The Tangent Line Trivia Quiz

Answer: 12

The derivative function gives the slope of the tangent line to a function at a given point. In this case, dy/dx = 3x^2, which when evaluated at x = 2 gives 3*2^2 = 12.

Answer: (3,16)

The derivative of y = x^2 + 4x - 5 is given by dy/dx = 2x + 4. We want the value of x for which the slope of the tangent line equals 10, so we set the derivative equal to 10 and solve for x:

2x + 4 = 10

2x = 6

x = 3.

Putting x = 3 into the equation of the parabola gives y = 16, so (3,16) is the desired point on the graph.

Answer: a = -6, b = 7

The derivative gives the slope of the tangent line. The derivative of y = x^2 + ax + b is given by:

dy/dx = 2x + a

The slope of the tangent line at the point (4,-1) is found by subsituting x = 4 into the derivative: We get 8 + a.

This time we know that y = 2x - 9 is the tangent line, which has slope 2. Hence we have the equation 8 + a = 2, thus a = -6.

To find b, note that the point (4,-1) lies on the graph of the parabola y = x^2 -6x + b, so substitute x = 4 and y = -1:

-1 = 4^2 - 6*4 + b
-1 = 16 - 24 + b
-1 = -8 + b
Thus b = 7. Note that this question is really backwards from the "usual" calculus questions!

Answer: 3x + 4y = 25

To find the equation of a line, you need a point and a slope. We have the point, (3,4), so all that is needed is the slope. There are two approaches to finding the slope:

Approach 1 (High school geometry): We know that a line through the center of the circle is perpendicular to the tangent line at a point on the circle. The line connecting the center of the circle, (0,0), to the point (3,4) has slope 4/3. Hence the tangent line has slope -3/4, since it is perpendicular to the line through the center of the circle and slopes of perpendicular lines are negative reciprocals of one another.

Approach 2 (Calculus): We first find dy/dx by implicit differentiation:

x^2 + y^2 = 25 defines a function y in terms of x implicitly. Differentiate everything with respect to x to obtain:

2x + 2y * (dy/dx) = 0. Solving for dy/dx gives dy/dx = -x/y. The slope of the tangent line is just dy/dx evaluated at the point (3,4), which means substituting x = 3 and y = 4 into the formula for dy/dx, giving -3/4.

So either method gives the slope of the tangent line is -3/4. A fast way of writing the equation of a line with slope -3/4 passing through (3,4) is by using point-slope form:

y - 4 = (-3/4)*(x - 3)

Multiply both sides by 4 to get 4y - 16 = -3x + 9. Add 3x to both sides and add 16 to both sides to get 3x + 4y = 25.

Answer: 4x + 7y = 25

This is a classic problem. My solution is based on a procedure from Cohen's "Algebra and Trigonometry" book that I'm recalling from memory. Let A have coordinates (x_0,y_0) and let B have coordinates (x_1,y_1). In the above problem I found the slope of the tangent line to the circle by implicit differentiation:

For x^2 + y^2 = 25, dy/dx = -x/y.

So the tangent line at point A has slope -x_0/y_0 and the tangent line at point B has slope -x_1/y_1. Let's use point-slope formula to find the equation of the tangent line through point A:

y - y_0 = (-x_0/y_0) * (x - x_0)

Multiply both sides by y_0 to get

y_0 * y - (y_0)^2 = -x_0 * x + (x_0)^2

Rearranging gives x_0 * x + y_0 * y = (x_0)^2 + (y_0)^2.

But (x_0,y_0) is a point on the circle, hence (x_0)^2 + (y_0)^2 = 25.

Therefore the tangent line through point A has equation x_0 * x + y_0 * y = 25.

The point (4,7) lies on this tangent line, so 4x_0 + 7y_0 = 25.

Similarly, the tangent line through point B has equation x_1 * x + y_1 * y = 25. The point (4,7) lies on this tangent line, so 4x_1 + 7y_1 = 25.

Finally, I claim that the line 4x + 7y = 25 passes through the points A and B. All that's needed to check is to see if A with coordinates (x_0,y_0) lies on this line and B with coordinates (x_1,y_1) lies on this line. But we already showed 4x_0 + 7y_0 = 25 and 4x_1 + 7y_1 = 25, so we're done. This solution is truly amazing - it is really elegant! It's also surprising - who would have thought that this line has such a nice equation?

Answer: (10/3, 8/3), (-10/3, -8/3)

First we need to find the derivative by implicit differentiation:

x^2 - y^2 = 4 Differentiate everything with respect to x:

2x - 2y*dy/dx = 0

Solving for dy/dx gives dy/dx = x/y.

Let (a,b) be a point on the graph where the tangent line has slope 5/4. Then a/b = 5/4 which means a = (5/4)b. (a,b) is a point on the hyperbola, so it satisfies the equation:

a^2 - b^2 = 4 Now substitute a = (5/4)b and solve for b:

(25/16)b^2 - b^2 = 4

(9/16)b^2 = 4

b^2 = 64/9 thus b = 8/3 or b = -8/3.

Now use a = (5/4)b to get the value of a in each case. So we get the two points are (10/3, 8/3), (-10/3, -8/3).

