Consider the function represented by the equation x - y = 3. What is the equation written in function notation, with x

Find the length of the following two-dimensional curve. r (t ) = (1/2 t^2, 1/3(2t+1)^3/2) for 0 < t < 16

andrezito [222]

Answer:

r = 144 units

Step-by-step explanation:

The given curve corresponds to a parametric function in which the Cartesian coordinates are written in terms of a parameter "t". In that sense, any change in x can also change in y owing to this direct relationship with "t". To find the length of the curve is useful the following expression;

$r(t)=\int\limits^a_b ({r`)^2 \, dt =\int\limits^b_a \sqrt{((\frac{dx}{dt} )^2 +\frac{dy}{dt} )^2)} dt$

In agreement with the given data from the exercise, the length of the curve is found in between two points, namely 0 < t < 16. In that case a=0 and b=16. The concept of the integral involves the sum of different areas at between the interval points, although this technique is powerful, it would be more convenient to use the integral notation written above.

Substituting the terms of the equation and the derivative of r´, as follows,

$r(t)= \int\limits^b_a \sqrt{((\frac{d((1/2)t^2)}{dt} )^2 +\frac{d((1/3)(2t+1)^{3/2})}{dt} )^2)} dt$

Doing the operations inside of the brackets the derivatives are:

1 ) $(\frac{d((1/2)t^2)}{dt} )^2= t^2$

2) $\frac{(d(1/3)(2t+1)^{3/2})}{dt} )^2=2t+1$

Entering these values of the integral is

$r(t)= \int\limits^{16}_{0} \sqrt{t^2 +2t+1} dt$

It is possible to factorize the quadratic function and the integral can reduced as,

$r(t)= \int\limits^{16}_{0} (t+1) dt= \frac{t^2}{2} + t$

Thus, evaluate from 0 to 16

$\frac{16^2}{2} + 16$

The value is $r= 144 units$

5 0

4 years ago

Which expression is equivalent to 5(4f+3)-2f

Ierofanga [76]

Answer:

f = 5/6

Step-by-step explanation:

5(4f + 3) - 2f

= 20f + 15 - 2f

= 18f + 15

18f = 15

f = 15/18

f = 5/6

6 0

3 years ago

Read 2 more answers

An article in Medicine and Science in Sports and Exercise "Maximal Leg-Strength Training Improves Cycling Economy in Previously

Hatshy [7]

Answer:

The 99% confidence interval for the mean peak power after training is [299.4, 330.6]

$299.4\leq\mu\leq 330.6$

Step-by-step explanation:

We have to construct a 99% confidence interval for the mean.

A sample of n=7 males is taken. We know the sample mean = 315 watts and the sample standard deviation = 16 watts.

For a 99% confidence interval, the value of z is z=2.58.

We can calculate the confidence interval as:

$M-z\sigma/\sqrt{n}\leq\mu\leq M+z\sigma/\sqrt{n}\\\\315-2.58*16/\sqrt{7}\leq\mu\leq 315+2.58*16/\sqrt{7}\\\\315-15.6\leq \mu\leq 315+15.6\\\\299.4\leq\mu\leq 330.6$