AI HL · 12.9
0 of 48 skills done
Gradient at a point and where the gradient equals a value
12.9.7
Drag the orange point P along the curve. The orange line is the tangent at P. Its gradient is f′(x) at that point.
The two axes have different scales, so the normal does not always look perpendicular on the screen.
Drag P along the curve.
Find the gradient of the curve \(y=2x^4-3x\) at the point where \(x=-2\).
Have a go first. Then open the working one line at a time.
\(\frac{dy}{dx}=8x^{3} - 3\)
The gradient of the curve is the derivative.
\(\text{When }x=-2:\ \frac{dy}{dx}=8(-2)^{3}-3=-67\)
Put the number in brackets so the sign is not lost.
If \(P=6-5r^{-2}\), find \(\frac{dP}{dr}\) when \(r=0.5\).
Have a go first. Then open the working one line at a time.
\(\frac{dP}{dr}=\frac{10}{r^{3}}\)
Multiply by the power, then take 1 off it: \(-2-1=-3\). The 6 gives 0.
\(\text{When }r=0.5:\ \frac{dP}{dr}=\frac{10}{(0.5)^{3}}=80\)
\(0.5^{-3}=\frac{1}{0.125}=8\).
Find the gradient of the tangent to the curve \(y=2x^2+\frac{3}{x}\) at the point where \(x=0.25\).
Have a go first. Then open the working one line at a time.
\(y=2x^2+3x^{-1}\)
The rule needs \(x^n\), so write \(\frac3x\) as \(3x^{-1}\).
\(\frac{dy}{dx}=4x - \frac{3}{x^{2}}\)
\(3x^{-1}\to-3x^{-2}\).
\(\text{When }x=0.25:\ \frac{dy}{dx}=4(0.25)-\frac{3}{(0.25)^{2}}=-47\)
\(0.25^2=0.0625\) and \(3\div0.0625=48\).
How quickly does \(N=\frac{6}{c}+2c\) change as c changes, when \(c=2\)?
Have a go first. Then open the working one line at a time.
\(\text{rate of change of }N\text{ with respect to }c=\frac{dN}{dc}\)
"How quickly does N change as c changes" means \(\frac{dN}{dc}\).
\(\frac{dN}{dc}=2 - \frac{6}{c^{2}}\)
Write \(\frac6c\) as \(6c^{-1}\) first.
\(\text{When }c=2:\ \frac{dN}{dc}=2-\frac{6}{(2)^{2}}=0.5\)
N is increasing by 0.5 for each 1 added to c, at this point.
Find the values of x at which the tangent to the curve \(y=x^3-9x^2+4\) has gradient 21.
Have a go first. Then open the working one line at a time.
\(\frac{dy}{dx}=3x^{2} - 18x\)
The gradient is given by the derivative.
\(3x^2-18x=21\)
You are told the gradient is 21.
\(x^2-6x-7=0\)
Move 21 across and divide by 3.
\((x-7)(x+1)=0\Rightarrow x=-1\text{ or }x=7\)
A quadratic gives up to two values of x.
Find the values of x at which the tangent to the curve \(y=4x+\frac{9}{x}\) has gradient 3.
Have a go first. Then open the working one line at a time.
\(y=4x+9x^{-1}\Rightarrow\frac{dy}{dx}=4-\frac{9}{x^2}\)
\(9x^{-1}\to-9x^{-2}\).
\(4-\frac{9}{x^2}=3\)
Rearrange to get \(x^2\) on its own.
\(\frac{9}{x^2}=1\Rightarrow x^2=9\)
Multiply both sides by \(x^2\).
\(x=3\text{ or }x=-3\)
Check both. Each has gradient 3, because \(x^2\) is the same.
The distance s metres that a cyclist has travelled t seconds after setting off is modelled by \(s=2t^2+3t\). (a) Find the average speed between t = 3 and t = 4, between t = 3 and t = 3.1, and between t = 3 and t = 3.01. (b) Find the speed of the cyclist at exactly t = 3.
Have a go first. Then open the working one line at a time.
\(s(3)=27,\ s(4)=44\). Average speed \(=\frac{44-27}{4-3}=17\) m/s
Distance gone divided by time taken. This is the gradient of the chord from t = 3 to t = 4.
t = 3 to 3.1: \(\frac{28.52-27}{0.1}=15.2\). t = 3 to 3.01: \(\frac{27.1502-27}{0.01}=15.02\)
The averages go 17, 15.2, 15.02. They are closing in on 15.
\(\frac{ds}{dt}=4t+3\), so at t = 3: \(4(3)+3=15\)
The derivative gives the number the averages are heading for.
The speed at t = 3 is 15 m/s
The average speed over a longer interval (17) is larger because the cyclist is speeding up.