Math Calculus Question

Question

Question: y=x/root(1-x) has a tangent with equation 5x+by=a at the point where x=-3. Find the values of a and b.

Answer: a=9, b=-16

Answer 1

desmos.com/calculator/qvernq1vjo

The above graph has a slider for any point of the graph of f along with the equation of the tangent line at

(x, f(x))

Answer 2

Hi Haru,

What we need to do in this problem is find a point on the graph AND its slope at that point.

With that information we can write the equation of the tangent line passing through that point.

The question also provides when written in the standard form that linear equation will appear as; 5x + by = a, and our task is to provide values for a and b.

First let's find the y-coordinate that pairs with x = -3 by substituting -3 for x in the function.

y = x/root(1-x) becomes y = -3/root(1-(-3)) = -3/2

So, our point of tangency is (-3, -3/2)

Next, we find the derivative of the function.

I chose to use the product rule.

Why? Because like everyone else, I HATE the quotient rules. ;)

To begin, I re-write the function as; y = x(1 - x)^-1/2 so we can see the function as a product of two others, x and (1 - x)^-1/2.

Recall that the product rule states that if a function is in the form uv, then its derivative is uv' + u'v.

Here, u = x, u' = 1, v = (1 -x)^-1/2 and v' = 1/2(1 - x)^-3/2.

Putting the pieces together, we get;

y' = x(1/2(1 - x)^-3/2 + (1 - x)^-1/2

While we might be tempted to clean this up, we don't need to.

We only need to find the slope when x = -3.

Making that substitution and evaluating, we find that at x = -3, the slope m = 5/16.

Lastly, we need to find the equation in the standard form of the line with slope 5/16 passing through the point (-3, -3/2).

I choose to use the slope-intercept form.

Why? We know m, x and y. We only need to find the y-intercept and rewrite the form.

-3/2 = 5/16(-3) + k (Can't use b for the y-intercept, because b is being used.)

We solve this to find k = -9/16.

y = 5/16x - 9/16

rewriting into the standard form we get;

5x -16y = 9

Therefore a = 9 and b = -16

I hope this helps.

Answer 3

Curve equation:

y=x/√(1-x)

y' = (2-x)/(2(1-x)^3/2)

y(-3) = -3/√4 = -3/2

y'(-3) = (2- -3)/(2(4)^3/2) = 5/16

Line equation

5x+by=a

y = (a-5x)/b

y' = -5/b

y(-3) = (a+15)/b

y'(-3) = -5/b

Setting y'(-3) values equivalent

-5/b = 5/16

b = -16

Setting y(-3) values equivalent:

(a+15)/-16 = -3/2

a+15 = 48/2 = 24

a = 24-15 = 9

a=9

b=-16

Answer 4

Hi Haru!

We're going to answer this similarly to the other tangent line equation you asked earlier. The key conceptual understandings are:

1) There will be one (and only one) point (m, n) that is a point on both the tangent line AND the original function.

2) The slope of the tangent line is equal to the value of the derivative of the function at x = m.

So, let's start with the first point:

Part 1: There is a point that is on both the tangent line and the function.

We know that the line passes through the point x = -3.

Let's substitute this into the equation of the original function: y=x/root(1-x)

y = -3/root(1+3) = -3/root(4) = -3/2

So we know that the tangent line passes through the point (-3, -3/2).

Let's substitute that in and see what we have:

5x + by = a

5(-3) + b (-3/2) = a

-15 + (-3/2) b = a

Hmmm let's just leave that for now.

Part 2

We also know that the slope of the tangent line needs to be the value of the derivative function at x = -3.

Our tangent line is 5x + by = a, so the slope of this line is -5/b. (Try changing the form to a slope-intercept form, y = mx + b, it that will help you identify the slope.)

We now need to take the derivative of the function and evaluate it at x = -3

y = x/√(1-x)

To take the derivative, we will use the quotient rule. [Remember also that √(1-x) is the same as (1 - x)^1/2]

So the derivative is:

y' = {(1 - x)^1/2 - x * (1/2)(1 - x)^-1/2 (-1)}/(1 - x)

y' = {(1 - x)^1/2 + (1/2)x(1 - x)^-1/2}/(1 - x)

*At this point, we can simplify this expression a bit by factoring out (1 - x)^-1/2

y' = (1 - x)^-1/2{(1 - x) + (1/2)x}/(1 - x)

y' = (1 - (1/2)x)/(1 - x)^3/2

y' = {1 - (1/2)x)}/(1 - x)^3/2

We know that at x = -3, y' = -5/b, so let's substitute that in:

-5/b = (1 + 3/2)/(1 + 3)^3/2

-5/b = (5/2)/4^3/2

-5/b = 5/16

b = -16

Part 3:

Now we can substitute our b value into the expression we found earlier to find a: -15 + (-3/2) b = a

-15 + (-3/2)(-16) = a

-15 + 24 = a

a = 9

If this answer was helpful, please feel free to reach out to schedule regular tutoring sessions!

Answer 5

Find the point on the function by plugging in x = -3:

y = (-3)/√(1 - (-3)) = -3/√4 = -3/2

So, a point on the function line is (-3, -3/2)

To find the slope of the tangent line, take the derivative.

For y = x/√(1-x) use the quotient rule (u/v)' = (u'v - uv')/v² where, in this case u = x, meaning u' = 1 and v = √(1 - x) meaning v' = 1/2(1 - x)^-1/2(-1) = -1/(2√(1 - x)) and v² = 1 - x. Put those together to get: the derivative. Then plug in x = -3 to find the slope at x = -3

Then use the slope you got plus the point on the function line (-3, -3/2) in the point slope form of a line: y - y₁ = m(x - x₁) to get the equation of the line