Review of Gradient and Original Functions
Notations for the derivative (investigation) LIVE
Rate of change functions
Deriving the Power Rule
Power Rule (x^n)
Derivative of a Sum (x^n)
Power rule (ax^n)
Derivative of a Sum (ax^n)
Power rule (x^n and ax^n)
Derivative of a Sum (x^n, ax^n)
Further derivatives using power rule
Derivative of a polynomial (expansion then power rule)
Evaluate Derivative at a point
Tangents (gradients and equations using power rule)
Simple applications (using power rule only)
Speed, displacement and velocity
Key Features of Graphs and the Derivative
Sketching functions using derivative information
Maximums and Minimums Using Calculus (power rule only)
Primitive Functions
Applications of primitive functions
Primitive functions are also known by the more descriptive term antiderivatives.
To be a primitive function, a function $F$F must be differentiable on an open interval. If it has derivative $F'=f$F′=f, then $F$F is said to be an antiderivative or primitive function of $f$f.
An antiderivative $F$F of a function $f$f is not unique since the derivative of $F+c$F+c where $c$c is any constant is also $f$f. We need further information in problems to determine the constant $c$c.
Generally, if we have a function that gives the rate of change of some quantity, often with respect to time, then an antiderivative of the function gives the quantity itself.
For example, velocity is the rate of change of displacement with respect to time. So, given velocity as a function of time, we obtain displacement as a function of time by antidifferentiation.
Similarly, acceleration is the time rate of change of velocity. So, a velocity function is obtained from an acceleration function by antidifferentiation.
The integral sign is used for antidifferentiation. Thus for displacement $s$s, velocity $v$v and acceleration $a$a, we have
$s=\int v(t)\ \mathrm{d}t+c$s=∫v(t) dt+c
$v=\int a(t)\ \mathrm{d}t+c$v=∫a(t) dt+c
Consider a body moving so that its velocity as a function of time is given by $v(t)=5t^2-2t+3$v(t)=5t2−2t+3. Given that its displacement at $t=0$t=0 is $-2$−2, find its displacement function $s(t)$s(t). What is the displacement at $t=3$t=3?
Since $s(t)=\int v(t)\ \mathrm{d}t+c$s(t)=∫v(t) dt+c, we have $s(t)=\int5t^2-2t+3\ \mathrm{d}t+c$s(t)=∫5t2−2t+3 dt+c. Hence, $s(t)=\frac{5t^3}{3}-t^2+3t+c$s(t)=5t33​−t2+3t+c. We know that $s(0)=-2$s(0)=−2. Therefore, $-2=c$−2=c and the function we seek is $s(t)=\frac{5t^3}{3}-t^2+3t-2$s(t)=5t33​−t2+3t−2.
From the displacement function, we have $s(3)=\frac{5\times27}{3}-9+9-2=43$s(3)=5×273​−9+9−2=43
A falling body near the surface of the earth has a constant acceleration of about $-9.8m/s^2$−9.8m/s2. If the body starts from rest at $t=0$t=0, find its velocity function. Given that the body begins its fall from a height of $200m$200m, find its displacement function.
Acceleration is the time rate of change of velocity. So, $v(t)=\int(-9.8)\ \mathrm{d}t+c$v(t)=∫(−9.8) dt+c. Hence, $v(t)=-9.8t+c$v(t)=−9.8t+c. We have $v(0)=-9.8\times0+c$v(0)=−9.8×0+c. Therefore $c=0$c=0 and the velocity function is $v(t)=-9.8t$v(t)=−9.8t.
Displacement is given by $s(t)=\int(-9.8t)\ \mathrm{d}t+k=\frac{-9.8t^2}{2}+k$s(t)=∫(−9.8t) dt+k=−9.8t22​+k. At $t=0$t=0, $s=200$s=200. So, $200=k$200=k and the displacement function is $s(t)=-4.9t^2+200$s(t)=−4.9t2+200.
The velocity of a particle moving in a straight line is given by $v\left(t\right)=6t+15$v(t)=6t+15, where $v$v is the velocity in metres per second and $t$t is the time in seconds.
The displacement after $2$2 seconds is $45$45 metres to the right of the origin.
Calculate the initial velocity of the particle.
Determine the function $x\left(t\right)$x(t) for the position of the particle. Use $C$C as the constant of integration.
Calculate the displacement of the particle after four seconds.
Find the total distance travelled between two and four seconds.
The velocity $v\left(t\right)$v(t) of an object travelling horizontally along a straight line after $t$t seconds is modelled by $v\left(t\right)=12t^2-48t$v(t)=12t2−48t, where $t\ge0$t≥0.
The object starts its movement $5$5 feet to the right of the origin. That is, $x\left(0\right)=5$x(0)=5.
State the displacement $x\left(t\right)$x(t) of the particle at time $t$t. Use $C$C as the constant of integration.
Solve for the times $t$t when the object is at rest.
Find the displacement at which the object is stationary other than its initial position.
The acceleration $a$a (in m/s2) of an object travelling horizontally along a straight line after $t$t seconds is modelled by $a\left(t\right)=6t-27$a(t)=6t−27, where $t\ge0$t≥0.
After $10$10 seconds, the object is moving at $90$90 metres per second in the positive direction.
State the velocity $v$v of the particle at time $t$t. Use $C$C as the constant of integration.
Solve for all times $t$t at which the particle is at rest. If there are multiple solutions write them on the same line separated by commas.
M7-10
Apply differentiation and anti-differentiation techniques to polynomials
Apply calculus methods in solving problems