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## Certified High School AP Calculus and Physics Teacher

### Scottsdale, AZ (85258)

 Travel Radius 30 miles Hourly Fee \$50.00 Discount \$42.50 - \$47.50 Save up to 15%
 Background Check Passed 10/17/2012 Your first hour with any tutor is always 100% refundable!

# Robert's Responses in WyzAnt Answers

#### Find dz/dy at (1, ln 2, ln 3) if z(x, y)?

Find dz/dy at (1, ln 2, ln 3) if z(x, y) is defined by the equation xe^y+ye^z+2lnx-2-3ln2=0.

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Asked by Sun from Los Angeles, CA
20

Differentiate both sides of xe^y+ye^z+2lnx-2-3ln2 = 0 with respect to y,

xe^y + e^z + ye^z ∂z/∂y = 0

Plug in (1, ln 2, ln 3),

2 + 3 + 3ln2 ∂z/∂y = 0

#### What is the rate of change of temperature?

Suppose the temperature in degrees Celsius at a point (x, y) is described by a function T(x, y) satisfying Tx(2, 7)=4, Ty(2, 7)=2. The position of a crawling ant after t seconds is given by x(t)=sqrt(1+t), y(t)=-2+3t. After 3 seconds, what is the rate of change of temperature along the ant's path in degrees Celsius per second?

I know that (x(3), y(3))=(2, 7)

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Asked by Sun from Los Angeles, CA
20

∂x/∂t = 1/[2sqrt(1+t)] = 1/4 at t = 3

∂y/∂t = 3

dT/dt = (∂T/∂x)(∂x/∂t) + (∂T/∂y)(∂y/∂t) = (4)(1/4) + 2(3) = 7 degrees of C/sec <==Answer

#### Find the maximal value of f(x, y)?

Find the maximal value of f(x, y)=3y+4x on the circle x^2+y^2=1.

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Asked by Sun from Los Angeles, CA
20

Method I. Using linear programming concept

Since the maximum value must be reached at the boundary of the circle,  draw a tangent line with the circle such that the tangent line has a slope of -4/3. Therefore, the point of tangency can be written as (4a, 3a), where a > 0. Plug into the equation of the circle: (3a)^2 + (4a)^2 = 1 => a = 1/5

fmax = f(4/5, 3/5) = 3(3/5) + 4(4/5) = 5 <==Answer

Method II. Using Lagrange multiplier

g(x, y) = 3y+4x + λ(x^2+y^2-1)

g'x = 4 + 2λx = 0 => x = -2/λ, (x > 0)

g'y = 3 + 2λy = 0 => y = -3/(2λ), (y > 0)

x^2+y^2 = (-2/λ)^2 + (-3/(2λ))^2 = 1 => λ = -2.5

x = 4/5, y = 3/5

fmax = f(4/5, 3/5) = 3(3/5) + 4(4/5) = 5 <==Answer

#### What was the ball's initial speed?

A spring gun at ground level fires a golf ball at an angle of 45 degrees. The ball lands 10 m away.

a) What was the ball's initial speed?

b) For the same initial speed, find the two firing angles that make the range 6 m.

Recall that the Ideal Projectile Motion Equation is

r=(vo*cos(theta))ti+((vo*sin(theta)t-1/2*g*t^2)j.

b) theta=1/2*arcsin(3/5),

theta=pi-1/2*arcsin(3/5).

And is arcsin the same thing as sin^-1?

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Asked by Sun from Los Angeles, CA
10

a)

Horizontal direction: vo cos45 t = 10 ......(1)

Vertical direction: vo sin45 t - (1/2)g t2 = 0 ......(2)

Since sin45 = cos45,

10 = (1/2)gt2 ==> t = √(20/g)

Plug in (1),

vo√(10/g) = 10

Solve for vo,

vo = √(10g)

b)

Horizontal direction: vo cosθ t = 6 ......(1)

Vertical direction: vo sinθ t - (1/2)g t2 = 0 =>Since t ≠ 0,  t = 2vo sinθ/g......(2)

Plug in (2) in (1),

vo2sin2θ/g = 6 => 10sin2θ = 6, since vo2 = 10g.

sin2θ = 3/5

θ = (1/2)arcsin(3/5)

or

2θ = pi - arcsin(3/5)

θ = pi/2 - (1/2)arcsin(3/5)

Yes, arcsin(x) = sin-1(x)

#### Find an equation of the plane?

Find an equation of the plane that passes through the point P(-1, 2, 1) and contains the line of intersection of the planes x+y-z=2 and 2x-y+3z=1.

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Asked by Sun from Los Angeles, CA
10

Concept: Three non-colinear points determine a plane.

Ideas of approach: Find the equation of the line first, and pick two points from the line plus the point P.

The equation of the line has a direction of <1, 1, -1> X <2, -1 ,3> = <2, -5, -3>

Pick one point from the line: <1, 1, 0>

So, the equation of the line is r = <1, 1, 0> + t<2, -5, -3>

Let t = 1 to get another point on the line <3, -4, -3>

With the three non-colinear points: <1, 1, 0>, <3, -4, -3> and <-1, 2, 1>, you can get the equation of the plane: x - 2y + 4z = -1.

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## Background Check Status for Robert J.

Robert J. passed a background check on 10/17/2012. You may run an updated background check on Robert once you send an email.