Applications of Differentiation Flashcards

All 27 cards in this deck

What are the steps to find the equation of the tangent to a curve at a given point, in the form $y = mx + c$?
e.g. the tangent to $y = x^2$ at $(3, 9)$

1. Differentiate: $\frac{dy}{dx} = 2x$
2. Substitute the $x$-coordinate to get the gradient: $m = 6$
3. Use $y - y_1 = m(x - x_1)$: $y = 6x - 9$

What is the normal to a curve at a point?

The straight line through that point perpendicular to the tangent there.

What are the steps to find the equation of the normal to a curve at a given point?
e.g. the normal to $y = x^2$ at $(1, 1)$

1. Differentiate and substitute $x$ for the tangent gradient: $m = 2$
2. Normal gradient $= -\frac{1}{m} = -\frac{1}{2}$
3. Use $y - y_1 = m(x - x_1)$: $y = -\frac{1}{2}x + \frac{3}{2}$

What are the steps to find the point on a curve where the tangent is parallel to a given line?
e.g. on $y = x^2 - 4x$, tangent parallel to $y = 2x + 1$

1. Differentiate: $\frac{dy}{dx} = 2x - 4$
2. Set $\frac{dy}{dx}$ equal to the line's gradient: $2x - 4 = 2$, so $x = 3$
3. Substitute into the curve for $y$: $(3, -3)$

What are the steps to find the area of the triangle a tangent forms with the coordinate axes?
e.g. the tangent $y = 2x - 6$

1. Put $y = 0$ for the $x$-intercept: $x = 3$
2. Put $x = 0$ for the $y$-intercept: $y = -6$
3. Area $= \frac{1}{2} \times$ base $\times$ height $= \frac{1}{2} \times 3 \times 6 = 9$

True or false? To find the gradient of the tangent at a point you take the negative reciprocal of $\frac{dy}{dx}$ at that point.

False. Substituting into $\frac{dy}{dx}$ gives the tangent gradient directly; the negative reciprocal gives the normal gradient.

What is true of $\frac{dy}{dx}$ where a function is increasing?

$\frac{dy}{dx} > 0$ (the gradient is positive).

What is true of $\frac{dy}{dx}$ where a function is decreasing?

$\frac{dy}{dx} < 0$ (the gradient is negative).

What are the steps to show a function is increasing at a given point?
e.g. show $y = x^2 + 3x$ is increasing at $x = 1$

1. Differentiate: $\frac{dy}{dx} = 2x + 3$
2. Substitute the $x$-value: $\frac{dy}{dx} = 5$
3. State the sign: $5 > 0$, so the function is increasing.

What are the steps to find the range of values of $x$ for which a function is increasing?
e.g. $y = x^2 - 6x$

1. Differentiate: $\frac{dy}{dx} = 2x - 6$
2. Set up $\frac{dy}{dx} > 0$: $2x - 6 > 0$
3. Solve the inequality: $x > 3$

What are the steps to find the values of a constant $k$ for which a function is increasing for all $x$?
e.g. $f'(x) = 3x^2 + kx + 3$

1. Increasing for all $x$ means $f'(x) > 0$ for all $x$
2. As the $x^2$ coefficient is positive, require discriminant $< 0$: $k^2 - 36 < 0$
3. Solve: $-6 < k < 6$

True or false? A single function can be increasing for some values of $x$ and decreasing for others.

True. The sign of $\frac{dy}{dx}$ can change, e.g. a quadratic decreases then increases.

What does $\frac{d^2y}{dx^2}$ mean?

The derivative of $\frac{dy}{dx}$ — differentiate the function twice.

What does the second derivative measure?

The rate of change of the gradient function.

What notation is used for the second derivative of $f(x)$?

$f''(x)$.

What are the steps to find $\frac{d^2y}{dx^2}$?
e.g. $y = x^3 - 3x^2$

1. Differentiate once: $\frac{dy}{dx} = 3x^2 - 6x$
2. Differentiate the result again: $\frac{d^2y}{dx^2} = 6x - 6$

What is a stationary point on a curve?

A point where $\frac{dy}{dx} = 0$ (the gradient is zero).

What are the steps to find the coordinates of the stationary points of a curve?
e.g. $y = x^2 - 4x + 1$

1. Differentiate: $\frac{dy}{dx} = 2x - 4$
2. Set $\frac{dy}{dx} = 0$ and solve: $x = 2$
3. Substitute into the curve equation for $y$: $(2, -3)$

How does the second derivative tell you the nature of a stationary point?

$\frac{d^2y}{dx^2} < 0$ gives a maximum; $\frac{d^2y}{dx^2} > 0$ gives a minimum.

How does the sign of the gradient either side of a stationary point tell you its nature?

$\frac{dy}{dx} > 0$ then $< 0$ gives a maximum; $\frac{dy}{dx} < 0$ then $> 0$ gives a minimum.

What are the steps to find an unknown coefficient from a given stationary point?
e.g. $y = x^3 + bx^2$ has a stationary point at $x = 2$

1. Differentiate: $\frac{dy}{dx} = 3x^2 + 2bx$
2. Put $\frac{dy}{dx} = 0$ at the given $x$: $12 + 4b = 0$
3. Solve: $b = -3$

True or false? At a maximum point on a curve the tangent is parallel to the $x$-axis.

True. The gradient there is zero.

On a sketch, what happens to the curve between a minimum point and the maximum point that follows it (as $x$ increases)?

The function is increasing (gradient positive).

What are the steps to use calculus to find the maximum value of a quantity?
e.g. $V = 450x^2 - 30x^3$

1. Differentiate: $\frac{dV}{dx} = 900x - 90x^2$
2. Set $\frac{dV}{dx} = 0$ and solve: $x = 10$
3. Substitute back into $V$: maximum value $= 15\,000$

What are the steps to set up and optimise a quantity in a practical problem?
e.g. maximise the area of a rectangle of perimeter 20 cm

1. Write the quantity in terms of one variable: $A = x(10 - x)$
2. Differentiate and set to $0$: $10 - 2x = 0$, so $x = 5$
3. Substitute back: maximum area $= 25$ cm$^2$

What are the steps to show that a turning value of $V$ is a minimum?
e.g. $\frac{dV}{dx} = 3x^2 - 12$ with turning point at $x = 2$

1. Differentiate again: $\frac{d^2V}{dx^2} = 6x$
2. Substitute the turning point: $\frac{d^2V}{dx^2} = 12$
3. It is positive, so $V$ has a minimum.

True or false? The value of $x$ found from solving $\frac{dV}{dx} = 0$ is the maximum value of $V$.

False. That $x$ must be substituted back into the expression for $V$ to give the maximum value.