Question

A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) At what rate is his distance from second base decreasing when he is halfway to first base? (Round your answer to one decimal place.) ft/s (b) At what rate is his distance from third base increasing at the same moment? (Round your answer to one decimal place.)

Answer 1

Answer:

Step-by-step explanation:

Given that :

the side of the square = 90ft

The speed of the runner = 31 ft/sec

By the time the runner is halfway to the first base;  the distance covered by the runner in time(t)  is (31 t) ft and the distance half the base = 90/2 = 45 ft

Thus; 31 t = 45

t = 45/31

From the second base ; the distance is given as:

P² = (90)² + (90 - 31t )²  

P = $\sqrt{(90)^2 + (90 - 31t )^2}$

By differentiation with time;

$\dfrac{dP}{dt} =\dfrac{1}{ 2 \sqrt{90^2 +(90-31t)^2} } *(0+ 2 (90-31t)(0-31))$

$\dfrac{dP}{dt} =\dfrac{1}{ 2 \sqrt{90^2 +(90-31t)^2} } * 2 (-31)(90-31t)$

At t = 45/31

$\dfrac{dP}{dt} =\dfrac{1}{ 2 \sqrt{90^2 +45^2} } * 2 (-31)(45)$

$\dfrac{dP}{dt} =\dfrac{-35*45}{100.623}$

= - 13.86 ft/sec

Hence, we can conclude that  as soon as the runner  is halfway to the first base, the distance to the second base is therefore decreasing by 13.86 ft/sec

b) The distance from third base can be expressed by the relation:

q² = (31t)² + (90)²

$q = \sqrt{(31t)^2+(90)^2}$

By differentiation with respect to time:

$\dfrac{dq}{dt} = \dfrac{1}{2\sqrt{90^2 + (31)t^2} } *(0+31^2 + 2t)$

At t = 45/31

$\dfrac{dq}{dt} = \dfrac{1}{2\sqrt{90^2 + 45^2} } *(0+31^2 + \frac{45}{31})$

$= \dfrac{31*45}{100.623}$

$= 13.86 \ ft/sec$

Thus, the rate at which the runner's distance is from the third base is increasing at the same moment of 13.86 ft/sec. So therefore; he is moving away from the third base at the same speed to the first base)

Answer 2

Answer:

a) -13.9 ft/s

b) 13.9 ft/s

Step-by-step explanation:

a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:

$D^{2} = (90 - x)^{2} + 90^{2}$   (1)

$\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}$

$2D\frac{d(D)}{dt} = \frac{d((90 - x)^{2})}{dt}$  

$D\frac{d(D)}{dt} = -(90 - x) \frac{dx}{dt}$   (2)

Since:

$D = \sqrt{(90 -x)^{2} + 90^{2}}$

When x = 45 (the batter is halfway to first base), D is:

$D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62$

Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:

$100.62 \frac{d(D)}{dt} = -(90 - 45)*31$          

$\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s$

Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.

b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:

$D^{2} = 90^{2} + x^{2}$  

$\frac{d(D^{2})}{dt} = \frac{d(90^{2} + x^{2})}{dt}$

$D\frac{dD}{dt} = x\frac{dx}{dt}$   (3)

We have that D is:

$D = \sqrt{x^{2} + 90^{2}} = \sqrt{(45)^{2} + 90^{2}} = 100.63$

By entering x = 45, dx/dt = 31 and D = 100.63 into equation (3) we have:

$\frac{dD}{dt} = \frac{45*31}{100.63} = 13.9 ft/s$

Therefore, the rate of the batter when he is from third base increasing at the same moment is 13.9 ft/s.

I hope it helps you!