Consider the curve given by x2 + sin(xy) + 3y = C, where Cis a constant. The point (1, 1) lies on this curve. Use the tangent line approximation to approximate the y-coordinate when x = 1.01

Question

round to the nearest thousanth

Answer 1

We have to construct the tangent to the curve x² + sin(xy) + 3y = Constant at the point (1, 1).

That means we need the slope (i.e., the derivative with respect to x) of the curve at this point.

Let's get this slope by doing an implicit differentiation, taking the derivative with respect to x (i.e, the d/dx) of both sides of the equation x² + sin(xy) + 3y = Constant:

(d/dx)(x² + sin(xy) + 3y) = (d/dx)Constant = 0

The right side is equal to zero because the derivative of a constant is always zero.

We can rewrite the left side as three separate terms, since "the derivative of a sum is the sum of the derivatives":

(d/dx)(x²) + (d/dx)(sin(xy)) + (d/dx)(3y) = 0

Of course, the derivative of x² with respect to x is 2x.

The next two terms are harder; we'll need the chain rule and the product rule:

2x + cos(xy)(x(dy/dx) + (dx/dx)y) + 3(dy/dx) = 0

Sine dx/dx = 1, we can simplify this to

2x + cos(xy)(x(dy/dx) + y) + 3(dy/dx) = 0

Now we'll "solve for dy/dx".

Rewrite using two applications of the distributive law so that dy/dx appears only once:

2x + (dy/dx)xcos(xy) + ycos(xy) + 3(dy/dx) = 0

2x + (dy/dx)(xcos(xy) + 3) + ycos(xy) = 0

Get the dy/dx term all by itself:

(dy/dx)(xcos(xy) + 3) = -2x - ycos(xy)

Divide both sides by xcos(xy) + 3 in order to get the dy/dx all by itself:

(dy/dx) = (-2x - ycos(xy))/(xcos(xy) + 3)

Now that we know the slope (dy/dx), let's find the slope at the point (1, 1):

(dy/dx) = (-2·1 - 1cos(1·1))/(1cos(1·1) + 3)

= (-2 - cos(1))/(cos(1) + 3)

(The slope we just got is a negative number, so the tangent line to the curve at the point (1, 1) goes through the point from upper left to lower right.)

Now that we know that the tangent line goes through the point (1, 1) and has the slope (-2 - cos(1))/(cos(1) + 3), we could write the equation of the tangent line using the "point-slope formula". Then using the equation of the tangent line, we could find the y coördinate of the point on the tangent line whose x coördinate is 1.01.

But let's get the answer in a simpler way.

The point (1, 1) is on the curve and also on the tangent line. How do we move from there to the point on the tangent line whose x coördinate is 1.01? We simply add .01 to the x coördinate of (1,1), and we add

((-2 - cos(1))/(cos(1) + 3))/100 (i.e., 1/100 of the slope of the tangent) to the y coördinate of (1, 1).

So the point we want on the tangent has the coördinates (1 + 1/100, 1 + ((-2 - cos(1))/(cos(1) + 3))/100), i.e.,

(1.01, 1 + ((-2 - cos(1))/(cos(1) + 3))/100). Use your calculator to approximate the value of the y coördinate of this point.

Hint: the y value should be a little less than 1, because the slope of the tangent is negative.)

Answer 2

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