) = \frac{4x^{2}-196}{4x-28} is undefined when the denominator is 0.

So, the function is undefined at x = 7.

\frac{4x^{2}-196}{4x-28} can be simplified by dividing numerator and denominator by 4, factoring the numerator, and dividing the common factor in the numerator and denominator.

\frac{4x^{2}-196}{4x-28} =  \frac{x^{2}- 49}{x-7}  = \frac{(x+7)(x-7)}{(x-7)} = x+7

The graphs of y = \frac{4x^{2}-196}{4x-28} and y = x+7 are the same expect that the graph of y = \frac{4x^{2}-196}{4x-28} has a hole at x = 7.

This hole can be eliminate by defining g(7) to be the value of x + 7 on the graph of y = x + 7.

Thus , g(7) should be define as 14.