In case you're wondering about the other answers, note that all of the other answers really are points on the hyperbola! They were obtained by finding a rational parameterization of the hyperbola usiing the following trick: Start with the point (-2,0) on the hyperbola and draw a line with slope t through it. The other point of intersection has x-coordinate equal to

x = 2(1 + t^2)/(1 - t^2).

Answer: (4,68)

First of all we need to find the equation of the tangent line to the graph of y = x^3 + x at the point where x = -2. The derivative of y = x^3 + x is given by

dy/dx = 3x^2 + 1

Evaluating this at x = -2 gives the slope of the tangent line is 13. Now use point-slope form to find the equation of the tangent line:

y + 10 = 13(x + 2)

Solving for y gives y = 13x + 16. Now we want the other point of intersection of this line with the cubic y = x^3 + x, so set the equations equal to each other and solve for x:

x^3 + x = 13x + 16

Putting everything on the left hand side gives x^3 - 12x - 16 = 0. Note that we know x = -2 is a solution since the tangent line passes through the point of the graph at x = -2. Therefore, x + 2 is a factor of the left hand side. Use synthetic division (or long division) to get the other factor:

(x + 2)(x^2 - 2x - 8) = 0

(x + 2)(x + 2)(x - 4) = 0

So we see that the other point of intersection is at x = 4. Putting x = 4 into the original equation gives y = 68, so the other point of intersection is at (4,68).

Interesting side note: Suppose instead the equation was y = x^3 + ax. If you draw the tangent line at x = -2, it will also intersect at x = 4!

Another side note: Note that -2 was a root of multiplicity 2 in the above equation. It should be expected that -2 has multiplicity greater than one because the line intersecting the curve at this point is tangent to the graph, it doesn't intersect transversally.

Answer: a = 3, b = 1, c = -2

We know the point (1,3) lies on the curve, so putting x = 1 and y = 3 gives:

3 = 1 + a + b + c

a + b + c = 2

Next, we know that (0,-2) lies on the curve, so putting x = 0 and y = -2 gives:

-2 = c

Substituting c = -2 above gives a + b - 2 = 2 or a + b = 4. We need one more equation. The derivative gives the slope of the tangent line:

dy/dx = 3x^2 + 2ax + b

Evaluating this at x = 0 gives b. But the slope of the tangent line y = x - 2 is 1. Thus b = 1. Substituting b = 1 into the equation a + b = 4 gives a = 3.

Answer: a = -48, b = 128

First of all, the derivative of y = x^3 + ax + b is given by

dy/dx = 3x^2 + a

The slope of the tangent line at (4,0) is found by substituting x = 4 into the derivative. We get the slope of the tangent line is 3 * 4^2 + a, or 48 + a. Since the curve is tangent to the x-axis at this point, the slope of the tangent line is 0. So set 48 + a equal to 0, which gives a = -48. Now we also know that (4,0) is a point on the graph of y = x^3 - 48x + b, so substitute x = 4 and y = 0:

0 = 4^3 - 48*4 + b

Simplifying gives 0 = -128 + b, hence b = 128.

Here's another solution: Since the graph is tangent to the x-axis at x = 4, (x + 4)^2 is a factor of x^3 + ax + b. We know that the sum of the roots of x^3 + ax + b is 0 (the coefficient of x^2), hence the third root of this equation is 8 (-4 has multiplicity 2). Therefore,

x^3 + ax + b = (x + 4)^2 * (x - 8)

Multiplying out the right hand side gives x^3 - 48x + 128. Equating coefficients gives a = -48, b = 128.

Answer: 2

First of all, the derivative gives the slope of the tangent line. Use the product rule to find the derivative:

dy/dx = x * d/dx(e^(-x)) + e^(-x) * d/dx(x) = -xe^(-x) + e^(-x)*1

Factoring gives dy/dx = (1 - x)e^(-x). This represents the slope of the tangent line at any value of x. We want to find the value of x that minimizes the slope, so we need to take the derivative of this function and set it equal to 0 to find the critical points of x. We use the product rule again to find the derivative:

d^2y/d^2x = (x - 2)e^(-x)

This is 0 only when x = 2 (exponential functions are never 0). Now we need to determine the nature of this critical value. I'll use the second derivative test, but note that the second derivative of the first derivative is of course the third derivative:

d^3y/dx^3 = (3 - x)e^(-x)

Evaluating this at x = 2 gives e^(-2), which is positive. Hence by the second derivative test, the derivative function dy/dx = (1 - x)e^(-x) has a local minimum at x = 2. Since there is only one local extrema, it must be a global one. Thus x = 2 minimizes dy/dx, thus at x = 2, the graph has a tangent line of minimum slope.

Now let me answer the question you may be thinking: Why isn't the answer x = 1? Setting the derivative equal to 0 and solving for x gives x = 1, and then evaluating the second derivative at x = 1 gives -e^(-1), which is negative. Thus y = xe^(-x) has a local maximum at x = 1. Since there is only one local extrema, it is a global one and y = xe^(-x) has a global maximum at x = 1. This means that at x = 1 we find the highest point on the graph. This is not what I'm asking - I'm asking at which point on the graph does the tangent line have the smallest slope? Note that this is a very different question. To solve this problem you need to minimize a different function than what is given - the derivative.

Notice that there is no maximum value to the slope of a tangent line - we see that dy/dx goes to infinity as x goes to minus infinity. Also observe how nice all the derivatives of this function are!

I hope you enjoyed my first calculus quiz! Thanks for playing!