10) Use the information to evaluate and compare delta y and dy. y=2-x^4, x=2 and delta x=dx=0.01
MY ANSWER: dy=-.32 and delta y=-.3224. Are they supposed to be negative, and how do I compare those in words? I don't even understand what I was supposed to do...
12) Find the differential dy of 3x^(2/3)
MY ANSWER: 9/(2cube root of x)
(Sorry, I don't know how to write a cube root thing)
20) Find the differential dy of the given function.
y=sec^2x/(x^2+1)
MY ANSWER: [tanx(x^2+1)-2sec^2x]dx/(x^2+1)^2
30) The measurement of the edge of a cube is found to be 15 inches, with a possible error of 0.03 inch. Use differentials to approximate the maximum possible propagated error in computing the volume of the cube.
MY ANSWER: 20.25
Please tell me if I got any of them wrong, and showing me what I'm supposed to be doing would be helpful! I had to teach this section to myself, so I'm struggling more than usual.
10) Use the information to evaluate and compare delta y and dy. y=2-x^4, x=2 and delta x=dx=0.01
MY ANSWER: dy=-.32 and delta y=-.3224. Are they supposed to be negative, and how do I compare those in words? I don't even understand what I was supposed to do...
-------------------------
dy = 0 - 4 x^3 dx
x = 2 and dx = .01
dy = -4 (8)(.01) = -.32 correct
ok but then x = 2 not 4 !!!!!
y at x = 2 is y = -14
y at x = 2.01 = -14.3224
delta y = -.3224
that is better :)
Never mind number 20! Someone already helped me with it!
I already did 20, scroll down
10) Use the information to evaluate and compare delta y and dy. y=2-x^4, x=2 and delta x=dx=0.01
MY ANSWER: dy=-.32 and delta y=-.3224. Are they supposed to be negative, and how do I compare those in words? I don't even understand what I was supposed to do...
-------------------------
dy = 0 - 4 x^3 dx
x = 2 and dx = .01
dy = -4 (8)(.01) = -.32 correct
y at x = 4 is y = -254
y at x = 4.01 = -256.56
delta y = -256.56 + 254 = -2.56
12) Find the differential dy of 3x^(2/3)
MY ANSWER: 9/(2cube root of x)
(Sorry, I don't know how to write a cube root thing)
------------------------
y = 3 x^(2/3)
dy/dx = (2) x^-(1/3)
dy = 2/(x^1/3)
20) Find the differential dy of the given function.
y=sec^2x/(x^2+1)
MY ANSWER: [tanx(x^2+1)-2sec^2x]dx/(x^2+1)^2
30) The measurement of the edge of a cube is found to be 15 inches, with a possible error of 0.03 inch. Use differentials to approximate the maximum possible propagated error in computing the volume of the cube.
MY ANSWER: 20.25
-----------
v = x^3
dv = 3 x^2 dx
dv = 3 (225)(.03) = 20.25 right
Thank you soooo much!
10) To evaluate and compare delta y and dy, we first need to understand what they represent. In calculus, delta y (Δy) represents the change in y, while dy represents the differential of y.
To find delta y, we can use the formula:
Δy = y(final) - y(initial)
In this case, y = 2 - x^4, x = 2, and delta x (dx) = 0.01. So, to find the initial y value, we substitute x = 2 into the equation:
y(initial) = 2 - (2^4) = -14.
Next, we need to find the final y value by substituting x = 2 + dx = 2 + 0.01 = 2.01 into the equation:
y(final) = 2 - (2.01^4).
By calculating both y(initial) and y(final), we can then find delta y by subtracting y(initial) from y(final).
To find dy, we can use the derivative of the function. Taking the derivative of y = 2 - x^4 with respect to x, we get:
dy/dx = -4x^3.
Now, to find dy, we substitute x = 2 into the derivative:
dy = -4(2)^3 = -32.
Comparing the values, we have dy = -32 and delta y = y(final) - y(initial). If you computed dy to be -0.32, it seems there may have been a calculation mistake.
In words, we can say that delta y represents the change in y when x increases by a small amount (dx), while dy represents the instantaneous change in y when considering only the derivative at a specific point.
12) To find the differential dy of the function 3x^(2/3), we can use the formula:
dy = (dy/dx) * dx.
Here, dy/dx is the derivative of 3x^(2/3) with respect to x. Taking the derivative, we get:
dy/dx = (2/3) * 3(x^(-1/3)).
Simplifying, we have:
dy/dx = 2x^(-1/3).
Now, we substitute this into the formula along with the given dx:
dy = (2x^(-1/3)) * dx.
In words, the differential dy represents the change in the function 3x^(2/3) due to a small change in x (dx).
20) To find the differential dy of the function y = sec^2(x) / (x^2 + 1), we apply a similar approach.
First, we'll find the derivative of the function. Taking the derivative with respect to x:
dy/dx = (d/dx)(sec^2(x) / (x^2 + 1)).
Applying the quotient rule and simplifying the expression, we get:
dy/dx = [tan(x)(x^2 + 1) - 2(sec^2(x))]/(x^2 + 1)^2.
Now, we can write the differential dy as:
dy = dy/dx * dx.
Substituting the derivative and dx into the equation, we get:
dy = [tan(x)(x^2 + 1) - 2(sec^2(x))]/(x^2 + 1)^2 * dx.
So, the differential dy represents the change in the function y = sec^2(x) / (x^2 + 1) due to a small change in x (dx).
30) To approximate the maximum possible propagated error in computing the volume of a cube, we can use differentials.
The volume of a cube is given by V = s^3, where s represents the edge length. Here, the edge length is measured as 15 inches, with a possible error of 0.03 inches.
We want to find the maximum possible propagated error in the volume, which can be approximated using differentials. The differential of the volume (dV) can be calculated as:
dV = (dV/ds) * ds,
where ds represents the change in the edge length.
Taking the derivative of V = s^3 with respect to s, we get:
dV/ds = 3s^2.
Now, we substitute the given values:
s = 15 inches,
ds = 0.03 inches.
Calculating the differential:
dV = 3(15^2) * 0.03 = 20.25.
Therefore, the maximum possible propagated error in computing the volume of the cube is approximately 20.25 cubic inches.