# Test help!!!

• Nov 13th 2007, 12:33 PM
jarny
Test help!!!
My teacher gave us this take home test for the final grade of the marking period. I am sincerely begging for you guys to check my work. I understand it is alot so i'd really appreciate this if you could help. Thanks!!! I am going to post the problems from greatest to least importance along with my answers:

1. Show that the slope dy/dx is defined at every point on the graph of 2y= x^2 + siny. I don't really know how to do this.

2. A 13 ft ladder is leaning against a house. when its bse starts to slide away. by the time the base is 12 ft. from the house, the base is moving at the rate of 5 ft/sec. a) how fast is the top of the ladder sliding down the wall at that moment? b) At what rate is the area of the triangle formed by the ladder, wall, and ground changing at that moment? c) At what rate is the angle theta between the ladder and the ground changing at that moment? i got -12ft/s;59.5 ft/s; and i am still working on c.

3. a point "p" moves from the left to right along the curve y=x^2 at a constant horizontal speed of 3 units/sec. How fast does the y coordinate of p increase at the moment when "p" passes through (1,1) and (2,4)? i got 6 u/s and 12u/s.

4. It's spring break at brown college; the annual migration to florida has begun. a state patrol helicopter hovers 1000 ft directly above the southbound highway, with officer roethlisberger at the radar gun. a car with brown decals passes underneath and continues south. Braylon and kellen, fresh from their math 095 midterm, aren't wathcing their speedometer. officer roethlisberger fires and results flashback. at the moment of the firing the car was 2000 ft from the helicopter and moving away at 85 ft/s. "got'em mutters roethlisberger. what happens next? Braylon and kellen deny all wrong doing. the speed limit is 65mph they tell the judge. We were clocked at 85ft/s. a simple calculation shows that 65 mph is 95ft/s. Clearly, we couldn't have been speeding. "what you say makes sense," says the judge. "explain yourself roethlisberger!" clearly show how braylon and kellen deserve a ticket from officer roethlisberger. I have they were going approximately 66.9201 mph and were 1.9201 mph over the speed limit.

5. Find the equation of the tangent line and normal line to the ellipse. x^2-x*y+y^2 at the point (-1, 2) i have the tang line is y=4/5x+14/5 and the normal being y= -5/4x +13/4.

6. The last question is to find the second derivative (implicitly) of 2x^3-3y^2=8. I got 2x/y-(x^4)/(y^3).

Again, can you please help me!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!also, could i request if the answer is right that you let me now it? i would appreciate that
• Nov 13th 2007, 01:20 PM
Jhevon
Quote:

Originally Posted by jarny
My teacher gave us this take home test for the final grade of the marking period. I am sincerely begging for you guys to check my work. I understand it is alot so i'd really appreciate this if you could help. Thanks!!! I am going to post the problems from greatest to least importance along with my answers:

1. Show that the slope dy/dx is defined at every point on the graph of 2y= x^2 + siny. I don't really know how to do this.

just find dy/dx (using implicit differentiation) and show that it continuous for all x and y
• Nov 13th 2007, 01:21 PM
jarny
how do i do that? Here's what i did. I showed dy/dx implicitly equalling dy/dx explicitly and showed there can't be any vertical tangents, but i can't figure out how to show that there's not a jump or how there's not a sharp turn at (0,0) even though i can tell there's not i need to prove it mathematically. and can you help with the 13ft ladder problem also. Thank you jhevon
• Nov 13th 2007, 01:30 PM
Jhevon
Quote:

Originally Posted by jarny
2. A 13 ft ladder is leaning against a house. when its bse starts to slide away. by the time the base is 12 ft. from the house, the base is moving at the rate of 5 ft/sec. a) how fast is the top of the ladder sliding down the wall at that moment? b) At what rate is the area of the triangle formed by the ladder, wall, and ground changing at that moment? c) At what rate is the angle theta between the ladder and the ground changing at that moment? i got -12ft/s;59.5 ft/s; and i am still working on c.

for part (a): use $\displaystyle y^2 = 13^2 - x^2$ .........(1) (by Pythagoras' theorem)

differentiate implicitly with respect to t and solve for $\displaystyle \frac {dy}{dt}$. at the instant you are interested in, $\displaystyle x = 12$ and $\displaystyle \frac {dx}{dt} = 5$ (you can find y by using equation (1))

for part (b): use $\displaystyle A = \frac 12 xy$

differentiate implicitly with respect to t (note, you will need the product rule). you have the values for $\displaystyle x,y,\frac {dx}{dt} \mbox{ and } \frac {dy}{dt}$ from the previous part, so just plug in the values to solve for $\displaystyle \frac {dA}{dt}$

for part (c): use $\displaystyle \sin \theta = \frac x{13}$ ..............(2)

differentiate implicitly with respect to t and plug in the values for $\displaystyle \cos \theta$ and $\displaystyle \frac {dx}{dt}$ to solve for $\displaystyle \frac {d \theta}{dt}$. (you can find $\displaystyle \theta$, and hence, $\displaystyle \cos \theta$, when $\displaystyle x = 12$ by equation (2))
• Nov 13th 2007, 01:31 PM
jarny
would i say there are no holes because nothing cancels out on the numerator and denominator of the derivative and there are no sharp turns because there is no absolute value?
• Nov 13th 2007, 01:33 PM
Jhevon
Quote:

Originally Posted by jarny
how do i do that? Here's what i did. I showed dy/dx implicitly equalling dy/dx explicitly and showed there can't be any vertical tangents, but i can't figure out how to show that there's not a jump or how there's not a sharp turn at (0,0) even though i can tell there's not i need to prove it mathematically. and can you help with the 13ft ladder problem also. Thank you jhevon

i will use $\displaystyle y'$ to mean $\displaystyle \frac {dy}{dx}$

we have $\displaystyle 2y = x^2 + \sin y$

differentiating implicitly we get:

$\displaystyle 2~y' = 2x + \cos y~y'$

$\displaystyle \Rightarrow (2 - \cos y)y' = 2x$

$\displaystyle \Rightarrow y' = \frac {2x}{2 - \cos y}$

now this function is continuous everywhere, since it is the quotient of two continuous functions and the denominator is never zero (since $\displaystyle -1 \le \cos y \le 1$ for all $\displaystyle y$)
• Nov 13th 2007, 01:36 PM
Jhevon
Quote:

Originally Posted by jarny
3. a point "p" moves from the left to right along the curve y=x^2 at a constant horizontal speed of 3 units/sec. How fast does the y coordinate of p increase at the moment when "p" passes through (1,1) and (2,4)? i got 6 u/s and 12u/s.

correct :) (Clapping)
• Nov 13th 2007, 01:38 PM
jarny
how come when i find the it with sinof theta (this is c) equation and the tan of theta eqn it comes out with a different answer? my eqn is tantheta= x/y and wrt dtheta/dt sec^2theta = (dx/dt y -dy/dt x)/ 25
• Nov 13th 2007, 02:05 PM
Jhevon
Quote:

Originally Posted by jarny
how come when i find the it with sinof theta (this is c) equation and the tan of theta eqn it comes out with a different answer? my eqn is tantheta= x/y and wrt dtheta/dt sec^2theta = (dx/dt y -dy/dt x)/ 25

how did you get 25 for the denominator?
• Nov 13th 2007, 02:07 PM
jarny
nevermind i have it