1) The sides of the triangle shown increase in such a way that (dz/dt=1) and (dx/dt=(3dy/dx))

At the instant when x = 12 and y = 5, what is the value of dx/dt?

2) Let f(x) = x^3 − 4.
Which of these is the equation for the normal line to this curve at the point (2,4).
A) y=1/2x-25/6
B) y=-1/12x+25/6
C) y= 12x-25/6
D) y=-12x-4
E) y= -12x-2

3) Which of the following could be the units for dy/dx if y is the surface area of a tumor and x is the radius of the tumor?

1) The sides of the triangle shown increase in such a way that (dz/dt=1) and (dx/dt=(3dy/dx))

At the instant when x = 12 and y = 5, what is the value of dx/dt?
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You do not say but I suspect this is a 5,12,13 right triangle. If I am guessing right about what you mean then

and x=12, y = 5, z = 13
x^2+y^2= z^2
2xdx+2ydy = 2zdz
dz= 1 dt
dx/dt = 3 dy/dx

24 dx +10 dy = 26 dz
24 + 10dy/dx = 26 dz/dx
24 + (10/3) dx/dt = 26 dt/dx
=26/dx/dt
(10/3)(dx/dt)^2+24 dx/dt -26 = 0

I get dx/dt = .956 or -8.16

3.

It is square millimeters per millimeter NOT meters per second

Prob. 1. Not sure...

Prob. 2.
The slope of the curve is df/dx = 3x^2.
At (x,y) = (2,4), df/dx = 12.
The normal line must have slope -1/12, which suggestscanswer b.

Prob. 3.
Area/Radius = meter^2/meter = meter
(Or cm, whatever units your book specifies).

1) Well, well, well, looks like we've got ourselves a triangle! Let's see what we have here. We know that dz/dt = 1, which means the height of our triangle is increasing at a rate of 1 unit per time unit. Now, we also have dx/dt = 3(dy/dx). Ooh, fancy differential equations! Don't worry, I'll break it down for you. This just means that the rate at which x is increasing is three times the rate at which y is changing with respect to x. Got it? Good.

Now, at the instant when x = 12 and y = 5, we need to find the value of dx/dt. Okay, let's use our super math skills. dx/dt = 3(dy/dx). We don't know dy/dx, but we can find it using the given information. Since y is the height of our triangle, and x is the base, we can use the good old Pythagorean theorem to relate them. We have x^2 + y^2 = z^2. Differentiating both sides with respect to x, we get 2x + 2yy' = 2zz'.

Now, plug in the values we know: x = 12 and y = 5. We get 24 + 10y' = 2z'. Since dz/dt = 1, we have z' = 1. Substituting that in, we have 24 + 10y' = 2. Solving for y', we find y' = -11/10.

Finally, we can find dx/dt = 3(dy/dx). Substituting y' = -11/10, we get dx/dt = 3(-11/10), which simplifies to dx/dt = -33/10. So, at the instant when x = 12 and y = 5, the value of dx/dt is -33/10. Voilà!

2) Ah, curves and lines. They're like yin and yang, always trying to balance each other out. Now, to find the equation for the normal line to the curve given by f(x) = x^3 - 4 at the point (2,4), we need to find the slope of the normal line. The slope of the normal line is the negative reciprocal of the derivative of the curve at that point.

So, let's differentiate f(x) = x^3 - 4 with respect to x. We get f'(x) = 3x^2. Now, let's find the slope at x = 2. Substituting x = 2 into f'(x), we get f'(2) = 3(2^2) = 12.

Now, the slope of the normal line is the negative reciprocal of 12, which is -1/12. So, we know the slope of the normal line is -1/12. But that's not enough, we need an equation!

Using the point-slope form, we have y - y1 = m(x - x1), where (x1, y1) is the point (2,4) and m is the slope we found earlier. Let's plug in the values and see what we get!

We have y - 4 = (-1/12)(x - 2). Simplifying that, we get y = (-1/12)x + 1/6 + 4. Tidying it up a bit, we have y = -1/12x + 25/6. Ah, option B) y = -1/12x + 25/6, you've won the title of the equation for the normal line to the curve at the point (2,4)!

3) Oh, tumor talk. Well, if y is the surface area of a tumor and x is the radius of the tumor, then the units for dy/dx, the derivative of y with respect to x, will be units of surface area per unit of radius.

To put it in simpler terms, let's say the surface area of the tumor is measured in square centimeters and the radius is measured in centimeters. This means that dy/dx will have units of square centimeters per centimeter. Get it? It's all about those wacky units!

So, the units for dy/dx could be square centimeters per centimeter, or whatever units are used for measuring the surface area and radius of the tumor. Hope that clears things up!

1) To find the value of dx/dt at the given point, we need to determine the relationship between x, y, and z in the triangle. We are given that dz/dt = 1, which means that the height of the triangle (z) is changing at a rate of 1 unit per unit of time. We are also given that dx/dt = 3(dy/dx).

Using the given information, we can set up a relationship between x, y, and z by integrating dx and dy.

∫dx = 3∫dy
x = 3y + C

We can differentiate both sides to find dx/dt in terms of dy/dt:

dx/dt = 3(dy/dt)

Now we need to determine the value of dy/dt at the specific point (x = 12, y = 5). Unfortunately, the information provided does not give us a direct value for dy/dt at that point. Thus, we cannot determine the exact value of dx/dt.

2) To find the equation for the normal line to the curve at a specific point, we need to find the derivative of the function f(x) and evaluate it at the given point. The derivative represents the slope of the tangent line to the curve, and the normal line will have a slope that is the negative reciprocal of the tangent line's slope.

Taking the derivative of f(x):

f'(x) = 3x^2

At the point (2, 4), we can evaluate the derivative:

f'(2) = 3(2)^2 = 12

The tangent line's slope at that point is 12. To find the slope of the normal line, we take the negative reciprocal:

m_normal = -1/12

Now, we can use the point-slope form of a line to find the equation of the normal line:

y - y1 = m(x - x1)

Using the point (2, 4):

y - 4 = (-1/12)(x - 2)

Simplifying the equation gives:

12y - 48 = -x + 2

12y = -x + 50

Dividing by 12:

y = (-1/12)x + (50/12)

Simplifying further:

y = (-1/12)x + (25/6)

Therefore, the equation for the normal line to the curve at the point (2, 4) is y = (-1/12)x + (25/6), which corresponds to option B.

3) When considering the relationship between a function and its input, the derivative represents the rate of change of the function with respect to the input variable. In this case, y represents the surface area of a tumor and x represents the radius of the tumor.

The units for dy/dx will depend on the units used to measure y and x. Since the surface area of a tumor is typically measured in square units (such as square centimeters) and the radius is measured in linear units (such as centimeters), the units for dy/dx would be square units per linear unit, or (unit^2/unit).

For example, if y is measured in square centimeters and x is measured in centimeters, then the units for dy/dx would be cm^2/cm.