Solutions | ||||
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D |
E |
P |
Q |
S_{2}a, b |

**Problem D**

Particle A moves along the positive x-axis
and particle B moves along the positive y-axis.

a) If $A$ is at the point (a,0) and B is at the point (0,b),
write an expression for the distance from A to B.

**Solution**

Expression for distance between A and B: Pythagorean Theorem gives... d^2 = a^2 + b^2 d = (a^2 + b^2)^(1/2)b) At a certain time, A is at the point (7,0) and moving away from the origin with a speed of 4 units/sec. At the same time B is at the point (0,5) and moving towards the origin at 2 units/sec. At what rate is the distance from A to B changing at that time? Are the particles moving towards each other or moving away from each other at that time?

**Solution**

Given: da/dt=-2 and db/dt=(t)=4 Chain rule... dd/dt=(1/2)[(a^2+b^2)^(-1/2)](2a(da/dt)+2b(db/dt)) a=7, b=5, a'=-2, b'=4 dd/dt=(1/2)[(7^2+5^2)^(-1/2)](2(7)(-2)+2(5)(4)) dd/dt=(1/2)[(74)^(-1/2)](12) dd/dt=6*sqrt(74) This value is positive so the particles are moving away from each other.

Matthew Daubert

**Problem E**

A particle is moving with the given data.
Find a formula for the position, s(t), of the particle at time t
if a(t)=cos(t)+sin(t), s(0)=0, and v(0)=5.

**Solution**

We know that (integral)a(t) = v(t) and (integral)v(t) = s(t) * a(t) = cos(t) + sin(t) (integral)a(t) = v(t) = sin(t) - cos(t) + C v(0) = 5 = 0 - 1 + C so C = 6 ** v(t) = sin(t) - cos(t) + 6 (integral)v(t) = s(t) = -cos(t) - sin(t) + 6t + C s(0) = 0 = -1 - 0 + 0 + C so C = 1 *** s(t) = -cos(t) - sin(t) + 6t + 1

Steven Emerson

**Problem P**

Compute the integral from 1 to 2 of (3x^2- {4\over x^2}) using calculus.

**Solution**

To compute the integral take the antiderivative of the function. 2 2 S 3x^2 - (4/x^2) dx = I x^3 + (4/x) 1 1 Plug the two bounds in for x and subtract the top from the bottom to find the value of the integral. This = 8 + 2 - 1 - 4 = 10 - 5 = 5

Michael Pandolfo

**Problem Q**

For the curve y^3+5xy^2+x^4=-16 find

a) dy/dx

b) An equation of the line tangent to the curve at (-2,2)

**Solution**

a) dy/dx (dy/dx both sides) y^3+5xy^2+x^4=-16 3y^2dy/dx+5y^2+10xydy/dx+4x^3=0 (solve for dy/dx) dy/dx(3y^2+10xy)= -5y^2-4x^3 dy/dx= (-5y^2-4x^3)/(3y^2+10xy) b) An equation of the line tangent to the curve at (-2,2) x=-2 and y=2. So: y=mx+b where m is the slope which we already have in form of dy/dx then dy/dx = (-5y^2-4x^3)/(3y^2+10xy) = m m= [-5(2)^2-4(-2)^3]/[3(2)^2+10(-2)(2)] = -3/7 then in the equation y=mx+b we will solve for the value of b 2=-(3/7)(-2)+b so 2-(6/7)=b and b=(8/7). Then the line tangent to the curve at (-2,2) is y=-(3/7)x+(8/7) ANSWER: a: dy/dx= (-5y^2-4x^3)/(3y^2+10xy) b: y=-(3/7)x+(8/7)

Mehreen Qureshi

We can ask

and we should get something like the picture displayed, which certainly appears that the line found by Ms. Qureshi is tangent!

**Problem S _{2}**

Compute the following limits. Give exact answers, as necessary in terms of well-known constants such as Pi, e, ln(3), sqrt{2}, etc.

a) lim

b) lim

c) lim

**Solution**

a)lim tan(17x)/sin(13x)= L'H x-0 0/0 lim [sec^2(17x)(17)] / [cos(13x)(13)] = 17/13 x-0 b) lim [(4+5x^4)^(1/2)]/[3+6x+7x^2] = x-00 (infinity) 00/00 divide everything by x^2 lim [{(4+5x^4)(1/x^4)}^(1/2)]/[(3+6x+7x^2)(1/x^2)] = [(5)^(1/2)]/7 x-00

Patricia Sandoval