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Find the tangent and normal to the curve y=(4-x)(x+2) at the point (2, 8)

-Tangent is a straight line that touches, but does not intersect, the curve at the point (2,8). We need to find the gradient of the curve at the point (2, 8). To do this, expand the equation, differentiate i...
SE
Answered by Sam E. • Maths tutor
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A curve C is defined by the parametric equations x=(4-e^(2-6t))/4 , y=e^(3t)/(3t), t doesnt = 0. Find the exact value of dy/dx at the point on C where t=2/3 .

To solve this we must use the chain rule which is dy/dt * dt/dx. Firstly, we differentiate dy/dt. For this we must use the quotient rule, this gives us dy/dt=(9te^3t - 3e^3t)/9t^2. Now for dx/dt, by substitu...
LK
Answered by Lauren K. • Maths tutor
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Find the integral of tan^2x dx

You can not integrate tan 2 x but you can integrate sec 2 x Since sec 2 x = 1 + tan 2 x Then tan 2 x = sec 2 x-1 so the intragral of tan 2 x dx = the integral of (sec 2 x-1) dx = intrgral of sec 2 x dx + int...
NP
Answered by Nandini P. • Maths tutor
21533 Views

How do i solve two linear simultaneous equations 2x+y=7 & 3x-y=8 ?

To start with, try and spot whether or not two of the coeffecients (numbers next to the letters) are the same for either question (i.e. could be a 3x in one equation and a 3x in the other). This also works i...
TB
Answered by Tom B. • Maths tutor
4563 Views

Find the equation of the tangent to the unit circle when x=sqrt(3)/2 (in the first quadrant)

Unit circle: x 2 + y 2 = 1 when x = sqrt(3)/2: y 2 = 1 - (sqrt(3)/2) 2 y 2 = 1 - 3/4 y 2 = 1/4 y = 1/2 or -1/2 (first quadrant, so y is positive, i.e. y = 1/2) find gradient at (sqrt(3)/2, 1/2): x 2 + y 2 = ...
KJ
Answered by Kiran J. • Maths tutor
4549 Views